Given: $\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\mathrm{o}}=0.34 \mathrm{~V}$
$\mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^0=0.8 \mathrm{~V}$
$\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V}$
$1$ Faraday electricity was passed through $\mathrm{Cu}^{2+}(1.5$ $\mathrm{M}, 1 \mathrm{~L}) / \mathrm{Cu}$…
Given: $\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\mathrm{o}}=0.34 \mathrm{~V}$
$\mathrm{E}_{\mathrm{Ag}^{+} / \mathrm{Ag}}^0=0.8 \mathrm{~V}$
$\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.06 \mathrm{~V}$
Solution
$\begin{aligned}& (1 \text { faraday }=\text { charge on } 1 \text { mole electron }) \\& \mathrm{t}=0 \quad 1.5 \quad 1 \mathrm{~mole} \\& \mathrm{t}=\mathrm{t} \quad 1 \quad-\quad 0.5 \text { mole }\end{aligned}$
$\left[\mathrm{Cu}^{+2}\right]=1 \mathrm{M}$ after electrolysis
* $\mathrm{Ag} \oplus+\mathrm{e}^{-} \quad \longrightarrow \mathrm{Ag}$
$\begin{array}{lll}\mathrm{t}=0 & 0.2 & 0.1 \text { mole } \\ \mathrm{t}=\mathrm{t} & 0.1 & - \qquad -\end{array}$
$\left[\mathrm{Ag}^{+}\right]=0.1 \mathrm{M}$ after electrolysis
Cell $\quad \mathrm{Cu}_{(\mathrm{s})}+2 \mathrm{Ag}_{(\mathrm{aq})}^{+} \rightarrow \mathrm{Cu}_{(\mathrm{aq})}^{+2}+2 \mathrm{Ag}_{(\mathrm{s})}$
reaction
$\begin{aligned} & \mathrm{E}=\mathrm{E}^{\circ}-\frac{0.06}{\mathrm{n}} \log \frac{\left[\mathrm{Cu}^{+2}\right]}{\left[\mathrm{Ag}^{+}\right]^2} \\ & \mathrm{E}=(0.8-0.34)-\frac{0.06}{2} \log \frac{1}{(0.1)^2}=0.4 \mathrm{~V}\end{aligned}$
Correct answer $=400 \mathrm{mV}$
Asked in: JEE Main 2025 (07 Apr Shift 1)