CBSE 2025 Β· Region 6 Β· Set 2 Β· Q32 Β· 5 marks
(i)Calculate $\displaystyle \mathrm{E}_{\text {cell }}$ of a galvanic cell in which the following reaction takes place at $\displaystyle 25^{\circ} \mathrm{C}$ : 
\[\mathrm{Zn}(\mathrm{~s})+\mathrm{Pb}^{2+}(0 \cdot 02 \mathrm{M}) \longrightarrow \mathrm{Zn}^{2+}(0 \cdot 1 \mathrm{M})+\mathrm{Pb}(\mathrm{~s}) \] [Given : $\displaystyle \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V}, \mathrm{E}_{\mathrm{Pb}^{2+} / \mathrm{Pb}}^{\circ}=-0.13 \mathrm{~V}$; $\displaystyle \log 2=0 \cdot 3010, \log 4=0 \cdot 6021, \log 5=0 \cdot 6990]$.(ii)State Faraday's first law of electrolysis. How much electricity, in terms of Faraday, is required to reduce one mol of $\displaystyle \mathrm{MnO}_{4}^{-}$to $\displaystyle \mathrm{Mn}^{2+}$ ion ?(i)The resistance of a conductivity cell containing $\displaystyle 0.001$ M KCl solution at $\displaystyle 298$ K is $\displaystyle 1000$ ohm. What is the cell constant if conductivity of $\displaystyle 0.001$ M KCl solution at $\displaystyle 298$ K is $\displaystyle 0 \cdot 125 \times 10^{-3} \mathrm{~S} \mathrm{~cm}^{-1}$ ?(ii)Calculate the $\displaystyle \mathrm{E}_{\mathrm{Mg}^{2+} / \mathrm{Mg}}$ potential for the following half cell at $\displaystyle 25$Β°C : $\displaystyle \mathrm{Mg} / \mathrm{Mg}^{2+}\left(1 \times 10^{-4} \mathrm{M}\right) ; \mathrm{E}_{\mathrm{Mg}^{2+} / \mathrm{Mg}}^{\circ}=+2 \cdot 36 \mathrm{~V}$ [Given : $\displaystyle \log 10=1$ ](iii)What is the effect of temperature on the electrical conductance of metallic conductor ?
(i)
Calculate $\displaystyle \mathrm{E}_{\text {cell }}$ of a galvanic cell in which the following reaction takes place at $\displaystyle 25^{\circ} \mathrm{C}$ : \[\mathrm{Zn}(\mathrm{~s})+\mathrm{Pb}^{2+}(0 \cdot 02 \mathrm{M}) \longrightarrow \mathrm{Zn}^{2+}(0 \cdot 1 \mathrm{M})+\mathrm{Pb}(\mathrm{~s}) \] [Given : $\displaystyle \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V}, \mathrm{E}_{\mathrm{Pb}^{2+} / \mathrm{Pb}}^{\circ}=-0.13 \mathrm{~V}$; $\displaystyle \log 2=0 \cdot 3010, \log 4=0 \cdot 6021, \log 5=0 \cdot 6990]$.
\[\mathrm{Zn}(\mathrm{~s})+\mathrm{Pb}^{2+}(0 \cdot 02 \mathrm{M}) \longrightarrow \mathrm{Zn}^{2+}(0 \cdot 1 \mathrm{M})+\mathrm{Pb}(\mathrm{~s}) \] [Given : $\displaystyle \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V}, \mathrm{E}_{\mathrm{Pb}^{2+} / \mathrm{Pb}}^{\circ}=-0.13 \mathrm{~V}$; $\displaystyle \log 2=0 \cdot 3010, \log 4=0 \cdot 6021, \log 5=0 \cdot 6990]$.(ii)
State Faraday's first law of electrolysis. How much electricity, in terms of Faraday, is required to reduce one mol of $\displaystyle \mathrm{MnO}_{4}^{-}$to $\displaystyle \mathrm{Mn}^{2+}$ ion ?
(i)
The resistance of a conductivity cell containing $\displaystyle 0.001$ M KCl solution at $\displaystyle 298$ K is $\displaystyle 1000$ ohm. What is the cell constant if conductivity of $\displaystyle 0.001$ M KCl solution at $\displaystyle 298$ K is $\displaystyle 0 \cdot 125 \times 10^{-3} \mathrm{~S} \mathrm{~cm}^{-1}$ ?
(ii)
Calculate the $\displaystyle \mathrm{E}_{\mathrm{Mg}^{2+} / \mathrm{Mg}}$ potential for the following half cell at $\displaystyle 25$Β°C : $\displaystyle \mathrm{Mg} / \mathrm{Mg}^{2+}\left(1 \times 10^{-4} \mathrm{M}\right) ; \mathrm{E}_{\mathrm{Mg}^{2+} / \mathrm{Mg}}^{\circ}=+2 \cdot 36 \mathrm{~V}$ [Given : $\displaystyle \log 10=1$ ]
(iii)
What is the effect of temperature on the electrical conductance of metallic conductor ?
Marking-scheme solution
(i)
ECell = (πΈππβ πΈππ) -
$\displaystyle 0$β
$\displaystyle 059$
πππ[
ππ$\displaystyle 2$+
ππ$\displaystyle 2$+]
= [(β $\displaystyle 0$ β
$\displaystyle 13$) β(β$\displaystyle 0$ β
$\displaystyle 76$)]β
$\displaystyle 0$β
$\displaystyle 059$
πππ
$\displaystyle 0$β
$\displaystyle 1$
$\displaystyle 0$β
$\displaystyle 02$
= $\displaystyle 0.63$ β $\displaystyle 0.0295$ log $\displaystyle 5$
=$\displaystyle 0.63$-$\displaystyle 0.0295$ X $\displaystyle 0.699$
= 0.61V
(ii)
The amount of chemical reaction which occurs at any electrode during electrolysis by a
current is proportional to the quantity of electricity passed through the electrolyte.
5F
(b) (i)
k= G*/R
G* = k X R =$\displaystyle 0.125$ X $\displaystyle 10$-$\displaystyle 3$ X $\displaystyle 1000$
=$\displaystyle 0.125$ cm-$\displaystyle 1$
(ii)
E Mg2+/Mg = E0 Mg2+/Mg β
$\displaystyle 0$β
$\displaystyle 059$
$\displaystyle 2$ log
[ππ$\displaystyle 2$+ ]
= $\displaystyle 2.36$ V β
$\displaystyle 0$β
$\displaystyle 059$
$\displaystyle 2$ log
$\displaystyle 10$β$\displaystyle 4$
= $\displaystyle 2.36$β $\displaystyle 0$Β·$\displaystyle 0295$ X $\displaystyle 4$ log $\displaystyle 10$
= $\displaystyle 2.242$ V
(iii)
It decreases with increase in temperature
ElectrochemistryNernst EquationApplylong_answerhard
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CBSE Class 12 Chemistry past-paper question from the 2025board exam, with the answer as CBSEβs own marking scheme gives it. Where our answers come from.