Question 12 Marks
Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.
AnswerA lead storage battery consists of anode of lead, cathode of a grid of lead packed with lead dioxide $\ce{(PbO_2)}$ and $38\%$ $\ce{H_2SO_4}$ solution as electrolyte. When the battery is in use, the reaction taking place are$:$
$\text{Anode}:\ \text{Pb(s)}+\text{SO}^{2-}_4\text{(aq)}\rightarrow\text{PbSO}_4\text{(s)}+2\text{e}^-$
$\text{Cathode}:\ \text{PbO}_2\text{(s)}+\text{SO}^{2-}_4\text{(aq)}+4\text{H}^+\text{aq}+2\text{e}^-$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \rightarrow\text{PbSO}_4\text{(s)}+2\text{H}_2\text{O}\text{l}$
$\overline{\text{Overall Reaction}:\ \text{Pb}\text{(s)}+\text{PbO}_2\text{(s)}+2\text{H}_2\text{SO}_4\text{(aq)}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \rightarrow2\text{PbSO}_4\text{(s)}+2\text{H}_2\text{O}\text{(l)}$
On charging the battery, the reverse reaction takes place, i.e., $\ce{PbSO_4}$ deposited on electrodes is converted back to $Pb$ and $\ce{PbO_2}$ and $\ce{H_2SO_4}$ is regenerated.
View full question & answer→Question 22 Marks
Predict the products of electrolysis in each of the following:
An aqueous solution of $CuCl_2$ with platinum electrodes.
AnswerAt cathode:
The following reduction reactions compete to take place at the cathode.
$Cu^{2+}(aq) + 2e^- \rightarrow Cu_{(s)} ; E^\circ = 0.34 V$
$\text{H}^+_{\text{(aq)}}+\text{e}^-\rightarrow\frac{1}{2}\text{H}_{2\text{(g)}}\ ;\ \text{E}^\circ=0.00\ \text{V}$
The reaction with a higher value of $E^\circ$ takes place at the cathode.
Therefore, deposition of copper will take place at the cathode.
Atanode:
The following oxidation reactions are possible at the anode.
$\text{Cl}^-_{\text{(aq)}}\rightarrow\frac{1}{2}\text{Cl}_{2\text{(g)}}+\text{e}^{-1}\ ;\ \text{E}^\circ=1.36\ \text{V}$
$2H_2O_{(I)} \rightarrow O_{2(g)} + 4H^+_{(aq)} + 4e^- ; E^\circ = +1.23 V$
At the anode, the reaction with a lower value of $E^\circ$ is preferred. But due to the over-potential of oxygen, $Cl^−$ gets oxidized at the anode to produce $Cl_2$ gas.
View full question & answer→Question 32 Marks
Predict the products of electrolysis in each of the following$:$ An aqueous solution of $\ce{AgNO_3}$ with silver electrodes.
AnswerAt cathode$:$
The following reduction reactions compete to take place at the cathode.
$\ce{Ag+ (aq) + e- \rightarrow Ag(s) ; E^\circ = 0.80V}$
$\text{H}^+_{\text{(aq)}}+\text{e}^-\rightarrow\frac{1}{2}\text{H}_{2\text{(g)}};\text{E}^\circ=0.00\ \text{V}$
The reaction with a higher value of $E^\circ$ takes place at the cathode.
Therefore, deposition of silver will take place at the cathode.
At anode$:$
The $Ag$ anode is attacked by $NO_3^−$ ions.
Therefore, the silver electrode at the anode dissolves in the solution to form $Ag^+.$
View full question & answer→Question 42 Marks
Using the standard electrode potentials given in Table $3.1,$ predict if the reaction between the following is feasible:$ Fe^{3+} (aq)$ and $Br^– (aq)$
Answer$\text{Fe}^{3+}_\text{(aq)}+\text{e}^-\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ \ } \text{Fe}^{2+}_\text{(aq)}\Big]\times2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ;\ \ \ \ \ \ \ \ \ \ \ \ \text{E}^\circ=+0.77\ \text{V}$
$2\text{Br}^-_\text{(aq)}\ \ \ \ \ \ \xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ \ } \text{Br}_{2\text{(I)}}+2\text{e}^-\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ;\ \ \ \ \ \ \ \ \ \ \ \text{E}^\circ=-1.09\ \text{V}$
$\overline{2\text{Fe}^{3+}_\text{(aq)}+2\text{Br}^-_\text{(aq)}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ \ }2\text{Fe}^{2+}_\text{(aq)}\ \text{and}\ \text{Br}_{2\text{(I)}} \ \ \ \ \ ;\ \ \ \ \ \ \ \ \ \ \ \text{E}^\circ=-0.32\ \text{V}}$
Since $E^\circ$ for the overall reaction is negative, the reaction between $Fe^{3+} (aq)$ and $Br^− (aq)$ is not feasible.
View full question & answer→Question 52 Marks
Predict the products of electrolysis in each of the following$:$ An aqueous solution of $\ce{AgNO_3}$ with platinum electrodes.
AnswerAt cathode$:$
The following reduction reactions compete to take place at the cathode.
$\ce{Ag^+_{(aq)} + e^- \rightarrow Ag_{(s)} ; E^\circ = 0.80V}$
$\text{H}^+_{\text{(aq)}}+\text{e}^-\rightarrow\frac{1}{2}\text{H}_{2\text{(g)}};\text{E}^\circ=0.00\ \text{V}$
The reaction with a higher value of $E^\circ$ takes place at the cathode.
Therefore, deposition of silver will take place at the cathode.
At anode$:$
Since $Pt$ electrodes are inert, the anode is not attacked by $NO_3^−$ ions.
Therefore, $OH^−$ or $NO_3^−$ ions can be oxidized at the anode.
