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Chemistry 5 Marks Questions

Electrochemistry · Rajasthan Board (RBSE) STD 12 Science (Rajasthan - English Medium). 15 questions.

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Question 15 Marks
Describe the structure of a fuel cell. Draw labelled diagram of a structure of fuel cell?
Answer
Fuel cells : The galvanic cell in which fuels like hydrogen $\left( H _2\right)$, methane $\left( CH _4\right)$, and methanol $\left( CH _3 OH \right)$ are present whose combustion energy is directly converted in to electrical energy which is called fuel cell. In the most successful fuel cell, the reaction of formation of water by the action of hydrogen and oxygen has been used.
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In the fuel cell, bubbles of $H _2$ and $O _2$ gases flow in to the concentrated aqueous NaOH solution through porous carbon electrodes. To increase the speed of electrode reaction, finely divided platinum or palladium metal (catalyst) is mixed in to the electrodes. Electrode reactions are as follows:
Cathode: $O _2(g)+2 H _2 O (l)+4 e ^{-} \rightarrow 4 OH ^{-}( aq )$
Anode: $2 H _2(g)+4 OH ^{-}( aq ) \rightarrow 4 H _2 O ( l )+4 e ^{-}$
The net reaction can be written as :
$
2 H_2(g)+O_2(g) \rightarrow 2 H_2 O(l)
$
As long as $H _2$ and $O _2$ gas continues to flow, electricity continues to be obtained, hence this cell never ends. Thermal plants have an efficiency of only $40 \%$ while fuel cells have an efficiency of $70 \%$.
Fuel cells were used to provide electricity in Apollo space program and water vapours formed by the reaction of $H _2$ and $O _2$ was condensed and used as drinking water for astronauts.
New electrode materials, catalysts and good electrolytes have been developed to increase the efficiency of fuel cells.
Due to being pollution free, fuel cells have been used in vehicles on trial basis.
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Question 25 Marks
Explain fuel cell in detail and explain the shortcomings of power generation from thermal plants.
Answer
self
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Question 35 Marks
Describe graphically the experimental method of measuring the conductivity of electrolyte solutions (ionic solution).
Answer
self
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Question 45 Marks
What is called a battery? Explain its characteristics and describe dry cell and mercury cell.
Answer
self
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Question 55 Marks
Derive the Nernst equation and explain how $E_{\text {cell }}^{\circ}$ depends on concentration of ions.
Answer
self
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Question 85 Marks
Explain the relationship between $\Delta G$ and $E _{\text {cell }}^{\circ}$ with the help of an example.
Answer
self
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Question 95 Marks
Explain relationship between $E _{\text {cell }}^{\circ}$ and equilibrium constant with the help of Nernst equation.
Answer
self
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Question 105 Marks
(a) Draw the structure of a galvanic cell and also describe the reaction taking place in it.
(b) What happens when opposite external potential is applied in the galvanic cell system? Explain.
Answer
self
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Question 115 Marks
Write the Nernst equation and emf of the following cells at 298 K :
(i) $Mg ( s )\left| Mg ^{2+}(0.001 M )|| Cu ^{2+}(0.0001 M )\right| Cu ( s )$
(ii) $Fe ( s )\left| Fe ^{2+}(0.001 M )|| H ^{+}(1 M )\right| H _2(g)(1 bar ) \mid Pt ( s )$
(iii) $Sn ( s )\left| Sn ^{2+}(0.050 M )||H ^{+}(0.020 M )\right| H _2(g)(1 bar ) \mid Pt ( s )$
(iv) $\operatorname{Pt}( s )|\overline{\operatorname{B}}r(0.010 M )|Br _{2(l)}||H ^{+}(0.030 M )|$ $H _2(g)(1 bar ) \mid Pt ( s )$.
