The cell potential for the given cell at 298 K
Pt| H₂ (g, 1 bar) | H⁺ (aq) || Cu²⁺ (aq) | Cu(s)
is 0.31 V. The pH of the acidic solution is found to be 3, whereas the concentration of Cu²⁺ is 10⁻ˣ M. The value of x is ___________.
(Given : E_(Cu²⁺/Cu)^Θ = 0.34 V and (2.303 RT)/F = 0.06 V)

Solution

Chemistry ma’am · 4 steps

Write the full cell reaction first, so you see that H⁺ appears twice and must be squared in Q.

1. Cell reaction. At the anode H₂ → 2H⁺ + 2e⁻, and at the cathode Cu²⁺ + 2e⁻ → Cu. Overall, H₂ + Cu²⁺ → 2H⁺ + Cu with n = 2 and E°_(cell) = 0.34 - 0 = 0.34 V.

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

2. Put in numbers. pH 3 gives [H⁺] = 10⁻³ M, so [H⁺]² = 10⁻⁶. With [Cu²⁺] = 10⁻ˣ, the log term is log 10⁻⁶⁺ˣ = x - 6. The H₂ pressure is 1 bar, so it drops out.
3. Solve for x. 0.31 = 0.34 - 0.03(x - 6), so 0.03(x - 6) = 0.03 and x - 6 = 1.
4. So. x = 7, meaning [Cu²⁺] = 10⁻⁷ M. The answer is 7.

The common slip: 4. Using [H⁺] instead of [H⁺]² in Q gives x - 3 = 1, so 4, not 7.

The idea to keep. In the Nernst equation, write Q from the balanced cell reaction. Every coefficient becomes a power, and n is the number of electrons that reaction transfers.

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