But $OH^−$ ions having a lower discharge potential and get preference and decompose to liberate $O_2.$
$\ce{OH^- \rightarrow OH^+e^-}$
$\ce{4OH^- \rightarrow 2H_2O + O_2}$
View full question & answer→Question 62 Marks
The resistance of a conductivity cell containing $0.001\ce{M KCl}$ solution at $298 K$ is $1500 \Omega .$ What is the cell constant if conductivity of $0.001\ce{M KCl}$ solution at $298 K$ is $0.146 \times 10^{–3} S \ cm^{–1}.$
AnswerGiven that,
Conductivity of the cell, $K = 0.146 \times 10^{-3} S \ cm^{-1}$
Resistance of the cell ,$ R = 1500 \Omega$
Formula of cell constant
Cell constant $= K \times R$
Plug the values we get
$= 0.146 \times 10^3 \times 1500$
$= 0.219 \ cm^{-1}$
View full question & answer→Question 72 Marks
Using the standard electrode potentials given in Table $3.1,$ predict if the reaction between the following is feasible$: Fe^{3+}(aq)$ and $I^– (aq)$
Answer$\text{Fe}^{3+}_\text{(aq)}+\text{e}^-\rightarrow\text{Fe}^{2+}_\text{(aq)}\Big]\times2;\ \ \ \ \ \ \ \ \ \ \ \text{E}^\circ=+0.77\ \text{V}$
$2\text{I}^-_\text{(aq)}\rightarrow2\text{I}_{2\text{(S)}}+2\text{e}^-;\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{E}^\circ=-0.54\ \text{V}$
$\overline{2\text{Fe}^{3+}_\text{(aq)}+2\text{I}^-\rightarrow2\text{Fe}^{2+}_\text{(aq)}+\text{I}_{2\text{(s)}}\ \ \ \ \text{E}^\circ=+0.23\ \text{V}}$
Since $E^\circ$ for the overall reaction is positive, the reaction between $Fe^{3+} (aq)$ and $I^- (aq)$ is feasible.
View full question & answer→Question 82 Marks
Using the standard electrode potentials given in Table $3.1,$ predict if the reaction between the following is feasible$: Br_2 (aq)$ and $Fe^{2+} (aq).$
Answer$\ \ \ \text{Br}_{2\text{(aq)}}+2\text{e}^-\ \ \xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ \ } 2\text{Br}^{-}_\text{(aq)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ;\ \ \ \ \ \ \ \ \ \ \ \ \text{E}^\circ=+1.09\ \text{V}$
$\text{Fe}^{2+}_\text{(aq)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ \ } \text{Fe}^{3+}_{\text{(aq)}}+\text{e}^-\big]\times2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ;\ \ \ \ \ \ \ \ \ \ \ \text{E}^\circ=-0.77\ \text{V}$
$\overline{\text{Br}_\text{2(aq)}+2\text{Fe}^{2+}_\text{(aq)}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ \ }2\text{Br}^{-}_\text{(aq)}+2\text{Fe}^{3+}_\text{(aq)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ;\ \ \ \ \ \ \ \ \ \ \ \ \text{E}^\circ=+0.32\ \text{V}}$
Since $E^\circ$ for the overall reaction is positive, the reaction between $Br^2 (aq)$ and $Fe^{2+} (aq)s$ is feasible.
View full question & answer→Question 92 Marks
A solution of $Ni(NO_3)_2$ is electrolysed between platinum electrodes using a current of $5$ amperes for $20$ minutes. What mass of $Ni$ is deposited at the cathode?
AnswerGiven,
Current $= 5A$
Time $= 20 \times 60 = 1200 s$
Therefore, Charge $=$ current $\times $ time
$= 5 \times 1200$
$= 6000 C$
According to the reaction,
$\text{Ni}^{2+}_\text{(aq)}+2\text{e}^-\rightarrow\text{Ni}_\text{(s)}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 58.7\text{g}$
Nickel deposited by $2 \times 96487\ C = 58.71g$
Therefore, nickel deposited by $6000\ C =\frac{58.71\times6000}{2\times96487}\text{g}$
$= 1.825g$
Hence, $1.825g$ of nickel will be deposited at the cathode.
View full question & answer→Question 102 Marks
The conductivity of $0.20 M$ solution of $\text{KCl}$ at $298 K$ is $0.0248\ S \ cm^{–1}.$ Calculate its molar conductivity.
AnswerWe have given that
Molarity of solution, $M = 0.20$
conductivity, i.e., specific conducitivity $= k = 0.248 s \ cm^{-1} = 2.48 \times 10^{-2} ohm^{-1 }cm^{-1}$
Molar conductivity, $\lambda_\text{m}=\frac{1000\text{k}}{\text{M}}\ \text{ohm}^{-1}\text{cm}^2\ \text{mol}^{-1}$
$=\frac{1000\times2.48\times10^{-2}}{0.20}$
$=124.0\ \text{s\cm}^2\ \text{mol}^{-1}$
Thus, molar conductivlty, $\lambda_\text{m} = 124.0 s\ cm^2 mol^{-1}$
View full question & answer→Question 112 Marks
Suggest a way to determine the $\wedge^\circ_\text{m}$ value of water.
AnswerBy using Kohlrausch's law, $\wedge^\circ_\text{m}$ for $H_2O$ can be calculated, we can write,
$\wedge^\circ_\text{m}=\wedge^\circ_\text{m}(\text{HCl})+\wedge^\circ_\text{m}(\text{NaOH})-\wedge^\circ_\text{m}(\text{NaCl})$
Being strong electrolytes, $\wedge^\circ_\text{m}$ values of $\ce{HCl},$
$\ce{NaOH}$ and $\ce{NaCl}$ are known.
By substituting their values, we can obtain $\wedge^\circ_\text{m}$ for $\ce{H_2O}.$
View full question & answer→Question 122 Marks
Calculate the potential of hydrogen electrode in contact with a solution whose $pH$ is $10.$
AnswerFor hydrogen electrode, $\text{H}^++\text{e}^-\rightarrow\frac{1}{2}\text{H}_2$ it is given that $pH = 10$
$\therefore\ [\text{H}_+]=10\ \text{M}$
Now, using Nernst equation$:$
$\text{H}_{\Big(\text{H}^+/\frac{1}{2}\text{H}_2\Big)}=\text{E}^\ominus_{\Big(\text{H}^+/\frac{1}{2}\text{H}_2\Big)}-\frac{\text{RT}}{\text{nF}}\text{In}\frac{1}{[\text{H}^+]}$
$\text{E}^\ominus_{\Big(\text{H}^+/\frac{1}{2}\text{H}_2\Big)}-\frac{0.0591}{1}\text{\log}\frac{1}{[\text{H}^+]}$
$=0-\frac{0.0591}{1}\text{\log}\frac{1}{[10^{-10}]}$
$= -0.0591 \log 10^{10}$
$= -0.591 V$
View full question & answer→Question 132 Marks
Using the standard electrode potentials given in Table $3.1,$ predict if the reaction between the following is feasible$: Ag(s)$ and $Fe^{3+} (aq)$
Answer$\ \ \ \text{Ag}_\text{(s)}\ \ \ \ \ \ \ \ \ \xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ \ } \text{Ag}^{+}_\text{(aq)}+\text{e}^- \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ;\ \ \ \ \ \ \ \ \ \ \ \ \text{E}^\circ=+0.80\ \text{V}$
$\text{Fe}^{3+}_\text{(aq)}+\text{e}^-\ \ \ \ \ \ \xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ \ } \text{Fe}^{2+}_{\text{(aq)}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ;\ \ \ \ \ \ \ \ \ \ \ \text{E}^\circ=+0.77\ \text{V}$
$\overline{\text{Ag}_\text{(s)}+\text{Fe}^{3+}_\text{(aq)}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ \ }\text{Ag}^{+}_\text{(aq)}+\text{Fe}^{2+}_\text{(aq)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ;\ \ \ \ \ \ \ \ \ \ \ \ \text{E}^\circ=-0.03\ \text{V}}$
Since $E^\circ E$ for the overall reaction is negative, the reaction between $Ag(s)$ and $Fe^{3+} (aq)$ is not feasible.
