Question

Consider the following cell reaction:
$2Fe\left(s\right)+O_2\left(g\right)+4H^{+}\left(aq\right)\to 2F{e}^{2+}\left(aq\right)+2H_{2}O\left(l\right),E^o=1.67V$
At $\left[F{e}^{2+}\right]={10}^{-3}M,P\left(O_2\right)=0.1$ atm and $pH=3$, the cell potential at ${25}^o$C is:

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

Answer 1

Answer: (C)

In the given reaction Fe is oxidised and O${}_2$ is reduced.

$2Fe\to 2F{e}^{2+}+4e^{-}$

$4e^{-}+O_2+4H^{+}\to 2H_{2}O$

$\therefore E_{cell}=E_{O{P}_{Fe}}^o-\frac{0.059}{4}log[F{e}^{2+}{]}^2+E_{R{P}_{O_2}}^o+\frac{0.059}{4}log{P}_{O_2}\times [H^{+}{]}^4$

$=E_{cell}^o+\frac{0.059}{4}log\frac{P_{O_2}\times [H^{+}{]}^4}{[F{e}^{2+}{]}^2}$

$=1.67+\frac{0.059}{4}log\frac{0.1\times ({10}^{-3}{)}^4}{({10}^{-3}{)}^2}$

$=1.67+\frac{0.059}{4}log{10}^{-7}$

$=1.67+\frac{0.059\times (-7)}{4}$

$=1.67-0.103=1.57V$