Question

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

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

Answer 1

Answer: (C)

Given that $E_{F{e}^{2+}/Fe}^o=-0.441V$
So, $Fe\to F{e}^{2+}+2e^{-},E{}^{\circ }=+0.441V$ ...(i)
and $E_{F{e}^{3+}/F{e}^{2+}}^o=0.771V$
So, $F{e}^{3+}+e^{-}\to F{e}^{2+},E{}^{\circ }=0.771V$ ...(ii)
Cell reaction,
(i) $Fe\to F{e}^{2+}+2e^{-}$
(ii) $\underline{2F{e}^{3+}+e^{-}\to 2F{e}^{2+},}$
$Fe+2F{e}^{3+}\to 3F{e}^{2+},$
So, on the basis of cell reaction following half-cell reactions are written At anode
(1) $Fe\to F{e}^{2+}+2e^{-}$ (oxidation)
At cathode
(2) $2F{e}^{3+}+2e^{-}\to 2F{e}^{2+}$ (reduction)
So, $E_{cell}^o=E_{cathode}^o-E_{anode}^o$ $=E_{F{e}^{3+}/F{e}^{2+}}^o-E_{F{e}^{2+}/Fe}^o$ $=(+0.771)-(-0.441)=+1.212V$