Question

Calculate the emf of the following concentration cell: Au(s) | Au3+ (0.10 M) || Au3+ (0.50 M) | Au(s)

Answer 1

The emf of the cell as written is calculated to be  0.014 V.

Using the Nernst equation;

Ecell = E°cell - 0.0592/n log Q

We must note that  E°cell = 0 because the anode and cathode are composed of the same type of metal.

Now;

Substituting values, we have;

Ecell = 0 -  0.0592/3 log (0.50 M)/(0.10 M)

Ecell = 0.014 V

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