At what pH, given half cell MnO₄⁻(0.1 M) | Mn²⁺(0.001 M) will have electrode potential of 1.282 V? ___________ (Nearest Integer)
Given E_(MnO₄ - |Mn²⁺)ᵒ=1.54 V,2.303RT/F=0.059V

Solution

Chemistry ma’am · 4 steps

H⁺ sits in this half reaction to the power 8, so E tells you the pH. Find [H⁺] from E.

1. Write the half reaction. MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Five electrons move, so n = 5. Eight H⁺ ions take part, so [H⁺] enters the Nernst equation to the power 8.

   E = E° - 0.059/5 log[Mn²⁺]/[MnO₄⁻][H⁺]⁸

2. Put in numbers. 1.282 = 1.54 - 0.0118 log 0.001/(0.1 [H⁺]⁸). So log 10⁻²/[H⁺]⁸ = 0.258/0.0118 = 21.86.
3. Bring in pH. The left side is -2 - 8 log[H⁺] = -2 + 8 pH. So 8 pH = 23.86, which gives pH = 2.98.
4. So. The pH is 2.98, which rounds to 3.

The common slip: 24. Leaving out the power 8 on [H⁺] gives -2 + pH = 21.86 and a pH of about 24, not 3.

The idea to keep. When H⁺ appears in a half reaction, it belongs in the Nernst quotient with its full stoichiometric power. That power becomes the coefficient of pH when you take the log.

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