The minimum voltage (in V) required to bring | vedclass

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(B) At the cathode,reduction of $Cu^{2+}$ ions occurs: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$; $E^{\circ}_{red} = 0.34 \ V$. At the anode,oxidation of

Vedclass · Accepted Answer

$0.89$

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(B) At the cathode,reduction of $Cu^{2+}$ ions occurs: $Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)$; $E^{\circ}_{red} = 0.34 \ V$. At the anode,oxidation of water occurs: $2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^-$; $E^{\circ}_{ox} = -1.23 \ V$. The standard cell potential is $E^{\circ}_{cell} = E^{\circ}_{cathode} + E^{\circ}_{anode} = 0.34 \ V + (-1.23 \ V) = -0.89 \ V$. Since the cell reaction is non-spontaneous $(E^{\circ}_{cell} < 0)$,the minimum external voltage required to drive the electrolysis is the magnitude of the cell potential,which is $0.89 \ V$.

The minimum voltage (in $V$) required to bring about the electrolysis of $1 \ M$ copper$(II)$ sulphate solution at $298 \ K$ is (Given $E^{\circ}_{Cu^{2+}/Cu} = 0.34 \ V$ and $E^{\circ}_{O_2/H_2O} = 1.23 \ V$)

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