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Chemistry Β· 2025 Β· 5 marks
CBSE 2025 Β· Region 1 Β· Set 2 Β· Q33
(A) (a) Write the cell reaction and calculate the e.m.f. of the following cell at 298 K: ( 3 + 2 = 5 ) Sn ( s )| Sn^2+(0.004 M )|| H^+(0.02 M )| H_2(β¦
(A)(a)
Write the cell reaction and calculate the e.m.f. of the following cell at $\displaystyle 298$ K : $\displaystyle \boldsymbol{(} \mathbf{3} \boldsymbol{+} \mathbf{2} \boldsymbol{=} \mathbf{5} \boldsymbol{)}$ $\displaystyle \mathrm{Sn}(\mathrm{s})\left|\mathrm{Sn}^{2+}(0.004 \mathrm{M})\right|\left|\mathrm{H}^{+}(0.02 \mathrm{M})\right| \mathrm{H}_{2}(\mathrm{~g})$ ($\displaystyle 1$ Bar) $\displaystyle \mid \mathrm{Pt}(\mathrm{s})$ (Given : $\displaystyle \mathrm{E}_{\mathrm{Sn}^{2+} / \mathrm{Sn}}^{\circ}=-0.14 \mathrm{~V}, \mathrm{E}^{\circ}{ }_{\mathrm{H}+\mid \mathrm{H}_{2}(\mathrm{~g}), \mathrm{Pt}}=0.00 \mathrm{~V}$ )
(b)
Account for the following ;
(i)
On the basis of $\displaystyle \mathrm{E}^{\circ}$ values, $\displaystyle \mathrm{O}_{2}$ gas should be liberated at anode but it is $\displaystyle \mathrm{C} l_{2}$ gas which is liberated in the electrolysis of aqueous NaCl.
(ii)
Conductivity of $\displaystyle \mathrm{CH}_{3} \mathrm{COOH}$ decreases on dilution.
OR 33. (B) (a) Write the anode and cathode reactions and the overall cell reaction occurring in a lead storage battery during its use. ( $\displaystyle \mathbf{2} \boldsymbol{+} \mathbf{3} \boldsymbol{=} \mathbf{5}$ )
(b)
Calculate the potential for half-cell containing $\displaystyle 0.01$ M $\displaystyle \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(\mathrm{aq}), 0.01 \mathrm{M} \mathrm{Cr}^{3+}(\mathrm{aq})$ and $\displaystyle 1.0 \times 10^{-4} \mathrm{M} \mathrm{H}^{+}(\mathrm{aq})$. The half cell reaction is \[\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(\mathrm{aq})+14 \mathrm{H}^{+}(\mathrm{aq})+6 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Cr}^{3+}(\mathrm{aq})+7 \mathrm{H}_{2} \mathrm{O}(l) \] and the standard electrode potential is given as $\displaystyle \mathrm{E}^{\circ}=1.33 \mathrm{~V}$. [Given : $\displaystyle \log 10=1$ ]
Marking-scheme solution
(A)(a)
The cell reaction is
Sn(s)+2H+(aq)βSn2+(aq)+H2(g)
ECell = (πΈππβ πΈππ) -
$\displaystyle 0$β
$\displaystyle 059$
$\displaystyle 2$ log
[ππ$\displaystyle 2$+ ]
[π»+ ].2
= [($\displaystyle 0$) β(β$\displaystyle 0$ β
$\displaystyle 14$)]β
$\displaystyle 0$β
$\displaystyle 059$
πππ
$\displaystyle 0$β
$\displaystyle 004$
($\displaystyle 0$β
$\displaystyle 02$).2
= $\displaystyle 0.14$ - $\displaystyle 0.0295$ log $\displaystyle 10$
= $\displaystyle 0.1105$ V
(b)(i)
overpotential of O2
(ii)
Number of ions carrying current per unit volume decreases on dilution
(B)
a) At anode:
Pb+SO4β$\displaystyle 2$βPbSO4+2eβ
At cathode:
PbO2+ SO4β$\displaystyle 2$+4H++2eββPbSO4+2H2O
Overall reaction:
Pb+PbO2+$\displaystyle 2$ SO4β$\displaystyle 2$+4H+ β2PbSO4+2H2O
(b)
ECell = EoCell -
πβ
πππ
π§
πππ[
[πͺππ+]π
[πͺπππΆππβ] [π―+]ππ]
Ecell = $\displaystyle 1.33$ β
$\displaystyle 6$ log ($\displaystyle 10$-$\displaystyle 2$)$\displaystyle 2$ /($\displaystyle 10$-$\displaystyle 2$)( $\displaystyle 1$ X10-$\displaystyle 4$ )$\displaystyle 14$
= $\displaystyle 1.33$ β
$\displaystyle 6$ ($\displaystyle 54$) log $\displaystyle 10$
= $\displaystyle 1.33$ β $\displaystyle 0.059$ x $\displaystyle 9$
= $\displaystyle 1.33$ β $\displaystyle 0.531$
= $\displaystyle 0.799$ V
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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.