Chemistry › Electrochemistry › Cells and Electrode Potential, Nernst Equation
The electrode potential for $$ \begin{aligned} & M^{2+}(a q)+e^{-} \longrightarrow M^{+}(a q) \\ & M^{+}(a…
The electrode potential for
$$
\begin{aligned}
& M^{2+}(a q)+e^{-} \longrightarrow M^{+}(a q) \\
& M^{+}(a q)+e^{-} \longrightarrow M(s)
\end{aligned}
$$
are $+0.15 \mathrm{~V}$ and $+0.50 \mathrm{~V}$ respectively. The value of $E^{\circ}{ }_{M^{2+} / M}$ will be
0.150 V 0.300 V 0.325 V 0.650 V Solution
Given,
$
\begin{aligned}
& M^{2+}(a q)+e^{-} \longrightarrow M^{+}(a q), E_1^{\circ}=+0.15 \mathrm{~V} \\
& M^{+}(a q)+e^{-} \longrightarrow M(s), E_2^{\circ}=0.5 \mathrm{~V}
\end{aligned}
$
$\because$ Free energy, i.e. $\left[\Delta G^{\circ}\right]=-n F E^{\circ}$
where, $n=$ number of electron involved
$
E^{\circ}=\text { standard electrode potential. }
$
$F=$ Faraday's constant
To find $E_3^{\circ}$ for $E_{\left(M^{2+} / M\right)}^{\circ}$
Thus,
$
\begin{array}{lr}
-\Delta G_1^{\circ}=n E_1^{\circ}=E_1^{\circ} & (\because n=1) \\
-\Delta G_2^{\circ}=n E_2^{\circ}=E_{{ }_2}{ }_2 & (\because n=1) \\
-\Delta G_3^{\circ}=n E_3=2 E_3^{\circ} & (\because n=2) \\
\text { [For reaction : } M^{2+}(a q)+2 e^{-} & \longrightarrow M(s) \text { ] }
\end{array}
$
Therefore,
$
\begin{aligned}
-\Delta G_3^{\circ} & =(-) \Delta G_1^{\circ}+(-) \Delta G_2^{\circ} \\
-2 E_3^{\circ} & =-\left(n E_1^{\circ}+n E_2^{\circ}\right) \\
2 E_3^{\circ} & =E_1^{\circ}+E_2^{\circ} \\
2 E_3^{\circ} & =0.15+0.5 \\
2 E_3^{\circ} & =0.65 \mathrm{~V} \\
E_3^{\circ} & =\frac{0.65}{2}=0.325 \mathrm{~V}
\end{aligned}
$
Hence, option (c) is the correct answer
Asked in: AP EAMCET 2019 (21 Apr Shift 1)
Practice more Electrochemistry questions on Aicharya