Chemistry - Electrochemistry Question with Solution | TestHub

Since $\mathrm{X}$ is getting dissolved, so reaction is

$\frac{\begin{array}{l}\mathrm{X}\to {\mathrm{X}}^{2+}+2{\mathrm{e}}^{-} \left[\mathrm{Anode}\right]\\ {\mathrm{Y}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Y} \left[\mathrm{cathode}\right]\end{array}}{{\mathrm{Y}}^{2+}+\mathrm{X}\to {\mathrm{X}}^{2+}+\mathrm{Y}}$

${\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{cell}}^{°}-\frac{0.06}{2}\log \frac{{10}^{-3}}{{10}^{-1}}$

${\mathrm{E}}_{\mathrm{cell}}={\mathrm{E}}_{\mathrm{cell}}^{°}+0.06$

(A) $\mathrm{Cd} \& \mathrm{Ni}$

$=0.16+0.05912\times 2$ $=.16+0.0591$ $=0.21(+\mathrm{ve})$

${\mathrm{E}}_{\mathrm{cell}}<0$ (Non-spontaneous)

(B) $\mathrm{Cd} \& \mathrm{Fe}$

${\mathrm{E}}_{\mathrm{cell}}=-0.044-\left(-0.40\right)+0.06$

$=-0.04+0.06$

$=0.02 \mathrm{V}$

${\mathrm{E}}_{\mathrm{cell}}>0$ (Spontaneous)

(C) $\mathrm{Ni} \& \mathrm{Pb}$

${\mathrm{E}}_{\mathrm{cell}}=\left(-0.13-\left(-0.24\right)\right)+0.06$

$=0.11+0.06$

$=0.17 \mathrm{V}$

${\mathrm{E}}_{\mathrm{cell}}>0$ (spontaneous)

(D) $\mathrm{Ni} \& \mathrm{Fe}$

${\mathrm{E}}_{\mathrm{cell}}=\left(-0.44-\left(-0.24\right)\right)+0.06$

$=-0.20+0.06$

$=-0.14 \mathrm{V}$

${\mathrm{E}}_{\mathrm{cell}}<0$ (Non-Spontaneous)