View full question & answer→Question 142 Marks
If a current of $0.5$ ampere flows through a metallic wire for $2$ hours, then how many electrons would flow through the wire?
Answer$I = 0.5 A$
$t = 2$ hours $= 2 \times 60 \times 60 s = 7200 s$
Thus, $Q = It$
$= 0.5 A \times 7200 s$
$= 3600 C$
We know that
$96487 C = 6.023 \times 10^{23}$
number of electrons.
Then,
$3600\ \text{C}=\frac{6.023\times10^{23}\times3600}{96487}\text{number of electrons}$
$= 2.25 \times 10^{22}$ number of electrons
Hence, $2.25 \times 10^{22}$ number of electrons will flow through the wire.
View full question & answer→Question 152 Marks
Predict the products of electrolysis in each of the following: A dilute solution of $\ce{H_2SO_4}$ with platinum electrodes.
AnswerAt the cathode, the following reduction reaction occurs to produce $H_2$ gas.
$\text{H}^+_{\text{(aq)}}+\text{e}^-\rightarrow\frac{1}{2}\text{H}_{2\text{(g)}}$
At the anode, the following processes are possible.
$2\text{H}_2\text{O}_\text{(I)}\rightarrow\text{O}_{2\text{(g)}}+4\text{H}^+_\text{(aq)}+4\text{e}^-;\text{E}^\circ=+1.23\ \text{V} ....(1)$
$2\text{SO}^{2-}_{4\text{(aq)}}\rightarrow\text{S}_2\text{O}^{2-}_{6\text{(aq)}}+2\text{e}^-\ ;\ \text{E}^\circ=+1.96\ \text{V} ....(2)$
For dilute sulphuric acid, reaction $(i)$ is preferred to produce $O_2$ gas.
But for concentrated sulphuric acid, reaction $(ii)$ occurs.
View full question & answer→Question 162 Marks
Calculate the emf of the cell in which the following reaction takes place $Ni(s) + 2Ag^+ (0.002 M) \rightarrow Ni^{2+} (0.160 M) + 2Ag(s)$
Given that $\text{E}^\ominus_{(\text{cell})} = 1.05 V$
Answer$E^(-)_{(cell) }= 1.05 V.$
Applying Nernst equation,
$\text{E}_{\text{call}}=\text{E}^\circ_{\text{call}}-\frac{0.0591}{\text{n}}\text{log}\frac{[\text{Ni}^{2+}]}{[\text{Ag}^+]^2}$
$=1.05\text{V}-\frac{0.0591}{2}\text{log}\frac{0.160}{(0.002)^2}$
$=1.05-\frac{0.0591}{2}\text{log}({4\times10}^4)$
$=1.05-\frac{0.0591}{2}(4.6021)$
$= 1.05 - 0.14 V$
$= 0.91 V$
View full question & answer→Question 172 Marks
Explain how rusting of iron is envisaged as setting up of an electrochemical cell.
AnswerAt a particular spot of an object made of iron oxidation takes place and that spot behaves as an anode.
At anode: $2Fe(s) \rightarrow 2Fe^{2+}+ 4e^−$
$E^\circ (Fe^{2+}, Fe) = -0.44V$
Electrons released at anode spot moves through the metal and go to another spot on the metal and reduce oxygen in presence of $H^+ ($which is believed to be available from $H_2CO_3$ formed due to dissolution of carbondioxide from air into water. Hydrogen ion in water may also be available due to dissolution of other acidic oxides from the atmosphere$).$ This spot behaves as a cathode with the reaction.
At cathode:
$O_2(g) + 4H^+(aq) + 4e^- \rightarrow 2H_2O(l)$
$[E^\circ = 1.23 V]$
The over reaction being
$2Fe(s) + O_2(g) + 4H^+(aq) \rightarrow 2Fe^{2+}(aq) + 2H_2O(l)$
$[E^\circ ($cell$) = 1.67 V]$
The ferrous ions are further oxidized by atmospheric oxygen to ferric ions which come out as rust in the form of hydrated ferric oxide $(Fe_2O_{3 }\times H_2O).$
View full question & answer→Question 182 Marks
How would you determine the standard electrode potential of the system $Mg^{2+}|Mg?$
AnswerThe standard electrode potential of $Mg^{2+} | Mg$ can be measured with respect to the standard hydrogen electrode,
represented by $Pt_{(s)}, H_{2(g)} (1$atm$) | H^{+(aq)} (1M).$
A cell, consisting of $Mg | MgSO_4 (aq\ 1M)$ as the anode and the standard hydrogen electrode as the cathode, is set up.
$\text{Mg}|\text{Mg}^{2+}(\text{aq, 1M})||\text{H}^+(\text{aq, 1M})|\text{H}_2(\text{g, 1 bar}),\text{Pt}_{(s)}$
Then, the emf of the cell is measured and this measured emf is the standard electrode potential of the magnesium electrode.
$\text{E}^\ominus=\text{E}^\ominus_\text{R}-\text{E}^\ominus_\text{L}$
Here, $\text{E}^\ominus_\text{R}$ for the standard hydrogen electrode is zero.
$\therefore\ \text{E}^\ominus=0-\text{E}^\ominus_\text{L}$
$=-\text{E}^\ominus_\text{L}$
View full question & answer→Question 192 Marks
Using the standard electrode potentials given in Table $3.1,$ predict if the reaction between the following is feasible$: Ag^+ (aq)$ and $Cu(s)$
Answer$\text{Ag}^{+}_\text{(aq)}+\text{e}^-\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ \ } \text{Ag}_\text{(s)}\Big]\times2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ;\ \ \ \ \ \ \ \ \ \ \ \text{E}^\circ=+0.80\ \text{V}$
$\text{Cu}_\text{(s)}\ \ \ \ \ \ \xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ \ } \text{Cu}^{2+}_\text{(aq)}+2\text{e}^-\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ;\ \ \ \ \ \ \ \ \ \ \ \text{E}^\circ=-0.34\ \text{V}$
$\overline{2\text{Ag}^{+}_\text{(aq)}+\text{Cu}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ \ }2\text{Ag}_\text{(s)}+\text{Cu}^{2+}\text{(aq)} \ \ \ \ \ \ ;\ \ \ \ \ \ \ \ \ \ \ \text{E}^\circ=+0.23\ \text{V}}$
Since $E^\circ$ for the overall reaction is positive, the reaction between $Ag^+ (aq)$ and $Cu(s)$ is feasible.
