Question

You are trying to determine what a CV will look like if you are analyzing an iron solution using an AgCl reference electrode. The reference solution is 1.0 M, while the iron solution is composed of 0.30 M Fe3+ and 0.20 M Fe2+ ions. The working electrode is platinum and has a diameter of 2 mm.

What is the E0 cell for this spontaneous reaction?

Answer 1

Ag(s) + Cl−(aq) ⇋ AgCl(s) +e−

The cathodic reaction, which is the right half-cell, is the reduction of Fe3+ to Fe2+.

Fe3+(aq) + e− ⇋ Fe2+(aq)

The overall cell reaction, therefore, is

Ag(s) + Fe3+(aq) + Cl−(aq) ⇋ AgCl(s) + Fe2+(aq)

The electrochemical cell’s shorthand notation is

Ag(s) | HCl(aq, aCl− = 0.100), AgCl(sat'd) || FeCl2(aq,aFe2+ = 0.0100), FeCl3(aq, aFe3+ = 0.0500) | Pt(s)

so the E0cell = E0 cathode -E0annode

= 0.77-0.22

= 0.55 V

using Nernts equation

Ecell = E0cell - (0.0591/n) X log(product /reactants)

= 0.55 - 0.00.0591 X log(0.2/0.3)

= 0.56 V