Question 12 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 22 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 32 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 42 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 $(PbO_2)$ and $38\%$ $H_2SO_4$ 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., $PbSO_4$ deposited on electrodes is converted back to Pb and $PbO_2$ and $H_2SO_4$ is regenerated.
View full question & answer→Question 52 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.
$Cu^{2+}(aq) + 2e^- → Cu_{(s)} ; E^\circ = 0.34\ V$
$\text{H}^+_{\text{(aq)}}+\text{e}^-\rightarrow\frac{1}{2}\text{H}_{2\text{(g)}}\ ;\ \text{E}^\circ=0.00\ \text{V}$
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.
$\text{Cl}^-_{\text{(aq)}}\rightarrow\frac{1}{2}\text{Cl}_{2\text{(g)}}+\text{e}^{-1}\ ;\ \text{E}^\circ=1.36\ \text{V}$
$2H_2O_{(I)} → O_{2(g)} + 4H^+_{(aq)} + 4e^- ; E^\circ = +1.23\ V$
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 62 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 $2 × 96487\ C = 58.71g$
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 72 Marks
The conductivity of $0.20\ M$ solution of $KCl$ 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 82 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 $HCl,$
NaOH and NaCl are known. By substituting their values, we can obtain $\wedge^\circ_\text{m}$ for $H_2O.$
View full question & answer→Question 92 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 102 Marks
Predict the products of electrolysis in each of the following:
An aqueous solution of $AgNO_3$ with platinum electrodes.
AnswerAt cathode:
The following reduction reactions compete to take place at the cathode.
$Ag^+_{(aq)} + e^- → Ag_{(s)} ; E^\circ = 0.80V$
$\text{H}^+_{\text{(aq)}}+\text{e}^-\rightarrow\frac{1}{2}\text{H}_{2\text{(g)}};\text{E}^\circ=0.00\ \text{V}$
The reaction with a higher value of E° 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 $O_2.$
$OH^- → OH^+e^-$
$4OH^- → 2H_2O + O_2$
View full question & answer→Question 112 Marks
Predict the products of electrolysis in each of the following:
An aqueous solution of $AgNO_3$ with silver electrodes.
AnswerAt cathode:
The following reduction reactions compete to take place at the cathode.
$Ag+ (aq) + e- → Ag(s) ; E^\circ = 0.80V$
$\text{H}^+_{\text{(aq)}}+\text{e}^-\rightarrow\frac{1}{2}\text{H}_{2\text{(g)}};\text{E}^\circ=0.00\ \text{V}$
The reaction with a higher value of E° 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 122 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 $Pt_{(s)}, H_{2( g )} (1atm) | H^{+(aq)} (1M).$
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 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
Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.
AnswerOxidation of ferrous ion means:
$\mathrm{Fe}^{2+}\rightarrow\mathrm{Fe}^{3+}+\mathrm{e}^{-} ; \mathrm{E}_{\mathrm{Fe} 3+/ \mathrm{Fe} 2+}^0=0.77 \mathrm{~V}$
Any substance which standard electrode potential is more than that of $\mathrm{Fe}^{+3} / \mathrm{F}^{+2}$ can oxidise ferrous ions. (refer to the table given in book)
The EMF of the substance whose reduction potentials greater than 0.77 V will oxidised ferrous ion.
For example $\mathrm{Br}_2, \mathrm{Cl}_2$, and $\mathrm{F}_2$
View full question & answer→Question 152 Marks
The resistance of a conductivity cell containing $0.001\ M\ KCl$ solution at $298$ K is $1500\ Ω.$ What is the cell constant if conductivity of $0.001\ M\ KCl$ 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 162 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 172 Marks
Predict the products of electrolysis in each of the following:
A dilute solution of $\text{H}_2\text{SO}_4$ with platinum electrodes.
AnswerAt the cathode, the following reduction reaction occurs to produce $\text{H}_2$ 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 $\text{O}_2$ gas. But for concentrated sulphuric acid, reaction (ii) occurs.