View full question & answer→Question 202 Marks
Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.
AnswerOxidation of ferrous ion means :
$Fe^{2+ \rightarrow } Fe^{3+} + e^- ; E^0_{Fe3+/Fe2+} = 0.77V$
Any substance which standard electrode potential is more than that of $Fe^{+3 }/F^{+2}$ can oxidise ferrous ions.
$($refer to the table given in book$)$
The $EMF$ of the substance whose reduction potentials greater than $0.77V$ will oxidised ferrous ion.
For example $Br_{2 },Cl_{2},$ and $F_2.$
View full question & answer→Question 212 Marks
Calculate the time to deposit $1.5\ g$ of silver at cathode when a current of $1.5\ A$ was passed through the solution of $AgNO_3. ($Molar mass of $Ag = 108\ g\ mol^{-1}, 1F = 96500\ C\ mol^{-1})$
AnswerWt. of $Ag = 1.5\ g $
Current $= i = 1.5\ amp$
Molecular mass $= 108\ g/mol$
$F = 96500\ C/mol$
$n =$ number of electron transferred $\text{W=}\frac{\text{M}\times\text{I}\times\text{t}}{\text{n}\times\text{F}}$
$\therefore\text{t=}\frac{\text{W}\times\text{n}\times\text{F}}{\text{M}\times\text{I}}$
$=\frac{\text{1.5}\times\text{1}\times\text{96500}}{\text{108}\times\text{1.5}} = 893.51 s$ or $14.89 \min$
Alternate Answer
At cathode: $Ag^+ +e^- \xrightarrow{\text{ }\ \ \ \ \ \ \ \ \ \ \ \ } Ag(s)$
$108g$ of $Ag$ require $1F$
$\therefore 1.5g$ of $Ag$ require $=\frac{\text{1.5}}{\text{108}}$
$\text{F}=\ \frac{\text{1.5}\times\text{96500}}{\text{108}} = 1340.27\ C$
$\text{t}=\frac{\text{Q}}{\text{i}}=\frac{\text{1340.27}}{\text{1.5}} = 893.51s$ or $14.89 \min$
View full question & answer→Question 222 Marks
Using $\text{IUPAC}$ norms write the formulae for the following $:$
- Potassium trioxalatoaluminate$(III)$
- Dichloridobis$($ethane$-1,2-$diamine$)$cobalt$(III)$
Answer
- $\ce{K_3[Al(C_2O_4)_3]}$
- $\ce{[Co Cl_2 (en)_2]^+}$
View full question & answer→Question 232 Marks
Write the name of the cell which is generally used in transistors. Write the reactions taking place at the anode and the cathode of this cell.
AnswerDry Cell $/$ Leclanche cell
Anode $: \ce{Zn_{(s)} \rightarrow Zn^{2+} + 2e^-}$
Cathode $:\ce{ MnO_2 + NH_4^+ +e^{- }\rightarrow MnO(OH) + NH_3}$
View full question & answer→Question 242 Marks
Define the following terms:
- Molar conductivity $(\wedge_{\text{m}})$
- Secondary batteries
Answer
- Molar conductivity is defined as the conductivity due to all the ions produced by dissolving one mole of an electrolyte in solution.
- In the secondary batteries, the reactions can be reversed by an external electrical energy source. These batteries can be recharged by passing electric current and used again and again.
View full question & answer→Question 252 Marks
Write the name of the cell which is generally used in inverters. Write the reactions taking place at the anode and the cathode of this cell.
AnswerLead storage battery
Anode $:\ce{ Pb_{(s)}+SO_4^{2-} _{(aq)} \rightarrow PbSO_{4(s)} + 2e^-}$
Cathode $: \ce{PbO_2+SO_4 _{(aq)}+ 4H^{+ }+2e- \rightarrow PbSO_{4(s)} + 2 H_20_{(l)}}$
View full question & answer→Question 262 Marks
Define the following terms:
- Fuel cell
- Limiting molar conductivity$(\wedge^\circ_\text{m}) $
Answer
- Galvanic cells that are designed to convert the energy of combustion of fuels (methane, methanol, etc.) directly into electrical energy are called fuel cells.
- Molar conductivity of electrolyte at infinite dilution or when concentration approaches zero.
View full question & answer→Question 272 Marks
Write the name of the cell which is generally used in hearing aids. Write the reactions taking place at the anode and the cathode of this cell.
AnswerMercury cell
Anode $: \ce{Zn(Hg) + 2OH^-$\rightarrow$ ZnO_{(s)} + H2O + 2e^-}$
Cathode $: \ce{HgO + H_2O + 2e^-$\rightarrow$ Hg(l) + 2OH^-}$
View full question & answer→Question 282 Marks
State Kohlrausch law of independent migration of ions. Why does the conductivity of a solution decrease with dilution?
AnswerKohlrausch law of independent migration of ions. The law states that limiting molar conductivity of an electrolyte can be stated as the sum of the individual contributions of the anion and cation of the electrolyte.
On dilution, the conductivity (κ) of the electrolyte decreases as the number of ions per unit volume of solution decreases.
View full question & answer→Question 292 Marks
The standard electrode potential $(E^\circ)$ for Deniell cell is $+1.1V.$ Calculated the $ʌG^\circ$ for the reaction
$Zn (s) + Cu^{2+ }(aq) \rightarrow Zn^{2 }(aq) + Cu (s)$
$(1 F = 96500\ C\ mol^{-1}).$
Answer$\triangle G^\circ = -n FE^\circ$ cell
$= -2 \times 96500\ C\ mol^{-1} \times 1.1\ V$
$= -212300\ J\ mol^{-1}$ or $-212.3\ k\ J\ mol^{-1}$
View full question & answer→Question 302 Marks
The molar conductivity of a $1.5M$ solution of an electrolyte is found to be $138.9 \ \ce{Scm^{2 }mol^{–1}}.$ Calculate the conductivity of this solution.
Answer$\Lambda\text{m} = 138.9 \ \text{S\cm}^2 \text{ mol}^{–1}$
$\text{M} = 1.5\text{M}$
$\text{K} = \ ?$$\Lambda_{M}\frac{1000K}{M}$
$\text{K}=\frac{\Lambda_{M}\times\text{M}}{1000}$
$\text{K}=\frac{{138.9}\times{1.5}}{1000}$
$=0.20835\text{ Scm}^{-1}$.
