$(i)\, Cu^{2+} + 2e^- \rightarrow Cu\,,$ $ E^o = 0.337\, V$
$(ii)\, Cu^{2+} + e^- \rightarrow Cu^+\,,$ $ E^o = 0.153\, V$
Electrode potential, $E^o$ for the reaction,
$Cu^+ + e^- \rightarrow Cu\,,$ will be ............ $V$.
For reaction, $C u^{2+}+2 e^{-} \rightarrow C u$
$\Delta G^{0}=-2 \times F \times 0.337 \ldots(i)$
For reaction, $C u^{+} \rightarrow C u^{2+}+e^{-}$
$\Delta G^{0}=-1 \times F \times 0.153 \quad \dots(i i)$
Adding Eqs. (i) and (ii), we get
$C u^{+}+e^{-} \rightarrow C u, \Delta G^{0}=-0.521 F$
$\Delta G^{0}=-n F E^{0}-0.521 F=-n F E^{0}$
$\therefore E^{0}=0.52 V$
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(Given : atomic number $Sm =62 ; Eu =63 ; Tb =65 ; Gd =64, Pm =61 \text { ) }$
$A.$ $Sm$ $B.$ $Eu$ $C.$ $Tb$ $D.$ $Gd$ $E.$ $Pm$
Choose the correct answer from the options given below:
$2 \mathrm{Fe}_{(\mathrm{s})}+\frac{3}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{Fe}_2 \mathrm{O}_{3(\mathrm{~s})}, \Delta \mathrm{H}^{\mathrm{o}}=-822 \mathrm{~kJ} / \mathrm{mol}$
$\mathrm{C}_{(\mathrm{s})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \rightarrow \mathrm{CO}_{(\mathrm{g})}, \Delta \mathrm{H}^{\mathrm{o}}=-110 \mathrm{~kJ} / \mathrm{mol}$
Then enthalpy change for following reaction
$3\mathrm{C}_{(\mathrm{s})}+\mathrm{Fe}_2 \mathrm{O}_{3(\mathrm{~s})} \rightarrow 2 \mathrm{Fe}_{(\mathrm{s})}+3 \mathrm{CO}_{(\mathrm{g})}$
Reason : In mercury cell, the electrolyte is a paste of $KOH$ and $ZnO$.
