Q. Consider the following cell reaction $2Fe(s) + \ O_{2}(g) + \ {4H}^{+}(aq) \longrightarrow$ ${2Fe}^{2 +}(aq) + \ {2H}_{2}O(l),E{^\circ} = 1.67\ V$ At[${Fe}^{2 +}\rbrack = \ 10^{- 3}\ M\ ,\ P\ {(O}_{2}) = 0.1$ atm and pH = 3, the cell potential at 25℃ is
A) 1.47 V
B) 1.77 V
C) 1.87 V
D) 1.57 V
Explanation
(d)
The half reactions are
$Fe(s)\ \rightarrow \ {Fe}^{2 +}(aq) + \ \ {2e}^{-}\ \times 2$
$O_{2}(g) + \ 4H^{+} + \ {4e}^{-} \longrightarrow \ {2H}_{2}O$
$2Fe(s)\ + \ O_{2}(g) + \ 4H^{+}\ \rightarrow \ 2{Fe}^{2 +}(aq)\ + \ {2H}_{2}O\ (l)$
$\ \ \ \ \ E\ = E{^\circ}\ - \ \frac{0.059}{4}\ \log\ \frac{(10^{- 3})^{2}}{(10^{- 3})^{4}(0.1)}\ \ \ \ = \ \ 1.57\ V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.