- A$Lu , Cu$
- B$Lu , Zn$
- C$Cu , Zn$
- ✓$Au , Cu$
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$\mathrm{M}(\mathrm{s}) \mid \mathrm{M}^{+}$(aq; $0.05$ molar) || $\mathrm{M}^{+}(\mathrm{aq}), 1$ molar) $\mid \mathrm{M}(\mathrm{s})$
For the above electrolytic cell the magnitude of the cell potential $\left|E_{\text {cell }}\right|=70 \mathrm{mV}$.
$1.$ For the above cell
$(A)$ $\mathrm{E}_{\text {cell }}<0 ; \Delta \mathrm{G}>0$ $(B)$ $\mathrm{E}_{\text {cell }}>0 ; \Delta \mathrm{G}<0$
$(C)$ $\mathrm{E}_{\text {cell }}<0 ; \Delta \mathrm{G}^{\circ}>0$ $(D)$ $\mathrm{E}_{\text {cell }}>0 ; \Delta \mathrm{G}^{\circ}>0$
$2.$ If the $0.05$ molar solution of $\mathrm{M}^{+}$is replaced by $0.0025$ molar $\mathrm{M}^{+}$solution, then the magnitude of the cell potential would be
$(A)$ $35 \mathrm{mV}$ $(B)$ $70 \mathrm{mV}$ $(C)$ $140 \mathrm{mV}$ $(D)$ $700 \mathrm{mV}$
Give the answer question $1,2.$
| List $I$ Oxide | List $II$ Type of Bond |
| $A$ $N _2 O _4$ | $I$. $1 N = O$ bond |
| $B$ $NO _2$ | $II$. $1 N - O - N$ bond |
| $C$ $N _2 O _5$ | $III$. $1 N - N$ bond |
| $D$ $N _2 O$ | $IV$. $1 N = N / N \equiv N$ bond |
Choose the correct answer from the options given below:
Reason : Zinc forms complex with $CuSO_4$

