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)

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, $E^{\circ}=E^{\circ}_{\text{cathode}}-E^{\circ}_{\text{anode}}$
$1.5=0.9-E^{\circ}_{\text{anode}}$
$\Rightarrow E^{\circ}_{\text{anode}}=0.9-1.5$
$=-0.6\,V$