Question

KMnO₄ is used as the titrant in a potentiometric titration of tin(II) to tin(IV), using a S.C.E. reference electrode. Calculate the cell potential in a pH =3.00 buffer solution at the equivalence point.

½ Hg₂Cl₂(s) + e = Hg(l) + Cl^-(aq)                        E(saturated KCl) = 0.241V

Sn⁴⁺ + 2e → Sn²⁺ (analyte)                                E° = 0.139V

MnO₄^- (titrant) + 8H⁺ + 5e → Mn²⁺ + 4H₂O          E⁰ = 1.507V

Answer 1

At equivalence point,

Eq 1: MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O

gives -5E1=-5E1o+0.0591log[MnO4-]/[Mn2+]+0.0591log[H+]8 using nerst equation.

Eq 2: Sn4+ + 2e- --> Sn2+

gives -2FE2=-2FE2o +0.0591log([Sn4+]/[Sn2+] using nerst equation

So, total net reaction: eqn 1 - eqn 2 = 2MnO4- + 16H+ + 5Sn2+ --> 2Mn2+ + 5Sn4+ + 8H2O

E total = [2x eqn 1] - [5 x(2)]:-7 E+ = 5 E°(MnO4-/Mn2+) + 2 E°(Sn4+/Sn2+) + 0.05916 log [H+]8

So, E1 = E°(MnO4-/Mn2+)= 1.5V

E2 = E°(Sn4+/Sn2+) = 0.139V

Given pH = 3

so, [H+] = 10^-3 = 1E-3M

So, 7 E+ = [5 x1.5] + [2 x0.139] + 0.059 log [1E-8]8

7E = 7.601

Ecell = 1.085V