At 298 K, a 1 litre solution containing 10 mmol of Cr₂O₇²⁻ and 100 mmol of Cr³⁺ shows a pH of 3.0.
Given : Cr₂O₇²⁻ → Cr³⁺ ; E° = 1.330V
and (2.303 RT)/F = 0.059 V
The potential for the half cell reaction is x×10⁻³ V. The value of x is __________

Solution

Chemistry ma’am · 4 steps

Balance the half-reaction first. The 14 H⁺ in it make pH the biggest term in the Nernst equation.

1. Balanced half-reaction. In acid, dichromate is reduced as Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. So n = 6, Cr³⁺ is squared in Q, and H⁺ is raised to the 14th power.

   E = E° - 0.059/6 log[Cr³⁺]²/[Cr₂O₇²⁻][H⁺]¹⁴

2. Concentrations. The volume is 1 L, so [Cr₂O₇²⁻] = 0.01 M and [Cr³⁺] = 0.1 M. pH 3 gives [H⁺] = 10⁻³ M.
3. Find log Q. Q = (0.1)²/(0.01 × (10⁻³)¹⁴) = 10⁻²/(10⁻² × 10⁻⁴²) = 10⁴², so log Q = 42.
4. So. E = 1.330 - (0.059 × 42)/6 = 1.330 - 0.413 = 0.917 V = 917 × 10⁻³ V, so x = 917.

The common slip: 907. Leaving out the square on [Cr³⁺] gives log Q = 43 and E = 0.907 V, so 907 instead of 917.

The idea to keep. For any half-cell with H⁺ in it, write the fully balanced equation before you touch Nernst. Every stoichiometric coefficient becomes a power in Q.

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