View full question & answer→Question 312 Marks
Express the relation among cell constant, resistance of the solution in the cell and conductivity of the solution. How is molar conductivity of a solution related to its conductivity?
Answer$k = 1/R \ (l/A)$
Where $k$ is conductivity, $\ce{Ris}$ resistance and $\ce{l/Ais}$ cell constant
$\Lambda_m = k/C$
Where $\Lambda m$ is molar conductivity $C$ is concentration.
View full question & answer→Question 322 Marks
Given that the standard electrode potentials $(E^o)$ of metals are$:$
$K^+/K = –2.93 V, Ag^+/Ag = 0.80 V, Cu^{2+}/Cu = 0.34 V,$
$Mg^{2+}/Mg = -2.37 V, Cr^{3+}/Cr = – 0.74 V, Fe^{2+}/Fe = – 0.44 V.$
Arrange these metals in an increasing order of their reducing power.
Answer$Ag^+ / Ag < Cu^{2+} / Cu < Fe^{2+} / Fe < Cr^{3+} / Cr < Mg^{2+} / Mg < K^+ / K.$
View full question & answer→Question 332 Marks
Express the relation among the cell constant, the resistance of the solution in the cell and the conductivity of the solution. How is the conductivity of a solution related to its molar conductivity?
Answerk = 1/R (l/A)
Where k is conductivity, R is resistance and 1/A is cell constant
Ëm = k/C
Where Ëm is molar conductivity and C is concentration of the solution.
View full question & answer→Question 342 Marks
Two half-reactions of an electrochemical cell are given below:
$MnO_4^- (aq) + 8 H^+ (aq) + 5e^– \rightarrow Mn^{2+} (aq) + 4H_2O (l), E^\circ = +1.51 V,$
$Sn^{2+} (aq) \rightarrow Sn^{4+} (aq) + 2e^–, E^\circ= +0.15 V.$
Construct the redox reaction equation from the two half-reactions and calculate the cell potential from the standard potentials and predict if the reaction is reactant or product favoured.
AnswerRedox Reaction
$2MnO_4^- + 5Sn^{2+} + 16H^{+}$
$\rightarrow 2Mn^{2+} + 5Sn^{4+} + 8H_2O.$
$E^\circ _{cell} = E^\circ _C-E^\circ _A$
$= (+1.51 – 0.15)V = +1.36V$
As $E^\circ _{cell}$ is positive, the reaction is product favoured.
View full question & answer→Question 352 Marks
Two half cell reactions of an electrochemical cell are given below$:$
$\ce{MnO_4^- (aq) + 8H^+ (aq) + 5e^– \rightarrow Mn^{2+ }(aq) + 4H_2O (l), E^o = + 1.51 V}$
$\ce{Sn^{2+} (aq) \rightarrow Sn^{4+} (aq) + 2e^–, E^o = + 0.15 V}$
Construct the redox equation from the two half cell reactions and predict if this reaction favours formation of reactants or product shown in the equation.
Answer$\text{MnO}_{4}^{-}+\text{8H}^{+}+\text{5e}^{-}\rightarrow\text{Mn}^{2+}+\text{4H}_{2}\text{O(I) E}^{o}=+1.51\text{V}$
$\frac{\text{Sn}^{2+}\rightarrow\text{Sn}^{4+}+\text{2e}^{-}\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{E}^{o}=+0.15\text{V}}{\text{2 MnO}_{4}^{-}+\text{5 Sn}^{2+}+16\text{H}^{+}\rightarrow\text{2 Mn}^{2+}+\text{5 Sn}^{4+}+\text{8H}_{2}\text{O (I)}}$
This reaction favors formation of products.
View full question & answer→Question 362 Marks
What type of cell is a lead storage battery? Write the anode and the cathode reactions and the overall cell reaction occurring in the use of a lead storage battery.
AnswerIt is secondary cell
Anode Reaction$:- \ce{Pb + $\text{SO}_{4}^{2-}$ \rightarrow PbSO_4(s) + 2e^–}.$
Cathode Reaction$:- \ce{PbO_2 + 4H^+ + $\text{SO}_{4}^{2-}$ + 2e^– \rightarrow PbSO_4 + 2H_2O}.$
Net reaction$:- \ce{Pb + PbO_2 + $2\text{SO}_{4}^{2-}$+ 4H^+ \rightarrow 2PbSO_4 + 2H_2O}.$
View full question & answer→Question 372 Marks
Depict the galvanic cell in which the reaction: $\ce{Zn(s) + 2Ag^+(aq) \rightarrow Zn^{2+}(aq) + 2Ag(s)}$ takes place.
Further indicate what are the carriers of the current inside and outside the cell. State the reaction at each electrode.
AnswerThe galvanic cell is depicted as$:$
$\ce{Zn(s)| Zn^{2+}(aq) || Ag^+(aq)|Ag (s)}$
- Zinc electrode is negatively charged.
- The ions formed i.e $Zn^{2+}$ and $Ag^+$ in the solution are the carriers of the current in the cell and electrons in the external circuit.
- At anode$: \ce{Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-}.$
At cathode$: \ce{2Ag^+(aq)+ 2e^{- } \rightarrow 2Ag(s)}.$ View full question & answer→Question 382 Marks
The resistance of a conductivity cell containing $0.001 \ce{M KCI}$ solution at $298 K$ is $1500 \Omega .$ What is the cell constant if the conductivity of $0.001 \ce{M KCI}$ solution at $298 K$ is $0.146 \times 10^{–3} \ce{S cm^{–1}}$?
Answer$R= \rho ( l/A )$ Cell constant, $l/A = R/\rho = Rκ=(1500 \Omega ) \times (0.146 \times 10^{-3} S \ cm^{-1})$
$= 0.219 \ cm^{-1}.$
View full question & answer→Question 392 Marks
Define conductivity and molar conductivity for the solution of an electrolyte.
AnswerConductivity: The conductivity of a solution at any given concentration is the conductance of unit volume of solution kept between two platinum electrodes with unit area of cross section and at a distance of unit length.
Molar conductivity: It can be defined as the conductance of the solution of an electrolyte kept between the electrodes of a conductivity cell at unit distance but area having cross section large enough to accommodate sufficient volume of solution that contains one mole of the electrolyte.
View full question & answer→Question 402 Marks
In a galvanic cell, the following cell reaction occurs:
$\ce{Zn (s)^+ 2 Ag^+ (aq) \longrightarrow Zn^{2+} (aq) + 2 Ag (s) E^o_{cell} = + 1.56 V}$
- Is the direction of flow of electrons from zinc to silver or silver to zinc?
