The minimum voltage (in V ) required to bring about the electrolysis of 1 M copper (II) sulphate solution at 298 ~K is (Given E_Cu^2+ / Cu^°=0.34 ~V and E_H_2 O / H^+^°=-1.23 ~V )

Correct answer: (B) +0.89. Oxidation : (anode) 2 H_2 O(I) arrow O_2(~g)+4 H^+ (aq.) ; E^°=+1.23 ~V Reduction : (cathode) Cu^2+(aq .)+2 e^- arrow Cu(s) ; E^°=+0.34 ~V E_cell^°=E_cathode^°-E_anode^°=(0.34)-(+1.23)=-0.89 ~V . Thus, a potential of minimum of +0.89 ~V would be required to carry out the given reaction.

Q24 / 160 · AP EAMCET 2023 (18 May Shift 2) · Chemistry

The minimum voltage (in $\mathrm{V}$ ) required to bring about the electrolysis of $1 \mathrm{M}$ copper (II) sulphate solution at $298 \mathrm{~K}$ is $$ \text { (Given } \mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\circ}=0.34 \mathrm{~V} \text { and } \mathrm{E}_{\mathrm{H}_2 \mathrm{O} / \mathrm{H}^{+}}^{\circ}=-1.23 \mathrm{~V} \text { ) } $$

A
+1.57
Correct answer
B
+0.89
C
-0.89
D
-1.57

Solution

Oxidation : (anode) $$ 2 \mathrm{H}_2 \mathrm{O}(\mathrm{I}) \rightarrow \mathrm{O}_2(\mathrm{~g})+4 \mathrm{H}^{+} \text {(aq.) } ; \mathrm{E}^{\circ}=+1.23 \mathrm{~V} $$ Reduction : (cathode) $$ \begin{aligned} & \mathrm{Cu}^{2+}(\mathrm{aq} .)+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{s}) ; \mathrm{E}^{\circ}=+0.34 \mathrm{~V} \\ & \mathrm{E}_{\text {cell }}^{\circ}=\mathrm{E}_{\text {cathode }}^{\circ}-\mathrm{E}_{\text {anode }}^{\circ}=(0.34)-(+1.23)=-0.89 \mathrm{~V} \text {. } \end{aligned} $$ Thus, a potential of minimum of $+0.89 \mathrm{~V}$ would be required to carry out the given reaction.

Hence, the correct answer is option: 'B'.

Solution

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