Answer
For the reaction :
$E _{ Mg ^{2+} / Mg }^{\circ}=-2.36 V \quad E _{ Cu ^{2+} / Cu }^{\circ}=+0.34 V . $
$E _{ Fe ^{2+} / Fe }^{\circ}=-0.44 V, \quad E _{ Sn ^{2+} / Sn }^{\circ}=-0.14 V . $
$E _{ Br _2 / Br ^{-}}^{\circ}=+1.09 V \quad E _{ H ^{+} / H }^{\circ}=0.0 V .$
(i) Cell reaction for the cell :
$Mg ( s )\left| Mg ^{2+}(0.001 M )\right|\left| Cu ^{2+}(0.0001 M )\right| Cu ( s )$
$Mg _{( s )}+ Cu _{( aq )}^{2+} \rightarrow Mg _{( aq )}^{2+}+ Cu \quad( n =2)$
On applying Nernst equation :
$E _{\text {cell }}= E _{\text {cell }}^{\circ}-\frac{0.059}{2} \log \frac{\left[ Mg ^{2+}\right]}{\left[ Cu ^{2+}\right]}$
$E _{\text {cell }}^{\circ}= E _{\text {cathode }}^{\circ}- E _{\text {anode }}^{\circ}$
In this reaction, Cu is of cathode and Mg is of anode.
$ E _{\text {cell }}^{\circ}=0.34-(-2.36) V $
$E _{\text {cell }}^{\circ}=0.34+2.36=2.7 V$
$ E _{\text {cell }}=2.7-\frac{0.059}{2} \log \left[\frac{(0.001)}{(0.0001)}\right] $
$E _{\text {cell }}=2.7-0.0295 \log 10 \quad(\log 10=1)$
$E _{\text {cell }}=2.7-0.0295 $
$E _{\text {cell }}=2.6705 V .$
(ii) $Fe ( s )\left| Fe ^{2+}(0.001 M )\right|\left| H ^{+}(1 M )\right| H _2(g)(1 bar ) \mid Pt ( s )$
Cell reaction for this cell is as follows :
$Fe _{( s )}+2 H _{( aq )}^{+} \rightarrow Fe _{( aq )}^{2+}+ H _{2(g)} \quad (n =2)$
$E _{\text {cell }}^{\circ}= E _{\text {cathode }}^{\circ}- E _{\text {anode }}^{\circ}$
Here, H is of cathode and Fe is of anode.
$E_{\text {cell }}^{\circ}=0-(-0.44)=+0.44 V$
According to Nernst equation,
$E _{\text {cell }}= E _{\text {cell }}^{\circ}-\frac{0.059}{2} \log \frac{\left[ Fe ^{2+}\right]}{\left[ H ^{+}\right]^2}$
$E _{ cell }^{\circ}=0.44-0.0295 \log \frac{(0.001)}{(1)^2}$
$E _{\text {cell }}=0.44-0.0295 \log 10^{-3}$
$E _{\text {cell }}=0.44-0.0295 \times-3 \log 10 \quad(\log 10=1)$
$E_{\text {cell }}=0.44-0.0295(-3)$
$E _{\text {cell }}=0.44+0.0885$
$E _{\text {cell }}=0.528=0.53 V$
(iii) $\operatorname{Sn}( s )\left| Sn ^{2+}(0.050 M )\right|\left| H ^{+}(0.020 M )\right| H _2(g)(1 bar ) \mid Pt ( s )$
For this cell, cell reaction :
$Sn _{( s )}+2 H _{( aq )}^{+} \rightarrow Sn _{( aq )}^{2+}+ H _{2(g)} \quad( n =2)$
$E _{\text {cell }}^{\circ}= E _{\text {cathode }}^{\circ}- E _{\text {anode }}^{\circ}=0-(-0.14)$
$E _{\text {cell }}^{\circ}=+0.14 V$
According to Nernst equation,
$E _{\text {cell }}= E _{\text {cell }}^{\circ}-\frac{0.059}{2} \log \frac{\left[ Sn ^{2+}\right]}{\left[ H ^{+}\right]^2}$
$E _{\text {cell }}=+0.14-0.0295 \log \frac{[0.050]}{[0.02]^2}$
$E _{\text {cell }}=+0.14-0.0295 \log \frac{0.050}{0.0004}$
$E_{\text {cell }}=+0.14-0.0295 \log 125 $
$E_{\text {cell }}=+0.14-0.0295(2.0969) $
$[\log 125=2.0969]$
$ E _{\text {cell }}=+0.14-0.06 $
