Question

Consider the cell
$Pt _{( s )}\left| H _2( g , 1 atm )\right| H ^{+}( aq , 1 M )|| Fe ^{3+}( aq ), Fe ^{2+}( aq ) Pt ( s )$
When the potential of the cell is $0.712 \, V$ at $298 \,K$, the ratio $\left[ Fe ^{2+}\right] /\left[ Fe ^{3+}\right]$ is ______(Nearest integer)
Given : $Fe ^{3+}+ e ^{-}- Fe ^{2+}, E ^{\circ}Fe ^{3+}, Fe ^{2+} Pt =0.771$
$\frac{2.303 RT }{ F }=0.06 \,V$

Answer 1

$ Pt _{( s )}\left| H _2( g , 1 atm )\right| H ^{+}( aq , 1 M ) \| Fe ^{3+}( aq ), Fe ^{2+}( aq )| Pt |( s ) $
$ \text { at anode } H _2 \longrightarrow 2 H ^{+}+2 e ^{-} $
$ \text {At cathode } Fe _{ aq }^{3+}+ e ^{-} \longrightarrow Fe _{ aq }^{2+}$
$E ^{\circ}= E _{ H _2 \mid H ^{+}}^{\circ}+ E _{ Fe ^{3+} \mid Fe ^{2+}}^{\circ}=0 \cdot 771 V $
$ E = E ^{\circ}-\frac{0 \cdot 06}{1} \log \frac{ Fe ^{2+}}{ Fe ^{3+}}$
$ 0 \cdot 712=(0+0 \cdot 771)-\frac{0 \cdot 06}{1} \log \frac{ Fe ^{2+}}{ Fe ^{3+}} $
$ \log \frac{ Fe ^{2+}}{ Fe ^{3+}}=\frac{0 \cdot 059}{0 \cdot 06} \approx 1 $
$ \frac{ Fe ^{2+}}{ Fe ^{3+}}=10$