Question

Calculate the emf of the following cell at $298K$?
$Fe\left(s\right)|F{e}^{2+}\left(0.001M\right)||H^{+}\left(1M\right)|H_2\left(g\right)\left(1bar\right),Pt\left(s\right)$
(Given: $E_{cell}^{\circ }=+0.44V$)

Answer 1

$Fe\left(s\right)+2H^{+}+\left(aq\right)\leftrightarrow F{e}^{+2}\left(aq\right)+H_2\left(g\right)$

Nernst equation

$E_{cell}=E_{cell}^0-\frac{0.0591}{n}log{k}_c$

There are two electron are transfering so that n=2

$E_{cell}^0=E_{right}^0-E_{left}^0$

$E_{cell}^0=0-(-0.44)V$

$E_{cell}^0=+0.44V$

Put the value we get

Concentration of Solid substance is 1 always so that $\left[Fe\right]=\left[H_2\right]=1$

$\Rightarrow E_{cell}=0.44-\frac{0.0591}{2}log\frac{\left[F{e}^{2+}\right]}{[H^{+}{]}^2}$

$\Rightarrow =0.44-\frac{0.0591}{2}log\left(\frac{0.001}{1^2}\right)$

$\Rightarrow -E_{cell}=0.44-\frac{0.0591}{2}log\left({10}^{-3}\right)$

$\Rightarrow E_{cell}=0.44-\frac{0.0591}{2}(-3)$

$\Rightarrow E_{cell}=0.44+0.089V$

$\Rightarrow E_{cell}=0.53V$