An oxidation-reduction reaction in which
3 electrons are transferred has a ΔGº of 17.37 kJ mol –1 at
25 ᵒC. The value of Eᵒ
cell (in V) is ______ × 10 –2 .
(1 F = 96,500 C mol –1 )

Solution

Chemistry ma’am · 4 steps

Link energy to voltage with Δ G° = -nFE°, and watch the minus sign.

1. The link. The free energy change and the cell potential are tied by one equation. Rearrange it to get E°_(cell).

   E°_(cell) = -(Δ G°)/nF

2. Put in numbers. Convert to joules first: Δ G° = 17.37 kJ mol⁻¹ = 17370 J mol⁻¹. With n = 3, nF = 3 × 96500 = 289500 C mol⁻¹.
3. Divide. E°_(cell) = -17370/289500 = -0.06 V. The value is negative because a positive Δ G° means the reaction is not spontaneous.
4. So. E°_(cell) = -6 × 10⁻² V, so the blank is −6.

The common slip: 6. Dropping the minus sign in Δ G° = -nFE° gives 6, but a positive Δ G° must give a negative cell potential.

The idea to keep. Δ G° and E° always have opposite signs. If Δ G° is positive, E°_(cell) must be negative. Change kJ to J before you divide by nF.

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