CBSE 2025 · Region 2 · Set 1 · Q32 · 5 marks
(a)Calculate the standard Gibbs energy $\displaystyle \left(\Delta_{\mathrm{r}} \mathrm{G}^{\circ}\right)$ of the following reaction at $\displaystyle 25$ °C : $\displaystyle 3$ + $\displaystyle 2$ $\displaystyle \mathrm{Au}(\mathrm{s})+\mathrm{Ca}^{2+}(1 \mathrm{M}) \rightarrow \mathrm{Au}^{3+}(1 \mathrm{M})+\mathrm{Ca}(\mathrm{s})$ $\displaystyle \mathrm{E}_{\mathrm{Au}^{3+} / \mathrm{Au}}^{\circ}=+1.5 \mathrm{~V}, \mathrm{E}_{\mathrm{Ca}^{2+} / \mathrm{Ca}}^{\circ}=-2.87 \mathrm{~V}$ Predict whether the reaction will be spontaneous or not at $\displaystyle 25$ $\displaystyle { }^{\circ} \mathrm{C}$. $\displaystyle \left[1 \mathrm{~F}=96500 \mathrm{C} \mathrm{mol}^{-1}\right]$(b)Tarnished silver contains $\displaystyle \mathrm{Ag}_{2} \mathrm{~S}$. Can this tarnish be removed by placing tarnished silverware in an aluminium pan containing an inert electrolytic solution such as NaCl ? The standard electrode potential for half reaction : $\displaystyle \mathrm{Ag}_{2} \mathrm{~S}(\mathrm{~s})+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Ag}(\mathrm{s})+\mathrm{S}^{2-}$ is -$\displaystyle 0.71$ V and for $\displaystyle \mathrm{A} l^{3+}+3 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Al}(\mathrm{s})$ is -$\displaystyle 1.66$ VOR 32. (B) (a) Define the following : $\displaystyle 2+3$(i)Cell potential(ii)Fuel cell(b)Calculate emf of the following cell at $\displaystyle 25$ °C : \[\mathrm{Zn}(\mathrm{~s})\left|\mathrm{Zn}_{(0.1 \mathrm{M})}^{2+}\right|\left|\mathrm{Cd}_{(0.01 \mathrm{M})}^{2+}\right| \mathrm{Cd}(\mathrm{~s}) \] Given : $\displaystyle \mathrm{E}_{\mathrm{Cd}^{2+} / \mathrm{Cd}}^{\circ}=-0.40 \mathrm{~V}$ \[\mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V} \] \[[\log 10=1] \]
(a)
Calculate the standard Gibbs energy $\displaystyle \left(\Delta_{\mathrm{r}} \mathrm{G}^{\circ}\right)$ of the following reaction at $\displaystyle 25$ °C : $\displaystyle 3$ + $\displaystyle 2$ $\displaystyle \mathrm{Au}(\mathrm{s})+\mathrm{Ca}^{2+}(1 \mathrm{M}) \rightarrow \mathrm{Au}^{3+}(1 \mathrm{M})+\mathrm{Ca}(\mathrm{s})$ $\displaystyle \mathrm{E}_{\mathrm{Au}^{3+} / \mathrm{Au}}^{\circ}=+1.5 \mathrm{~V}, \mathrm{E}_{\mathrm{Ca}^{2+} / \mathrm{Ca}}^{\circ}=-2.87 \mathrm{~V}$ Predict whether the reaction will be spontaneous or not at $\displaystyle 25$ $\displaystyle { }^{\circ} \mathrm{C}$. $\displaystyle \left[1 \mathrm{~F}=96500 \mathrm{C} \mathrm{mol}^{-1}\right]$
(b)
Tarnished silver contains $\displaystyle \mathrm{Ag}_{2} \mathrm{~S}$. Can this tarnish be removed by placing tarnished silverware in an aluminium pan containing an inert electrolytic solution such as NaCl ? The standard electrode potential for half reaction : $\displaystyle \mathrm{Ag}_{2} \mathrm{~S}(\mathrm{~s})+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Ag}(\mathrm{s})+\mathrm{S}^{2-}$ is -$\displaystyle 0.71$ V and for $\displaystyle \mathrm{A} l^{3+}+3 \mathrm{e}^{-} \longrightarrow 2 \mathrm{Al}(\mathrm{s})$ is -$\displaystyle 1.66$ V
OR 32. (B) (a) Define the following : $\displaystyle 2+3$
(i)
Cell potential
(ii)
Fuel cell
(b)
Calculate emf of the following cell at $\displaystyle 25$ °C : \[\mathrm{Zn}(\mathrm{~s})\left|\mathrm{Zn}_{(0.1 \mathrm{M})}^{2+}\right|\left|\mathrm{Cd}_{(0.01 \mathrm{M})}^{2+}\right| \mathrm{Cd}(\mathrm{~s}) \] Given : $\displaystyle \mathrm{E}_{\mathrm{Cd}^{2+} / \mathrm{Cd}}^{\circ}=-0.40 \mathrm{~V}$ \[\mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V} \] \[[\log 10=1] \]
Marking-scheme solution
(a)
Eocell= Eocathode−Eoanode
= - $\displaystyle 2.87$ − $\displaystyle 1.5$ V
= - $\displaystyle 4.37$ V
△G0 = - nF Eocell
= -$\displaystyle 6$ x $\displaystyle 96500$ X (-$\displaystyle 4.37$)
= $\displaystyle 2530.230$ kJ/mol
Reaction is non-spontaneous.
(b)
Yes, the tarnish can be removed.
Aluminium has more negative standard electrode potential than silver so will reduce silver
sulphide to silver, tarnish will be removed. /
Eocell= Eocathode−Eoanode
= - $\displaystyle 0.71$ −(-$\displaystyle 1.66$) V
= $\displaystyle 0.95$ V
This indicates that the reaction is feasible and tarnish can be removed.
(a)
(i)
Potential difference between two electrodes of a galvanic cell.
(ii)
The galvanic cell in which combustion energy of fuels is directly converted into electrical
energy.
(b)
n =$\displaystyle 2$
Eocell= Eocathode−Eoanode
= - $\displaystyle 0.40$ −(-$\displaystyle 0.76$) V
= $\displaystyle 0.36$ V
ECell = Eoceli -
𝟎⋅𝟎𝟓𝟗
𝒍𝒐𝒈[
𝒁𝒏𝟐+
𝑪𝒅
𝟐+]
= [𝟎. 𝟑𝟔]–
𝟎⋅𝟎𝟓𝟗
𝒍𝒐𝒈
𝟎⋅𝟏
𝟎⋅𝟎𝟏
= (𝟎⋅𝟑𝟔– 𝟎⋅𝟎𝟐𝟗𝟓)
= $\displaystyle 0$⋅$\displaystyle 3305$ V
ElectrochemistryNernst EquationApplylong_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.