Question

The cell voltage for the following galvanic cell is 1.27V. Calculate the pH of the solution at 25^oC: 2Ag⁺(0.010 M) + H₂(1 atm) → 2Ag(s) + 2H⁺(pH = ?)

Answer 1

Ag+ is reduced to Ag while H2 is oxidized to H+; the half reactions and the standard electrode potentials are given as

Reduction: Ag+ (aq) + e- --------> Ag (s); E0red = 0.80 V

Oxidation: H2 (g) --------> 2 H+ (aq) + 2 e-‘; E0ox = 0.00 V

E0cell = E0red + E0ox = (0.80 V) + (0.00 V) = 0.80 V

The balanced chemical equation for the cell is given as

2 Ag+ (aq) + H2 (g) -------> Ag (s) + 2 H+ (aq)

The Nernst equation for the cell is given as

Ecell = E0cell – RT/2F*ln [H+]2/[Ag+]2PH2 where 2 denotes the number of electrons transferred.

Given Ecell = 1.27 V and PH2 = 1 atm, we have,

1.27 V = (0.80 V) – (8.314 J/mol.K)*(298 K)/(2*96485 J/V.mol)*ln [H+]2/[Ag+]2 - (8.314 J/mol.K)*(298 K)/(2*96485 J/V.mol)*ln 1/PH2

====> 0.47 V = -(0.0128 V)*ln{[H+]/[Ag+]}2 – 0

====> 0.47 V = -2*(0.0128 V)*ln [H+]/[Ag+]

====> ln [H+]/[Ag+] = -(0.47 V)/(2*0.0128 V) = -18.3594

====> [H+]/[Ag+] = exp^(-18.3594) = 1.0632*10-8

====> [H+] = 1.0632*10-8*(0.010 M) = 1.0632*10-10 M

Therefore, pH = -log [H+] = -log (1.0632*10-10) = 9.9734 ≈ 9.97 (ans).