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(A) The given reactions are:$(1)$ $Cu^{2+} + e^- \rightarrow Cu^+$,$E_1^{\circ} = 0.15 \ V$$(2)$ $Cu^+ + e^- \rightarrow Cu$,$E_2^{\circ} = 0.5 \ V$$(

Vedclass · Accepted Answer

$0.325$

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(A) The given reactions are:$(1)$ $Cu^{2+} + e^- ightarrow Cu^+$,$E_1^{\circ} = 0.15 \ V$$(2)$ $Cu^+ + e^- ightarrow Cu$,$E_2^{\circ} = 0.5 \ V$$(3)$ $Cu^{2+} + 2e^- ightarrow Cu$,$E_3^{\circ} = x \ V$Using the relationship $\Delta G_3^{\circ} = \Delta G_1^{\circ} + \Delta G_2^{\circ}$:$-n_3 F E_3^{\circ} = -n_1 F E_1^{\circ} - n_2 F E_2^{\circ}$$n_3 E_3^{\circ} = n_1 E_1^{\circ} + n_2 E_2^{\circ}$$2 	imes x = 1 	imes 0.15 + 1 	imes 0.5$$2x = 0.65$$x = 0.325 \ V$

The value of $x$ from the given data is

Similar Questions

Calculate the cell potential in volts for the cell reaction given below:
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If $E^0_{Fe^{3+}/Fe} = x \ V$ and $E^0_{Fe^{2+}/Fe} = y \ V$,then what will be the value of $E^0_{Fe^{3+}/Fe^{2+}}$?

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