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Chemistry Β· 2025 Β· 5 marks
CBSE 2025 Β· Region 7 Β· Set 1 Β· Q33
(a) (i) Calculate the emf of the following cell at 25Β°C: Zn ( s )| Zn^2+(0.1 M )|| H^+(0.01 M )| H_2( g ) (1 bar), Pt ( s ) [ Given: E_Zn^2+ /β¦
(i)
Calculate the emf of the following cell at $\displaystyle 25$Β°C : $\displaystyle \mathrm{Zn}(\mathrm{s})\left|\mathrm{Zn}^{2+}(0.1 \mathrm{M})\right|\left|\mathrm{H}^{+}(0.01 \mathrm{M})\right| \mathrm{H}_{2}(\mathrm{~g})$ ($\displaystyle 1$ bar), $\displaystyle \mathrm{Pt}(\mathrm{s})$ [ Given : $\displaystyle \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V}, \mathrm{E}_{2 \mathrm{H}^{+} / \mathrm{H}_{2}}^{\circ}=0.00 \mathrm{~V}, \log 10=1$ ]
(ii)
State Faraday's second law of electrolysis. How much electricity is required in terms of Faraday for the reduction of $\displaystyle 1$ mol of $\displaystyle \mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-}$ to $\displaystyle \mathrm{Cr}^{3+}$ ?
Answer the following questions :
(i)
The conductivity of $\displaystyle 0.20$ M solution of KCl is $\displaystyle 2.48 \times 10^{-2} \mathrm{~S} \mathrm{~cm}^{-1}$. Calculate its molar conductivity and degree of dissociation ( $\displaystyle \alpha$ ). \[\begin{aligned} {\left[\text { Given : } \lambda_{\left(\mathrm{K}^{+}\right)}^{\circ}\right.} & =73 \cdot 5 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} \\ \lambda_{\left(\mathrm{Cl}^{-}\right)}^{\circ} & \left.=76 \cdot 5 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\right] \end{aligned} \]
(ii)
Calculate $\displaystyle \Delta_{\mathrm{r}} \mathrm{G}^{\circ}$ of the following cell : \[\mathrm{Mg}(\mathrm{~s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{Mg}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{~s}) \] [Given : $\displaystyle \mathrm{E}_{\mathrm{Mg}^{2+} / \mathrm{Mg}}^{\circ}=-2 \cdot 37 \mathrm{~V}, \quad \mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=+0 \cdot 34 \mathrm{~V}$ \[\left.1 \mathrm{~F}=96500 \mathrm{C} \mathrm{~mol}^{-1}\right] \]
(iii)
What type of cell is mercury cell ? Why is it more advantageous than dry cell ? \[\]
Marking-scheme solution
(a)(i)
ECell = (πΈππβ πΈππ) -
$\displaystyle 0$β
$\displaystyle 059$
$\displaystyle 2$ log
[ππ$\displaystyle 2$+ ]
[π»+ ].2
= [($\displaystyle 0$) β(β$\displaystyle 0$ β
$\displaystyle 76$)]β
$\displaystyle 0$β
$\displaystyle 059$
πππ
$\displaystyle 0$β
$\displaystyle 1$
($\displaystyle 0$β
$\displaystyle 01$).2
= $\displaystyle 0.76$ - $\displaystyle 0.0295$ log $\displaystyle 103$
= $\displaystyle 0.76$ β $\displaystyle 0.0885$
= $\displaystyle 0.6715$ V or $\displaystyle 0.67$ V
(ii)
The amounts of different substances liberated by the same quantity of electricity
passing through the electrolytic solution are proportional to their chemical equivalent
weights.
6F
(b) (i)
π¬π =
πΎ
πΆΓ $\displaystyle 1000$
π¬π =
$\displaystyle 2.48$ Γ $\displaystyle 10$β$\displaystyle 2$
= $\displaystyle 124$ S cm2 molβ$\displaystyle 1$
= $\displaystyle 150$ S cm2 molβ$\displaystyle 1$
(ii)
E0cell= E0cathode- E0 anode
= $\displaystyle 0.34$-(-$\displaystyle 2.37$)
=2.71V
Ξ rG0= -nFE0cell
= -$\displaystyle 2$ X $\displaystyle 96500$ X $\displaystyle 2.71$
= -$\displaystyle 523030$ Jmol-$\displaystyle 1$ or -$\displaystyle 523.03$ kJmol-$\displaystyle 1$
(iii)
Primary cell
Maintains constant potential throughout its usage/ longer lifespan
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CBSE Class 12 Chemistry past-paper question from the 2025 board exam, with the answer as CBSEβs own marking scheme gives it. Where our answers come from.