- How will concentration of $Zn^{2+}$ ions and $Ag^+$ ions be affected when the cell functions?
Answer
- Zinc to silver.
- Concentration of $Zn^{2+}$ ions will increase and $Ag^+$ ions will decrease.
View full question & answer→Question 412 Marks
Calculate the degree of dissociation $(\alpha)$ of acetic acid if its molar conductivity $(\wedge_m)$ is $39.05\ S \ cm^2mol^{–1}.$
Given $\lambda^0(H^+) = 349.6\ S \ cm^2 mol^{–1}$
$\lambda^0(CH_3COO^–) = 40.9\ S \ cm^2 mol^{–1}$
Answer$\wedge^0 CH_3COOH = \lambda_0$
$CH3COO^- + \lambda_0 H^+$
$= 40.9 + 349.6 = 390.5 S \ cm_2/mol$
Now, $\alpha = Λ_m /Λ^o m$
$= 39.05 / 390.5 = 0.1$
View full question & answer→Question 422 Marks
Calculate the degree of dissociation $(\alpha)$ of acetic acid if its molar conductivity $(\wedge_m)$ is $39.05 S \ cm^2\ mol^{–1}.$
Given$\lambda^0 (H^+) = 349.6 S \ cm^2\ mol^{–1}$ and $\lambda^0 (CH_3COO^–) = 40.9 S \ cm^2\ mol^{–1}$
Answer$\wedge^0\ CH_3COOH =\lambda_0$
$CH3COO^- + \lambda_0 H^+$
$= 40.9 + 349.6 = 390.5 S \ cm_2/mol$
Now, $\alpha = Λ_m /Λ^o m$
$= 39.05 / 390.5 = 0.1$
View full question & answer→Question 432 Marks
- Following reactions occur at cathode during the electrolysis of aqueous silver chloride solution$:$
- $\ce{Ag^+(aq) + e^- \rightarrow Ag(s) E^{o }= +0.80 V}$
$\ce{H^+(aq) + e^- \rightarrow H_2(g) E^o = 0.00 V}$
On the basis of their standard reduction electrode potential $(E^o)$ values, which reaction is feasible at the cathode and why?
- Define limiting molar conductivity. Why conductivity of an electrolyte solution decreases with the decrease in concentration?
Answer
- $\ce{Ag+ (aq) + e^- \rightarrow Ag (s)}$
Reaction with higher $\ce{E^0 value/ \Delta G^0}$ negative.
- Molar conductivity of a solution at infinite dilution or when concentration approaches zero.
Number of ions per unit volume decreases. View full question & answer→Question 442 Marks
The Conductivity of $0.20 M$ solution of $ \ce{KCl}$ at $298 K$ is $0.025 \ \ce{S cm^{-1}}.$ Calculate its moral conductivity.
Answer$\Lambda m = k/C$
$\Lambda\text{m}=\frac{\text{0.025 S \ cm}^{-1}}{\text{0.20 mol L}^{-1}}$
$\Lambda m =125 \ \ce{S cm^2 mol^{-1}}$
View full question & answer→Question 452 Marks
The chemistry of corrosion of iron is essentially an electrochemical phenomenon. Explain the reactions occurring during the corrosion of iron in the atmosphere.
Answer$\text{Oxidation:}\text{ Fe (s)}\rightarrow\text{Fe}^{2+}\text{(aq)}+\text{2e}^{+}$.$\text{Reduction: O}_{2}\text{(g)}+\text{4H}^{+}\text{(aq)}+\text{4c}^{-}\rightarrow\text{2H}_{2}\text{O(1)}$.
$\text{Atmospheric oxidation: 2Fe}^{2+}\text{(aq)}+\text{2H}_{2}\text{O(l)}+\frac{1}{2}\text{O}_{2}\text{(g)}^{+}\text{(aq)}\rightarrow+\text{Fe}_{2}\text{O}_{3}\text{(s)}+\text{4H}^{+}\text{(aq)}$.
View full question & answer→Question 462 Marks
Determine the values of equilibrium constant $(K_C)$ and $\Delta G^{\circ}$ for the following reaction:
$\text{Ni (s)}+\text{2Ag}^{+}\text{(aq)}\rightarrow\text{Ni}^{2+}+\text{2Ag(s)},\text{E}^{\circ}=1.05\text{V}\ \ \ (\text{IF} = 96500 \text{C mol}^{–1}).$
Answer$\Delta \text{rG}^\circ = -\text{nFE}^\circ$
$= -\ 2 \times (96500\text{C mol }^{-1} \times 1.05\text{V} )$
$= -\ 202650\ \text{J mol}^{–1} \Rightarrow -\ 202.6\ \text{kJ mol}^{–1}$
$\text{log K}_{\text{c}}=\frac{\text{nE}^{\circ}}{0.0591}$
$=\frac{2\ \times\ 1.05\text{V}}{0.0591}$
$\text{K}_{\text{c}} = 3.412 \times 1035.$
View full question & answer→Question 472 Marks
Note: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
- Both assertion and reason are true and the reason is the correct explanation of assertion.
- Both assertion and reason are true and the reason is not the correct explanation of assertion.
- Assertion is true but the reason is false.
- Both assertion and reason are false.
- Assertion is false but reason is true.
Assertion: Conductivity of all electrolytes decreases on dilution.
Reason: On dilution number of ions per unit volume decreases.
Answer
- Both assertion and reason are true and the reason is the correct explanation of assertion.
Explanation:
Conductivity always decreases with decrease in concentration both, for weak and strong electrolytes. This can be explained by the fact that the number of ions per unit volume that carry the current in a solution decreases on dilution.
View full question & answer→Question 482 Marks
Why on dilution the $m Λ_m$ of $\ce{CH_3COOH}$ increases drastically, while that of $\ce{CH_3COONa}$ increases gradually?
AnswerIn the case of $\ce{CH_3COOH},$ which is a weak electrolyte, the number of ions increase on dilution due to an increase in degree of dissociation.
$\text{CH}_3\text{COOH}+\text{H}_2\text{O}\rightleftharpoons\text{CH}_3\text{COO}^{-}(\text{aq}.)+\text{H}_3\text{O}^+$
In the case of strong electrolyte such as $\ce{CH_3COONa},$ the number of ions remains the same but the interionic attraction decreases.
View full question & answer→Question 492 Marks
What advantage do the fuel cells have over primary and secondary batteries?
AnswerPrimary batteries contain a limited amount of reactants and are discharged when the reactants have been consumed. Secondary batteries can be recharged but take a long time to recharge. Fuel cell runs continuously as long as the reactants are supplied to it and products are removed continuously.
View full question & answer→Question 502 Marks
How will the $pH$ of brine $(\ce{aq. NaCl}$ solution$)$ be affected when it is electrolysed?
