From the information in the question, the E° and E for the cell is 0.00 V and 0.12 V.
Using the Nernst equation;
Ecell = E°cell - 0.0592/n log Q
We know that E°cell = 0.00 V since the anode and cathode are both made up of cadmium.
Substituting the given values into the Nernst equation;
Ecell = 0.00 V - 0.0592/2 log (1.0 × 10-5 M/0.100 M)
Ecell = 0.00 V - 0.0296 log(1 × 10^-4)
Ecell = 0.12 V
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