$E _{\text {cell }}=0.08 V$
(iv) $Pt _{( s )}\left| Br ^{-}(0.010 M )\right|\left| Br _{2(l)}\right|\left| H ^{+}(0.020 M )\right|H _2(g)(1 \text { bar})| Pt ( s )|$
For this cell, cell reaction,
$2 Br _{( aq )}^{-}+2 H _{( aq )}^{+} \rightarrow Br _{2(l)}+ H _{2(g)}(n =2)$
$ E _{\text {cell }}^{\circ}= E _{\text {cathode}}^{\circ}- E _{\text {anode }}^{\circ} $
$E _{\text {cell }}^{\circ}=0.0-1.09 V$
$E _{ cell }^{\circ}=-1.09 V$
According to Nernst equation,
$E _{\text {cell }}= E _{\text {cell }}^{\circ}-\frac{0.059}{2} \log \frac{1}{\left[ Br ^{-}\right]^2\left[ H ^{+}\right]^2}$
Because, $\left[ H _2\right]_{ g }=1$ and $\left[ Br _2\right]_l=1$
$E _{\text {cell }}=-1.09-0.0295 \log \frac{1}{(0.01)^2(0.03)^2}$
$E _{\text {cell }}=-1.09-\left(0.0295 \log \frac{1}{9 \times 10^{-8}}\right)$
$E _{\text {cell }}=-1.09-\left[0.0295\left(\log 1.1 \times 10^7\right)\right]$
$E_{\text {cell }}=-1.09-(0.0295 \times 7.0414)$
$(\log 1.1=0.0414)$
$ E _{\text {cell }}=-1.09-0.2077 $
$E _{\text {cell }}=-1.297=-1.3 V .$
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Question 125 Marks
Calculate the standard cell potentials of galvanic cell in which the following reactions take place :
(i) $2 Cr ( s )+3 Cd ^{2+}( aq ) \rightarrow 2 Cr ^{3+}( aq )+3 Cd ( s )$
(ii) $Fe ^{2+}$ (aq) $+ Ag ^{+}$(aq) $\rightarrow Fe ^{3+}$ (aq) + Ag (s)
Calculate the $\Delta_r G^{\circ}$ and equilibrium constant of the reactions.
Answer
From active category :
$\begin{array}{ll} E _{ Cr ^{3+} / Cr }^{\circ}=-0.74 V, & E _{ Cd ^{2+} / Cd }^{\circ}=-0.40 V \\ E _{ Ag ^{+} / Ag }^{\circ}=0.80 V, & E _{ Fe ^{3+} / Fe ^{2+}}^{\circ}=0.77 V\end{array}$
(i) $E _{\text {cell }}^{\circ}= E _{\text {cathode }}^{\circ}- E _{\text {anode }}^{\circ}$
In this reaction, Cd is cathode and Cr at anode.
Hence, $E _{\text {cell }}^{\circ}=-0.40-(-0.74)$
$\qquad\qquad\qquad\qquad$$=-0.40+0.74=0.34 V$
$\Delta G ^{\circ}=- n F E _{\text {cell }}^{\circ}$
n = 6 because Cr3+ is being formed from Cr and there are 2 Cr in reaction.
F = 96500 Coulomb,$E_{\text {cell }}^{\circ}=+0.34 V$
Hence,$\Delta G ^{\circ}=-6 \times 96500 \times 0.34$
$=-196860 C V mol^{-1}$ or $-196860 J mol ^{-1}$
$\Delta G ^{\circ}=-196.86 kJ mol ^{-1}$
$\Delta G^{\circ}=-2.303 R T \log K$
$R =8.314$ Joule, $T =298 K$ (standard temperature)
$\begin{array}{l}-196860=-2.303 \times 8.314 \times 298 \log K \\ -196860=-5705.8 \log K\end{array}$
$\log K=\frac{-196860}{-5705.8}$
$\log K=34.5017$
$K=\operatorname{Antilog}(34.5017)=3.1 \times 10^{34}$
Hence, equilibrium constant = 3.1 × 1034
(ii)$E _{\text {cell }}^{\circ}= E _{\text {cathode }}^{\circ}- E _{\text {anode }}^{\circ}$
In this reaction, cathode is of Ag and anode is of Fe.
