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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