- A$107.88$
- B$1.6$
- C$0.8$
- ✓$21.6$
At anode $2 \mathrm{OH}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}+\frac{1}{2} \mathrm{O}_{2}+2 \mathrm{e}^{-}$
$E_{A g}=\frac{108}{1}$ ; $E_{O_{2}}=\frac{\frac{1}{2} \times 32}{2}=8$
$\frac{{{\text{W}}_{\text{Ag}}}}{{{\text{E}}_{\text{Ag}}}}=\frac{{{\text{W}}_{{{\text{O}}_{2}}}}}{{{\text{E}}_{{{\text{O}}_{2}}}}}\Rightarrow $ ${{\text{W}}_{\text{Ag}}}=\frac{1.6\times 108}{8}=21.6\,\text{gm}$
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$\frac{2.303 RT }{ F }=0.059\,V$
The potential for the half cell reaction is $x \times 10^{-3}\,V$. The value of $x$ is $........$
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