Question 13 Marks
Calculate the emf of the following cell at $25^{\circ} C$.$Fe \left| Fe ^{2+}(0.001 M )\right|\left| H ^{+}(0.01 M )\right| H _2(g)(1 bar ) \mid Pt ( s )$
Given that, $E_{ Fe ^{2+} / Fe }^o=-0.44 V ; E_{H^{+} / H_2}^o=0.00 V$
Given that, $E_{ Fe ^{2+} / Fe }^o=-0.44 V ; E_{H^{+} / H_2}^o=0.00 V$
Answer
View full question & answer→At anode, $Fe ( s ) \rightarrow Fe ^{2+}( aq )+2 e ^{-}$
At cathode, $\left[ H ^{+}( aq )+ e ^{-} \rightarrow \frac{1}{2} H _2(g)\right] \times 2$
Overall reaction:
$Fe(s)+2 H^{+}(aq) \rightarrow Fe^{2+}(aq)+H_2(g)$
Given, $E_{ Fe ^{2+} / Fe }^o=-0.44 V$ and $E_{H^{+} / H_2}^o=0.00 V$
$\left.E^0(\text { cell })=E^0 \text { (cathode) }-E^0 \text { (anode }\right)=[0.00-(-0.44)]=0.44 V$
Now, Applying Nernst equation:
$E_{\text {cell }}=E_{\text {cell }}^o-\frac{0.00591}{2} \log \frac{\left[Fe^{2+}\right]}{\left[H^{+}\right]^2}=0.44-\frac{0.00591}{2} \log \frac{10^{-3}}{\left(10^{-2}\right)^2}$
$=0.44-\frac{0.00591}{2}=0.4104 V$
At cathode, $\left[ H ^{+}( aq )+ e ^{-} \rightarrow \frac{1}{2} H _2(g)\right] \times 2$
Overall reaction:
$Fe(s)+2 H^{+}(aq) \rightarrow Fe^{2+}(aq)+H_2(g)$
Given, $E_{ Fe ^{2+} / Fe }^o=-0.44 V$ and $E_{H^{+} / H_2}^o=0.00 V$
$\left.E^0(\text { cell })=E^0 \text { (cathode) }-E^0 \text { (anode }\right)=[0.00-(-0.44)]=0.44 V$
Now, Applying Nernst equation:
$E_{\text {cell }}=E_{\text {cell }}^o-\frac{0.00591}{2} \log \frac{\left[Fe^{2+}\right]}{\left[H^{+}\right]^2}=0.44-\frac{0.00591}{2} \log \frac{10^{-3}}{\left(10^{-2}\right)^2}$
$=0.44-\frac{0.00591}{2}=0.4104 V$
