Question

The $EMF$ of a galvanic cell by coupling two electrodes $M_1\left|M_1^{2+}\left(0.1M\right)\right|\left|M_2^{2+}\left(0.01M\right)\right|M_2$ is $+1.47V$ . If the $E^{\circ }$ value (reduction potential) of $M_2$ electrode is $0.9V,E^{\circ }$ (reduction potential) value of $M_1$ electrode in volts would be
$\left[Assume\frac{2.303RT\left(T=298k\right)}{F}=0.06\right]$

• A. $-0.57$
• B. $-0.60$
• C. $+0.57$
• D. $+0.60$

Answer 1

Answer: (B)

The overall cell reaction can be represented as:
$M_1+M_2^{2+}\to M_1^{2+}+M_2$
$E_{cell}=E^{\circ }-\frac{2.303RT}{nF}log\frac{M_1^{2+}}{M_2^{2+}}$
$1.47=E^{\circ }-\frac{0.06}{2}log\frac{0.1}{0.01}$
$\Rightarrow 1.47=E^{\circ }-0.03$
$\Rightarrow E^{\circ }=1.47+0.03=1.5V$
Now, $E^{\circ }=E_{\text{cathode}}^{\circ }-E_{\text{anode}}^{\circ }$
$1.5=0.9-E_{\text{anode}}^{\circ }$
$\Rightarrow E_{\text{anode}}^{\circ }=0.9-1.5$
$=-0.6V$