View full question & answer→Question 182 Marks
Calculate the emf of the cell in which the following reaction takes place $Ni(s) + 2Ag^+ (0.002 M) → Ni^{2+} (0.160 M) + 2Ag(s)$
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 192 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: $2Fe(s) → 2Fe^{2+}+ 4e^−$
$E^\circ(Fe^{2+}, Fe) = -0.44V$
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:
$O_2(g) + 4H^+(aq) + 4e^- → 2H_2O(l)$
$[E^\circ = 1.23\ V]$
The over reaction being
$2Fe(s) + O_2(g) + 4H^+(aq) → 2Fe^{2+}(aq) + 2H_2O(l)$
$[E^\circ(cell) = 1.67\ V]$
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 202 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 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 $Ag = 1.5g$ Current $= i = 1.5amp$
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
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: $\mathrm{Zn}_{(\mathrm{s})} \rightarrow \mathrm{Zn}^{2+}+2 \mathrm{e}^{-}$
Cathode: $\mathrm{MnO}_2+\mathrm{NH}_4^{+}+\mathrm{e}^{-} \rightarrow \mathrm{MnO}(\mathrm{OH})+\mathrm{NH}_3$
View full question & answer→Question 232 Marks
Using IUPAC norms write the formulae for the following :
- Potassium trioxalatoaluminate(III)
- Dichloridobis(ethane-1,2-diamine)cobalt(III)
Answera. $\mathrm{K}_3\left[\mathrm{Al}\left(\mathrm{C}_2 \mathrm{O}_4\right)_3\right]$
b. $\left[\mathrm{CoCl}_2(\mathrm{en})_2\right]^{+}$
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: $\mathrm{Pb}_{(\mathrm{s})}+\mathrm{SO}_4{ }^{2-}(\mathrm{aq}) \rightarrow \mathrm{PbSO}_{4(\mathrm{~s})}+2 \mathrm{e}^{-}$
Cathode: $\mathrm{PbO}_2+\mathrm{SO}_{4(\mathrm{aq})}+4 \mathrm{H}^{+}+2 \mathrm{e}-\rightarrow \mathrm{PbSO}_{4(\mathrm{~s})}+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})}$
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: $Zn(Hg) + 2OH^-\rightarrow ZnO_{(s)} + H2O + 2e^-$
Cathode: $HgO + H_2O + 2e^-\rightarrow Hg(l) + 2OH^-$
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?
Answer Kohlrausch 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 $(E^o)$ for Deniell cell is $+1.1V.$ Calculated the $ʌG^o$ for the reaction
$Zn (s) + Cu^{2+}\ (aq) \rightarrow Zn^2(aq) + Cu\ (s)$
$(1\ F = 96500\ C\ mol^{-1}).$
Answer$\triangle G^o = -n FE^o$ 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 $1.5\ M$ solution of an electrolyte is found to be $138.9\ Scm^2\ mol^{–1}$. 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,Ris resistance and $l/A$ is cell constant
$\Lambda_m = k/C$
Where $\Lambda\ m$ is molar conductivity
$C$ is concentration.
View full question & answer→Question 322 Marks
Two half-reactions of an electrochemical cell are given below:
$MnO_4^- (aq) + 8\ H^+\ (aq) + 5e^– → Mn^{2+} (aq) + 4H_2O\ (l), E^o = +1.51\ V,$
$Sn^{2+}\ (aq) → Sn^{4+}\ (aq) + 2e^–, E^o= +0.15\ V.$
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
${2MnO_4}^- + 5Sn^{2+} + 16H^+\rightarrow 2 Mn^{2+} + 5Sn^{4+} + 8H_2O.$
${E^\circ}_{cell} = {E^\circ}_C-{E^\circ}_A$
$= (+1.51 – 0.15)V = +1.36V$
As ${E^\circ}_{cell}$ is positive, the reaction is product favoured.
View full question & answer→Question 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
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$Ag^+ / Ag < Cu^{2+} / Cu < Fe^{2+} / Fe < Cr^{3+} / Cr < Mg^{2+} / Mg < K^+ / K.$
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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:- $Pb + \text{SO}_{4}^{2-} → PbSO_4(s) + 2e^–.$
Cathode Reaction:- $PbO_2 + 4H^+ + \text{SO}_{4}^{2-} + 2e^– → PbSO_4 + 2H_2O.$
Net reaction:- $Pb + PbO_2 + 2\text{SO}_{4}^{2-}+ 4H^+ → 2PbSO_4 + 2H_2O.$
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Two half cell reactions of an electrochemical cell are given below:
$MnO_4^- (aq) + 8H^+ (aq) + 5e^– → Mn^{2+}(aq) + 4H_2O (l), E^o = + 1.51\ V$
$Sn^{2+} (aq) → Sn^{4+} (aq) + 2e^–, E^o = + 0.15\ V$
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.