Hence, $E _{\text {cell }}^{\circ}=0.80-0.77 V=+0.03 V$.
$E _{\text {cell }}^{\circ}=0.03 V$
$\Delta G ^{\circ}=- n F E _{\text {cell }}^{\circ}$
Here, n = 1 because Ag is being formed from Ag+.
$\Delta G^{\circ}=-1 \times 96500 \times 0.03$
$=-2895 CV mol ^{-1}=-2895 J mol ^{-1}$
$\Delta G ^{\ominus}=-2.895 kJ mol ^{-1}$
$\Delta G ^{\ominus}=-2.303 R T \log K$
$\begin{array}{l}-2895=-2.303 \times 8.314 \times 298 \log K \\ -2895=-5705.8 \log K\end{array}$
$\log K=\frac{-2895}{-5705.8}$
$\log K =0.5074$
$K=\operatorname{Antilog}(0.5074)=3.217$
Hence, equilibrium constant = 3.22
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Question 135 Marks
Predict the products of electrolysis in each of the following : 
(i) An aqueous solution of AgNO3 with silver electrodes.
(ii) An aqueous solution of AgNO3 with platinum electrodes.
(iii) An aqueous solution of H2SO4 with platinum electrodes.
(iv) An aqueous solution of CuCl2 with platinum electrodes.
Answer
(i) On electrolysis of aqueous solution of AgNO3 with silver electrodes, the silver electrode at anode will take part in reaction because it is active electrode, hence.
At cathode, Ag will be formed from reduction of Ag+ 
$Ag ^{+}( aq )+ e ^{-} \rightarrow Ag ( s )$ and
At anode : Ag will be oxidised to Ag+
$Ag ( s ) \rightarrow Ag ^{+}( aq )+ e ^{-}$
(ii) On electrolysis of aqueous solution of AgNO3 with platinum Pt electrodes, Pt will not participate in the reaction because it is an inert electrode, hence the reaction.
At cathode : $\quad Ag _{( aq )}^{+}+ e ^{-} \rightarrow Ag _{( s )}$ and
At anode : $\quad H _2 O \rightarrow 2 H ^{+}+\frac{1}{2} O _2+2 e ^{-}$
Because the electrical activity of H2O is more than of NO3- hence H2O will get oxidized.
(iii) On electrolysis of dilute solution of H2SO4 with platinum electrodes, the following reactions will occur because SO42- ion is cell reactive than H2O. 
At cathode : $2 H ^{+}+2 e ^{-} \rightarrow H _2$ (reduction)
At anode : $H _2 O \rightarrow 2 H ^{+}+\frac{1}{2} O _2+2 e ^{-}$(oxidation)
(iv) On electrolysis of aqueous solution of CuCl2 with platinum electrode the following reaction will occur-
At cathode : $Cu ^{2+}+2 e ^{-} \rightarrow Cu$
$\left[ Cu ^{2+}\right.$ is more reactive than $\left.H _2 O \left( H ^{+}\right)\right]$
At anode : $2 Cl ^{-}-2 e ^{-} \rightarrow Cl _2$
[ $Cl ^{-}$is more reactive than $H _2 O \left( OH ^{-}\right)$]
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Question 145 Marks
Three electrolytic cells A, B, C containing solutions of ZnSO4, AgNO3 and CuSO4, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?
Answer
The following reaction is used to deposit silver on the cathode of cell B :
$Ag ^{+}+ e ^{-} \rightarrow Ag$
Atomic mass of 1 mole of Ag = 108 g
Amount of electricity required deposition of 108 g Ag = 96500 C
Therefore, for deposition of 1.45 g silver Ag
$=\frac{96500}{108} \times 1.45$
$=1295.6 C$ amount of electricity is required.
