Calculate the cell voltage for the following reaction:
Cu2+ (0.010 M) + H2(1 atm) —> Cu(s) + 2H+ (pH = 7.0)
First, we need to find the reduction potential (E°) for each half-reaction:
Cu2+ + 2e- → Cu E° = 0.34 V (from standard reduction potential table)
H+ + e- → 1/2 H2 E° = 0.00 V (from standard reduction potential table)
Since we're interested in the cell voltage (Ecell), we need to add the reduction potential of the reduction half-reaction to the oxidation potential of the oxidation half-reaction:
Ecell = E°(reduction) + (- E°(oxidation))
Ecell = 0.00 V + (-0.34 V)
Ecell = -0.34 V
Therefore, the cell voltage for this reaction is -0.34 V.