Question

$E_{F{e}^{2+}/Fe}^0=-0.441V$ and $E_{F{e}^{3+}/F{e}^{2+}}^0=0.771V$, the standard emf of the reaction $Fe+2F{e}^{3+}\to 3F{e}^{2+}$ will be :

• A. 0.111 V
• B. 0.330 V
• C. 1.653 V
• D. 1.212 V

Answer 1

Answer: (D)

$Fe\to F{e}^{2+}+2e^{-}$..........(i)
$E_{FeF{e}^{2+}}^0=-E_{F{e}^{2+}/Fe}^{\circ }=+0.441$
$F{e}^{3+}+e^{-}\to F{e}^{2+}$
$2F{e}^{3+}+2e^{-}\to 2F{e}^{2+}$............(ii)
Adding (i) and (ii) we get,
$Fe+2F{e}^{3+}\to 3F{e}^{2+}$
For this reaction, no. of electron used in oxidation is equal to no. of electron used for reduction. So,
$E_{\text{cell }}^0=E_{OPFeF{e}^{2+}}^0+E_{ROF{e}^{3+}/F{e}^{2+}}^0$
$E^0=0.441+0.771=1.212V$