Charge $Q = It , t =\frac{ Q }{ I }=\frac{1295.6}{1.5}=863.73$ second.
Hence, time in minutes $=\frac{863.73}{60}=14.39$
$=14.40$ minutes.
Hence, electric current flowed for 14.40 minutes.
The reaction for deposition of Cu :
$Cu ^{2+}+2 e ^{-} \rightarrow Cu$
Atomic mass of 1 mole Cu = 63.5 g
Hence, copper obtained from 2 × 96500 C = 63.5 g
Copper obtained from 1295.6 C electricity
$\begin{array}{l}=\frac{63.5 \times 1295.6}{2 \times 96500} \\ =0.426 g\end{array}$
Hence, mass of deposited copper = 0.426 g
Reaction required for deposition of Zn :
$Zn ^{2+}+2 e ^{-} \rightarrow Zn$
Atomic mass of 1 mole Zn = 65 g
Zinc obtained from 2 × 96500C = 65 g
Zn obtained from 1295.6 C electricity
$=\frac{65 \times 1295.6}{2 \times 96500}=0.436 g$
Therefore, mass of deposited zinc = 0.436 g.
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Question 155 Marks
The conductivity of sodium chloride at 298 K has been determined at different concentration and the results are given below :
Concentration/M0.0010.0100.0200.0500.100
$10^2 \times \kappa / S m ^{-1}$1.23711.8523.1555.53106.74
Calculate $\Lambda_{ m }$ for all concentrations and draw a plot between $\Lambda_{ m }$ and $c^{1 / 2}$. Find the value of $\Lambda_{ m }{ }^{\circ}$.
Answer
(i) Concentration (c) = 0.001M
$c^{\frac{1}{2}}=\sqrt{0.001}=0.0316 mol^{\frac{1}{2}} L^{\frac{1}{2}}$
$\kappa\left( S m ^{-1}\right)=1.237 \times 10^2$
Molar conductivity $\Lambda_{ m }=\frac{\kappa\left( Sm ^{-1}\right)}{10^3 \times M \left( molL ^{-1}\right)}$
$\Lambda_{ m }=\frac{1.237 \times 10^2}{10^3 \times 0.001}=1.237 \times 10^2$
(ii) $c =0.010, \sqrt{ c }=\sqrt{0.01}=0.1 \quad \kappa=11.85 \times 10^2$
$\Lambda_{ m }=\frac{11.85 \times 10^2}{10^3 \times 0.01}=1.185 \times 10^2$
(iii) $c=0.020, \sqrt{c}=\sqrt{0.020}=0.141$
$\kappa=23.15 \times 10^2$
$\Lambda_{ m }=\frac{23.15 \times 10^2}{10^3 \times 0.020}=1.157 \times 10^2$
(iv) c $=0.050, \sqrt{c}=\sqrt{0.05}=0.224$
$ \kappa  =55.53 \times 10^2 $ \Lambda_{ m }  =\frac{55.53 \times 10^2}{10^3 \times 0.05}=1.110 \times 10^2$
(v) $c=0.10, \sqrt{c}=\sqrt{0.10}=0.316$
$\kappa=106.74 \times 10^2$
$\Lambda_{ m }=\frac{106.74 \times 10^2}{10^3 \times 0.1}=1.067 \times 10^2$
Table
S.No$\sqrt{c}$ or $c^{1 / 2}$$\Lambda_{ m }$
(i)0.03161.237 × 102
(ii)0.101.185 × 102
(iii)0.1411.157 × 102
(iv)0.2241.110 × 102
(v)0.3161.067 × 102
Plotting $\Lambda_{ m }$ through $c^{1 / 2}$ results in straight line in whih the value of $\Lambda_{ m }$ decreases with $c^{1 / 2}$.
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The value of $\Lambda_{ m }{ }^{\circ}$ is calculated by extrapolating the graph to zero concentration which is approximately 1.255 × 102m2mol-1.
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