Question

The $EMF$ of a galvanic cell by coupling two electrodes $M_{1} \left|M_{1}^{2+}\left(0.1 \, M\right)\right|\left|M_{2}^{2+}\left(0.01\,M\right)\right|M_{2}$ is $+1.47\,V$ . If the $E^{\circ}$ value (reduction potential) of $M_{2}$ electrode is $0.9\, V, E^{\circ}$ (reduction potential) value of $M_{1}$ electrode in volts would be $\left[Assume \frac{2.303RT\left(T=298\,k\right)}{F}=0.06\right]$

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

Answer 1

Answer: (B)

$-0.60$

Answer 2

The overall cell reaction can be represented as:
$M_{1}+M^{2+}_{2} \to M^{2+}_{1}+M_{2}$
$E_{cell}=E^{\circ}-\frac{2.303 \,RT}{nF} log \frac{M^{2+}_{1}}{M^{2+}_{2}}$
$1.47=E^{\circ}-\frac{0.06}{2} log \frac{0.1}{0.01}$
$\Rightarrow 1.47=E^{\circ}-0.03$
$\Rightarrow E^{\circ} =1.47+0.03=1.5\,V$
Now, $E^{\circ}=E^{\circ}_{\text{cathode}}-E^{\circ}_{\text{anode}}$
$1.5=0.9-E^{\circ}_{\text{anode}}$
$\Rightarrow E^{\circ}_{\text{anode}}=0.9-1.5$
$=-0.6\,V$