CBSE 2025 · Region 5 · Set 1 · Q31 · 5 marks
(i)For a galvanic cell, the following half reactions are given. Decide, which will remain as reduction reaction and which will be reversed to become an oxidation reaction. Give reason for your answer.(I)$\displaystyle \mathrm{Cr}^{3+}+3 \mathrm{e}^{-} \rightarrow \operatorname{Cr}(\mathrm{s}) ; \mathrm{E}^{\circ}=-0.74 \mathrm{~V}$(II)$\displaystyle \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe}(\mathrm{s}) ; \mathrm{E}^{\circ}=-0.44 \mathrm{~V}$(ii)Represent the cell in which the following reaction takes place : $\displaystyle \mathrm{Mg}(\mathrm{s})+2 \mathrm{Ag}^{+}(0 \cdot 001 \mathrm{M}) \rightarrow \mathrm{Mg}^{2+}(0 \cdot 100 \mathrm{M})+2 \mathrm{Ag}(\mathrm{s})$ Calculate $\displaystyle \mathrm{E}_{\text {cell }}$ if $\displaystyle \mathrm{E}_{\text {cell }}^{\circ}=3 \cdot 17 \mathrm{~V} .(\log 10=1)$(i)State Kohlrausch's law. Give any two applications of it.(ii)$\displaystyle \mathbf{\Lambda}_{\mathrm{m}}^{\circ} \mathrm{NH}_{4} \mathrm{Cl}, \quad \mathbf{\Lambda}_{\mathrm{m}}^{\circ} \mathrm{NaOH}$ and $\displaystyle \mathbf{\Lambda}_{\mathrm{m}}^{\circ} \mathrm{NaCl}$ are $\displaystyle 129$‧$\displaystyle 8$, $\displaystyle 217.4$, and $\displaystyle 108 \cdot 9 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ respectively. Molar conductivity of $\displaystyle 1 \times 10^{-2} \mathrm{M}$ solution of $\displaystyle \mathrm{NH}_{4} \mathrm{OH}$ is $\displaystyle 9.33 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$. Calculate the degree of dissociation $\displaystyle (\alpha)$ of $\displaystyle \mathrm{NH}_{4} \mathrm{OH}$ solution at this concentration.
(i)
For a galvanic cell, the following half reactions are given. Decide, which will remain as reduction reaction and which will be reversed to become an oxidation reaction. Give reason for your answer.
(I)
$\displaystyle \mathrm{Cr}^{3+}+3 \mathrm{e}^{-} \rightarrow \operatorname{Cr}(\mathrm{s}) ; \mathrm{E}^{\circ}=-0.74 \mathrm{~V}$
(II)
$\displaystyle \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe}(\mathrm{s}) ; \mathrm{E}^{\circ}=-0.44 \mathrm{~V}$
(ii)
Represent the cell in which the following reaction takes place : $\displaystyle \mathrm{Mg}(\mathrm{s})+2 \mathrm{Ag}^{+}(0 \cdot 001 \mathrm{M}) \rightarrow \mathrm{Mg}^{2+}(0 \cdot 100 \mathrm{M})+2 \mathrm{Ag}(\mathrm{s})$ Calculate $\displaystyle \mathrm{E}_{\text {cell }}$ if $\displaystyle \mathrm{E}_{\text {cell }}^{\circ}=3 \cdot 17 \mathrm{~V} .(\log 10=1)$
(i)
State Kohlrausch's law. Give any two applications of it.
(ii)
$\displaystyle \mathbf{\Lambda}_{\mathrm{m}}^{\circ} \mathrm{NH}_{4} \mathrm{Cl}, \quad \mathbf{\Lambda}_{\mathrm{m}}^{\circ} \mathrm{NaOH}$ and $\displaystyle \mathbf{\Lambda}_{\mathrm{m}}^{\circ} \mathrm{NaCl}$ are $\displaystyle 129$‧$\displaystyle 8$, $\displaystyle 217.4$, and $\displaystyle 108 \cdot 9 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$ respectively. Molar conductivity of $\displaystyle 1 \times 10^{-2} \mathrm{M}$ solution of $\displaystyle \mathrm{NH}_{4} \mathrm{OH}$ is $\displaystyle 9.33 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}$. Calculate the degree of dissociation $\displaystyle (\alpha)$ of $\displaystyle \mathrm{NH}_{4} \mathrm{OH}$ solution at this concentration.
Marking-scheme solution
(i)
(II)
will remain as reduction reaction / (II)
(I)
will be reversed to become an oxidation reaction
Due to low reduction potential of Cr
(ii)
Cell representation Mg(s)/Mg2+ (aq,0.100M)‖Ag+(aq,0.001M)/Ag(s)
n=$\displaystyle 2$
Ecell = E°cell −
2.303R⊤
nF
log
[Mg2+]
[Ag+]$\displaystyle 2$
= $\displaystyle 3.17$ −
log
= $\displaystyle 3.17$ −
log $\displaystyle 105$
= $\displaystyle 3.17$ −$\displaystyle 0.0295$ × $\displaystyle 5$
= $\displaystyle 3.17$ −$\displaystyle 0.1475$
= $\displaystyle 3.0225$ V or $\displaystyle 3.02$ V
(i)
Limiting molar conductivity of an electrolyte can be represented as the sum of the
individual contributions of the anion and cation of the electrolyte.
To determine -1. Limiting molar conductivity of an electrolyte.
2.Dissociation constant of a weak electrolyte
(ii)
Λ°mNH4OH = Λ°mNH4Cl +Λ°mNaOH −Λ°mNaCl
= $\displaystyle 129.8$ + $\displaystyle 217.4$ −$\displaystyle 108.9$
=$\displaystyle 238.3$ Scm2mol-$\displaystyle 1$
𝛼=
Λ𝑚𝑐
Λ°𝑚
=$\displaystyle 0.039$ /$\displaystyle 3.9$%
ElectrochemistryGalvanic CellsApplylong_answerhard
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CBSE Class 12 Chemistry past-paper question from the 2025board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.