The cell potential for the following cell
Pt |H₂(g)|H⁺ (aq)|| Cu²⁺ (0.01 M)|Cu(s)
is 0.576 V at 298 K. The pH of the solution is __________. (Nearest integer)
(Given : E_(Cu²⁺/Cu)ᵒ = 0.34 V and (2.303 RT)/F = 0.06 V)

Solution

Chemistry ma’am · 4 steps

Write the cell reaction first; the Nernst term then turns into a straight line in pH.

1. Cell reaction. Hydrogen is oxidised at the anode and copper is reduced at the cathode: H₂ + Cu²⁺ → 2H⁺ + Cu, with n = 2. Since E°_(H+/H₂) = 0, the standard potential is E°_(cell) = 0.34 V. Take p_(H₂) = 1 bar.

   E_(cell) = E°_(cell) - 0.06/2 log[H⁺]²/[Cu²⁺]

2. Bring in pH. log [H⁺]² = -2 pH and log 0.01 = -2, so log[H⁺]²/0.01 = -2 pH + 2. Multiplying by -0.03 gives E_(cell) = 0.34 - 0.06 + 0.06 pH = 0.28 + 0.06 pH.
3. Put in numbers. 0.576 = 0.28 + 0.06 pH, so pH = 0.296/0.06 = 4.93.
4. So. The pH is 4.93, which rounds to 5.

The common slip: 3. Using 0.06 instead of 0.06/2, that is forgetting n = 2, gives 0.576 = 0.22 + 0.12 pH and a pH of about 3.

The idea to keep. Always balance the cell reaction before using Nernst. The electron count n and the power on [H⁺] both come from it, and with a hydrogen electrode the cell potential becomes linear in pH.

Answer from the NTA final answer key. Solution by PrepWiser.