Cu(s) + Sn²⁺ (0.001M) → Cu²⁺ (0.01M) + Sn(s)
The Gibbs free energy change for the above reaction at 298 K is x × 10⁻¹ kJ mol⁻¹. The value of x is __________. [nearest integer]
[Given : E_(Cu²⁺/Cu)^Θ = 0.34 V ; E_(Sn²⁺/Sn)^Θ = - 0.14 V ; F = 96500 C mol⁻¹]

Solution

Chemistry ma’am · 4 steps

Copper is being oxidised here, so this cell runs backwards: find a negative E first, then Δ G.

1. Standard cell potential. Cu loses electrons, so it is the anode. Sn²⁺ gains electrons, so Sn is the cathode. That gives E°_(cell) = E°_(cathode) - E°_(anode) = -0.14 - 0.34 = -0.48 V, with n = 2.
2. Reaction quotient. Solids are left out, so Q = [Cu²⁺]/[Sn²⁺] = 0.01/0.001 = 10.

   E_(cell) = E°_(cell) - 0.0591/n log Q = -0.48 - 0.0591/2 log 10 = -0.50955 V

3. Gibbs energy. Δ G = -nFE_(cell) = -2 × 96500 × (-0.50955) = 98343 J mol⁻¹ ≈ 98.34 kJ mol⁻¹. It is positive, so the reaction is not spontaneous as written.
4. So. 98.34 kJ mol⁻¹ = 983.4 × 10⁻¹ kJ mol⁻¹, so x is 983.

The common slip: 869. Writing Q upside down as 0.001/0.01 makes E = -0.45045 V and gives 869, not 983.

The idea to keep. Read the anode and cathode from the reaction as written, not from which metal is the better reductant. In Q, products go on top and solids drop out. Then Δ G = -nFE with E carrying its sign.

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