Question

Consider the following cell reaction:
$2 Fe ( s )+ O _{2}( g )+4 H ^{+}( aq ) \rightarrow 2 Fe ^{2+}( aq )+2 H _{2} O ( l ) ; E^{\circ}=1.67\, V$
At $\left[ Fe ^{2+}\right]=10^{-3} M , p\left( O _{2}\right)=0.1 atm$ and $pH =3$, the cell potential at $25^{\circ} C$ is

• A. $1.47\, V$
• B. $1.77\, V$
• C. $1.87 \,V$
• D. $1.57\, V$

Answer 1

Answer: (D)

The half reactions are
$ Fe(s) \longrightarrow Fe^{2+}(aq) + 2e^- \times 2$
$O_2(g) + 4H^{+} + 4e^- \longrightarrow 2H_2 O$
$2Fe(s)+O_2(g) + 4H^{+} \longrightarrow + 2Fe^{2+}(aq)+ 2H_2 O (l);$
$E=E^\circ-\frac{0.059}{4}\log\frac{(10^{-3})^2}{(10^{-3})^4 (0.1)}=1.57\,V$