AnswerAqueous solution of brine contains $Na^+, CP, H^+$ and $H^–.$
Electrode process are given as follows$:$
$2\text{H}^++2\text{e}^{-}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ }\text{H}_2\ \ \ \ (\text{cathode})$
$2\text{Cl}^-\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ }\text{Cl}_2+2\text{e}\ \ \ \ (\text{Anode})$
Remaining solution will contain $\ce{NaOH}$ which is base.
therefore, $pH$ will increase.
View full question & answer→Question 512 Marks
Which reference electrode is used to measure the electrode potential of other electrodes?
AnswerStandard hydrogen electrode is the reference electrode whose electrode potential is taken to be zero. The electrode potential of other electrodes is measured with respect to it.
View full question & answer→Question 522 Marks
Solutions of two electrolytes $'A\ ’$ and $'B\ ’$ are diluted. The $Λ_m$ of $'B\ ’$ increases $1.5$ times while that of $A$ increases $25$ times. Which of the two is a strong electrolyte? Justify your answer.
AnswerElectrolyte $'B\ ’$ is strong as on dilution the number of ions remains the same, only interionic attraction decreases.
therefore increase in $∧_m$ is small.
View full question & answer→Question 532 Marks
Note: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
- Both assertion and reason are true and the reason is the correct explanation of assertion.
- Both assertion and reason are true and the reason is not the correct explanation of assertion.
- Assertion is true but the reason is false.
- Both assertion and reason are false.
- Assertion is false but reason is true.
Assertion: For measuring resistance of an ionic solution an AC source is used.
Reason: Concentration of ionic solution will change if DC source is used.
Answer
- Both assertion and reason are true and the reason is the correct explanation of assertion.
Explanation:
DC current can change the composition of electrolytic solution.
View full question & answer→Question 542 Marks
A galvanic cell has electrical potential of 1.1V. If an opposing potential of 1.1V is applied to this cell, what will happen to the cell reaction and current flowing through the cell?
AnswerWhen the opposing potential becomes equal to electrical potential, the cell reaction stops and no current flows through the cell. Thus, there is no chemical reaction.
View full question & answer→MCQ 552 Marks
Note: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: Current stops flowing when $E_{Cell} = 0.$
Reason: Equilibrium of the cell reaction is attained.
- ✓
Both assertion and reason are true and the reason is the correct explanation of assertion.
- B
Both assertion and reason are true and the reason is not the correct explanation of assertion.
- C
Assertion is true but the reason is false.
- D
Both assertion and reason are false.
AnswerCorrect option: A. Both assertion and reason are true and the reason is the correct explanation of assertion.
At equilibrium $E_{Cell }= 0$ and therefore current stops flowing.
View full question & answer→Question 562 Marks
Write the Nernst equation for the cell reaction in the Daniel cell. How will the $E_{Cell}$ be affected when concentration of $Zn^{2+}$ ions is increased?
AnswerDaniell cell
$\text{Zn(s)}|\text{Zn}^{2+}||\text{Cu}^{2+}|\text{Cu}|\text{Cu(s)}$
$\text{Anode}:\ \ \ \ \ \ \ \ \ \text{Zn(s)}\xrightarrow{\ \ \ \ \ \ \ \ \ \ }\text{Zn}^{2+}+2\text{e}^{-}\\ \text{Cathode}:\ \ \ \ \ \text{Cu}^{2+}+2\text{e}^{-}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ }\text{Cu(s)}\\\overline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\\ \text{Cell reaction}\ \ \text{Zn(s)}+\text{Cu}^{2+}\rightleftharpoons\text{Zn}^{2+}+\text{Cu(s)}\\\overline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }$
$\text{Q}=\frac{\big[\text{Zn}^{2+}\big]}{\big[\text{Cu}^{2+}\big]}$
$\text{E}_{\text{Cell}}=\text{E}^\circ_{\text{cell}}-\frac{0.059}{2}\log\text{Q}=\frac{\big[\text{Zn}^{2+}\big]}{\big[\text{Cu}^{2+}\big]}$
Above equation shows that the cell potential will decrease with increase in the concentration of $Zn^{2+}$ ion.
View full question & answer→Question 572 Marks
Tarnished silver contains $\ce{Ag_2S}.$ Can this tarnish be removed by placing tarnished silver ware in an aluminium pan containing an inert electrolytic solution such as $\ce{NaCl}$? The standard electrode potential for half reaction$:$
$\text{Ag}_2\text{S(s) + 2e}^-\xrightarrow{\ \ \ \ \ \ \ }2\text{Ag(s) + S}^{2-}\text{is}-0.71\text{V}$
and for $\text{Al}^{3+}+3\text{e}^{-}\xrightarrow{ \ \ \ \ \ \ \ \ \ \ }2\text{Al(s) is}-1.66\text{V}$
Answer$E^\circ _{cell}$ for reaction of tarnished silver ware with aluminium pan is $(-0.71V) - (-1.66V)$
i.e., $+ 095V$ Tarnished silver ware,
therefore, can be cleaned by placing it in an aluminium pan as $E^\circ _{cell}$ is positive.
View full question & answer→Question 582 Marks
State two advantages of $H_2-O_2$ fuel cell over ordinary cell.
Answer
- Advantages of fuel cell: a. It is a pollution$-$free device since no harmful products are formed.
- Its efficiency is about $75\%$ which is considerably higher than conventional cells.
- It is a continuous source of energy if the supply of gases is maintained.
View full question & answer→MCQ 592 Marks
Note: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: $E_{Cell }$ should have a positive value for the cell to function.
Reason: $\text{E}_{\text{cathode}}<\text{E}_{\text{anode}}$
- A
Both assertion and reason are true and the reason is the correct explanation of assertion.
- B
Both assertion and reason are true and the reason is not the correct explanation of assertion.
- ✓
Assertion is true but the reason is false.
- D
Both assertion and reason are false.
AnswerCorrect option: C. Assertion is true but the reason is false.
$\text{E}_{\text{Cell}}=\text{E}_{\text{cathode}}-\text{E}_{\text{anode}}$
To have positive value of $\text{E}_{\text{Cell}}=\text{E}_{\text{cathode}}<\text{E}_{\text{anode}}$
View full question & answer→MCQ 602 Marks
Note: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: Electrolysis of $\ce{NaCl}$ solution gives chlorine at anode instead of $O_2.$
Reason: Formation of oxygen at anode requires overvoltage.
- ✓
Both assertion and reason are true and the reason is the correct explanation of assertion.
- B
Both assertion and reason are true and the reason is not the correct explanation of assertion.
- C
Assertion is true but the reason is false.
