Explanation:
$\text{Cu}+2\text{H}_2\text{SO}_4\rightarrow\text{CuSO}_4+\text{SO}_2+2\text{H}_2\text{O}$
$3\text{Cu}+8\text{H}\text{NO}_3\rightarrow3\text{Cu(NO}_3)_2+2\text{NO}+4\text{H}_2\text{O}$
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$(i)\,\,(CH_3)_2CH-CH_2Br \xrightarrow{{{C_2}{H_5}OH}} (CH_3)_2CH-CH_2OC_2H_5 + HBr$
$(ii)\,\,(CH_3)_2CH-CH_2-Br \xrightarrow{{{C_2}{H_5}{O^- }}} (CH_3)_2CH-CH_2OC_2H_5 + Br^-$
The mechanisms of reaction $(i)$ and $(ii)$ are respectively
$\mathrm{M}(\mathrm{s}) \mid \mathrm{M}^{+}$(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 $1,2.$
