A solution of $\mathrm{Fe}^{3+}$ is titrated potentiometrically using $\mathrm{Ce}^{3+}$ solution at 25°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 $\mathrm{E}_{\mathrm{Fe}^{3+} \mathrm{Fe}^{2+}}^{\mathrm{o}}=0.77 \mathrm{V}, \log 2=0.301$ )
(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
At 50 % titrated [Fe³⁺] = [Fe²⁺], E = 0.77 V. At 80 %, [Fe²⁺]/[Fe³⁺] = 4: E = 0.77 − 0.0591 log 4 = 0.734 V.