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The resistance of a conductivity cell containing $0.001\ M\ KCI$ solution at $298\ K$ is $1500\ Ω$. What is the cell constant if the conductivity of $0.001\ M\ KCI$ solution at $298\ K$ is $0.146 \times 10^{–3}\ S\ cm^{–1}?$
Answer$R= ρ( l/A )$
Cell constant, $l/A = R/ρ = Rκ$
$=(1500 Ω) \times (0.146 \times 10^{-3}\ S\ cm^{-1})$
$= 0.219\ cm^{-1}.$
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Depict the galvanic cell in which the reaction: $Zn(s) + 2Ag^+(aq) → Zn^{2+}(aq) + 2Ag(s)$ 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:
$Zn(s)| Zn^{2+}(aq) || Ag^+(aq)|Ag\ (s)$
- 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: $Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-.$
At cathode: $2Ag^+(aq)+ 2e^-\rightarrow 2Ag(s).$
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Define conductivity and molar conductivity for the solution of an electrolyte.
Answer Conductivity: 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.
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In a galvanic cell, the following cell reaction occurs:
$Zn\ (s)^+ 2 Ag^+ (aq) \longrightarrow Zn^{2+} (aq) + 2 Ag\ (s) {E^o}_{cell} = + 1.56\ V$
- 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.
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Calculate the degree of dissociation $(\alpha)$ of acetic acid if its molar conductivity $(\wedge_m)$ is $39.05\ S\ cm^2\ mol^{–1}.$
Given $\lambda^0(H^+) = 349.6\ S\ cm^2\ mol^{–1}$
$\lambda^0(CH_3COO^–) = 40.9\ S\ cm^2\ mol^{–1}$
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, $\alpha = Λ_m /Λ^o m$
$= 39.05 / 390.5 = 0.1$
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- Following reactions occur at cathode during the electrolysis of aqueous silver chloride solution:
$Ag^+(aq) + e^- \rightarrow Ag(s) E^o= +0.80\ V$
$H^+(aq) + e^- \rightarrow H_2(g) E^o = 0.00\ V$
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
- $Ag+ (aq) + e^- \rightarrow Ag\ (s)$
Reaction with higher $E^0$ 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 $KCl$ at $298\ K$ is $0.025\ S\ cm^{-1}.$ Calculate its moral conductivity.
Answer$\Lambda m = k/C$
$\Lambda\text{m}=\frac{\text{0.025 S cm}^{-1}}{\text{0.20 mol L}^{-1}}$
$\Lambda m =125\ S\ cm^2\ mol^{-1}$
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Determine the values of equilibrium constant $(K_C)$ and $\Delta G^o$ for the following reaction:
$\text{Ni (s)}+\text{2Ag}^{+}\text{(aq)}\rightarrow\text{Ni}^{2+}+\text{2Ag(s)},\text{E}^{o}=1.05\text{V}\ \ \ (\text{IF} = 96500 \text{C mol}^{–1}).$
Answer$\Delta \text{rG}^\circ = -\text{nFE}^\circ$
$= -\ 2 \times (96500\text{C mol }^{-1} \times 1.05\text{V} )$
$= -\ 202650\text{J mol}^{–1} \Rightarrow -\ 202.6\text{kJ mol}^{–1}$
$\text{log K}_{\text{c}}=\frac{\text{nE}^{\circ}}{0.0591}$
$=\frac{2\ \times\ 1.05\text{V}}{0.0591}$
$\text{K}_{\text{c}} = 3.412 \times 1035.$
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The chemistry of corrosion of iron is essentially an electrochemical phenomenon. Explain the reactions occurring during the corrosion of iron in the atmosphere.
Answer$\text{Oxidation:}\text{ Fe (s)}\rightarrow\text{Fe}^{2+}\text{(aq)}+\text{2e}^{+}$.$\text{Reduction: O}_{2}\text{(g)}+\text{4H}^{+}\text{(aq)}+\text{4c}^{-}\rightarrow\text{2H}_{2}\text{O(1)}$.
$\text{Atmospheric oxidation: 2Fe}^{2+}\text{(aq)}+\text{2H}_{2}\text{O(l)}+\frac{1}{2}\text{O}_{2}\text{(g)}^{+}\text{(aq)}\rightarrow+\text{Fe}_{2}\text{O}_{3}\text{(s)}+\text{4H}^{+}\text{(aq)}$.
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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→MCQ 482 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→MCQ 492 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: $Cu$ is less reactive than hydrogen.
Reason: $\text{E}^{\ominus}_{\text{Cu}^{2+}/\text{Cu}}$ is negative.
- 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.
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.
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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.
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