A solution of $\mathrm{Fe^{3+}}$ is titrated potentiometrically using $\mathrm{Ce^{3+}}$ solution at $25\,{}^{\circ}\mathrm{C}$. The emf (in V) of the redox system thus formed when, (i) $50\%$ of $\mathrm{Fe^{3+}}$ and (ii) $80\%$ of $\mathrm{Fe^{3+}}$ are titrated, would respectively be (Given $E^{o}_{\mathrm{Fe^{3+}/Fe^{2+}}}=0.77\,V$, $\log 2=0.301$)
(a)0.734 and 0.77
(b)0.77 and 0.385
(c)0.77 and 0.734
(d)0.385 and 0.367
Answer
Answer: C ✓ checked by 4AB · confidence high
Explanation
Ce³⁺ reduces Fe³⁺ to Fe²⁺. At 50 % titrated [Fe³⁺] = [Fe²⁺], so E = E° = 0.77 V. At 80 % titrated [Fe²⁺]/[Fe³⁺] = 4: E = 0.77 − 0.0591·log 4 = 0.77 − 0.036 = 0.734 V.