- D
Both assertion and reason are false.
AnswerCorrect option: A. Both assertion and reason are true and the reason is the correct explanation of assertion.
Formation of oxygen has lower value of $E^\circ$ than formation of chlorine even then it is not formed because it requires overvoltage.
View full question & answer→MCQ 612 Marks
Note: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: Copper sulphate can be stored in zinc vessel.
Reason: Zinc is less reactive than copper.
- A
Both assertion and reason are true and the reason is the correct explanation of assertion.
- B
Both assertion and reason are true and the reason is not the correct explanation of assertion.
- ✓
Assertion is true but the reason is false.
- D
Both assertion and reason are false.
AnswerCorrect option: C. Assertion is true but the reason is false.
Zinc will get dissolved in $\ce{CuSO_4}$ solution since zinc is more reactive than copper.
View full question & answer→MCQ 622 Marks
Note: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: $Λ_m$ for weak electrolytes shows a sharp increase when the electrolytic solution is diluted.
Reason: For weak electrolytes degree of dissociation increases with dilution of solution.
- ✓
Both assertion and reason are true and the reason is the correct explanation of assertion.
- B
Both assertion and reason are true and the reason is not the correct explanation of assertion.
- C
Assertion is true but the reason is false.
- D
Both assertion and reason are false.
AnswerCorrect option: A. Both assertion and reason are true and the reason is the correct explanation of assertion.
Weak electrolytes dissociate partially in concentrated solution. On dilution, their degree of dissociation increases hence their $Λ_m$ increases sharply.
View full question & answer→Question 632 Marks
Define the following terms:
- Limiting molar conductivity.
- Fuel cell.
Answer
- The molar conductivity when concentration approaches to zero is called limiting molar conductivity.
- A fuel cell is a device which converts energy produced during the combustion of fuels like hydrogen, methane, methyl alcohol etc. directly into electrical energy. One such successful fuel cell is hydrogen-oxygen fuel cell.
View full question & answer→MCQ 642 Marks
Note: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion: $E_{Ag+/Ag}$ increases with increase in concentration of $Ag^+$ ions.
Reason: $E_{Ag+/Ag}$ has a positive value.
- A
Both assertion and reason are true and the reason is the correct explanation of assertion.
- ✓
Both assertion and reason are true and the reason is not the correct explanation of assertion.
- C
Assertion is true but the reason is false.
- D
Both assertion and reason are false.
AnswerCorrect option: B. Both assertion and reason are true and the reason is not the correct explanation of assertion.
$\text{Ag}^-+\text{e}\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ }\text{Ag}$
$\text{E}_{\text{Ag}^+/\text{Ag}}=\text{E}^\circ_{\text{Ag}^+/\text{Ag}}-\frac{0.059}{1}\log\frac{1}{\big[\text{Ag}^+\big]}$
$=\text{E}^\circ_{\text{Ag}^+/\text{Ag}}+0.059\big[\text{Ag}^+\big]$
On increasing $[Ag],.E_{Ag^+/Ag }$ will increase and it has a positive value.
View full question & answer→Question 652 Marks
Note: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
- Both assertion and reason are true and the reason is the correct explanation of assertion.
- Both assertion and reason are true and the reason is not the correct explanation of assertion.
- Assertion is true but the reason is false.
- Both assertion and reason are false.
- Assertion is false but reason is true.
Assertion: Cu is less reactive than hydrogen.
Reason: $\text{E}^{\ominus}_{\text{Cu}^{2+}/\text{Cu}}$ is negative. Answer
- Assertion is true but the reason is false.
Explanation:
Standard electrode potential of $\text{E}^{0}_{\text{Cu}^{2+}/\text{Cu}}=0.34\text{V}$ and $\text{E}^{0}_{\text{H}^{\pm}/\text{H}}=0.00\text{V}.$ This shows that copper is less reactive than hydrogen. View full question & answer→Question 662 Marks
Note: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
- Both assertion and reason are true and the reason is the correct explanation of assertion.
- Both assertion and reason are true and the reason is not the correct explanation of assertion.
- Assertion is true but the reason is false.
- Both assertion and reason are false.
- Assertion is false but reason is true.
Assertion: Mercury cell does not give steady potential.
Reason: In the cell reaction, ions are not involved in solution.
Answer
- Assertion is false but reason is true.
Explanation:
Correct assertion is mercury cell gives steady potential.
View full question & answer→Question 672 Marks
Silver is electrodeposited on a metallic vessel of total surface area $500\ cm^2$ by passing a current of $0.5 \ \ce{amp}$ $n$ for two hours. Calculate the thickness of silver deposited.
$[$Given$:$ Density of silver $= 10.5\ce{g cm^{-3}},$ Atomic mass of silver $= 108 \ \ce{amu}, F = 96,500 \ \ce{C mol^{-1}}]$
Answer$\text{m}=\text{Z I t}=\frac{108}{96500}\times0.5\times2\times3600=4.029\text{g}$
$\text{d}=\frac{\text{m}}{\text{V}}$
$\Rightarrow\text{V}=\frac{\text{m}}{\text{d}}$
$\text{V}=\frac{4.029}{10.5\text{g cm}^{-3}}$
$=0.3837\ \text{cm}^3$
Let the thickness of silver deposited be $x \ cm.$
$\therefore\text{V = A}\times\text{x}$
$\text{x}=\frac{\text{V}}{\text{A}}$
$\text{x}=\frac{0.3837}{500}$
$=7.67\times10^{-4}\text{cm}$
View full question & answer→Question 682 Marks
Under what condition is $\text{E}_{\text{Cell}}=0\text{ or }\Delta_\text{r}\text{G}=0?$
AnswerAt equilibrium i.e., when the cell is completely discharged,
$\text{E}_{\text{Cell}}=0\text{ and }\Delta_\text{G}=0$
$\therefore\ \Delta_\text{G}=-\text{nFE}$
$\therefore\text{ when E}=0,\Delta_\text{G}$ is also equal zero.
View full question & answer→Question 692 Marks
Write the cell reaction of a lead storage battery when it is discharged. How does the density of the electrolyte change when the battery is discharged?
Answer$\text{Pb}+\text{PbO}_2+2\text{H}_2\text{SO}_4\xrightarrow{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }2\text{PbSO}_4+2\text{H}_2\text{O}$
Density of electrolyte decreases as water is formed and sulphuric acid is consumed as the product during discharge of the battery.
View full question & answer→Question 702 Marks
Unlike dry cell, the mercury cell has a constant cell potential throughout its useful life. Why?
AnswerElectrolyte is not consumed in the cell process of mercury cell hence it will deliver the current at constant potential throughout its life.
View full question & answer→