The Gibbs change (in J) for the given reaction at
[Cu²⁺] = [Sn²⁺] = 1 M and 298K is :
Cu(s) + Sn²⁺(aq.) → Cu²⁺(aq.) + Sn(s);
(E_(Sn²⁺|Sn)⁰ = - 0.16 V,
E_(Cu²⁺|Cu)⁰ = 0.34 V)
Take F = 96500 C mol –1 )

Solution

Chemistry ma’am · 3 steps

Copper gives electrons to tin here, so E_(cell) turns negative and Δ G positive.

1. Pick the electrodes. In the reaction as written, Cu loses electrons and becomes Cu²⁺, so Cu is the anode. Sn²⁺ gains electrons and becomes Sn, so Sn is the cathode. Two electrons move, so n = 2.
2. Find the cell EMF. E⁰_(cell) = E⁰_(cathode) - E⁰_(anode) = -0.16 - 0.34 = -0.50 V. Both ion concentrations are 1 M, so Q = 1 and the Nernst log term is zero. That gives E_(cell) = E⁰_(cell) = -0.50 V.

   Δ G = -nFE_(cell) = -2 × 96500 × (-0.50)

3. So. Δ G = +96500 J, which is 96500. The positive sign tells you the reaction is not spontaneous as written.

The common slip: -96500. Taking E⁰_(cell) as +0.50 V by putting Cu at the cathode flips the sign and gives -96500.

The idea to keep. Read the anode and cathode off the reaction as it is written, not off the larger E⁰. A negative E_(cell) is allowed, and it just makes Δ G positive.

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