Question

For given cell, at T K
Pt | H2(g) | H+ || Fe3+ ; Fe2+ | Pt
( 1 bar ) ( 1 M )
Ecell = .712 V
E°cell = .770 V
if $\frac{[Fe^{2+}]}{[Fe^{3+}]}$ is t $(\frac{2.303\ RT}{F}=.058)$
then find $(\frac{t}{5})$

Answer 1

The correct answer is : 2
.712 = .770 - $\frac{.058}{2}\log[\frac{Fe^{2+}}{Fe^{3+}}]^2$
-.058 = .-058 $\log\frac{[Fe^{2+}]}{[Fe^{3+}]}$
$\frac{Fe^{2+}}{Fe^{3+}}=10=t$
$\frac{t}{5}=2$