Calculate Ecell for the following electrochemical cell:
Ni(s) | Ni2+(aq) || Hg22+(aq) | Hg(l) | Pt(s)
To calculate Ecell, we need to first write the half-reactions and their respective standard reduction potentials:
Ni2+(aq) + 2e- → Ni(s) E°red = -0.26 V
Hg22+(aq) + 2e- → 2Hg(l) E°red = +0.79 V
Next, we need to determine which half-reaction will be oxidized and which will be reduced based on their standard reduction potentials. The half-reaction with the more positive E°red will be reduced, and the half-reaction with the less positive E°red will be oxidized. In this case, the reduction half-reaction is Hg22+(aq) + 2e- → 2Hg(l) and the oxidation half-reaction is Ni(s) → Ni2+(aq) + 2e-
We can now add the two half-reactions and cancel out the electrons to get the overall balanced redox reaction:
Ni(s) + Hg22+(aq) → Ni2+(aq) + 2Hg(l)
The cell diagram can be written as:
Ni(s) | Ni2+(aq) || Hg22+(aq) | Hg(l) | Pt(s)
Using the standard reduction potentials and the Nernst equation, we can calculate Ecell:
E°cell = E°reduction + E°oxidation
E°cell = (+0.79 V) + (-0.26 V)
E°cell = +0.53 V (at standard conditions)
Since the reaction does not involve gases or non-standard concentrations of solutions, we do not need to apply the Nernst equation to calculate Ecell. Therefore, the final answer is Ecell = +0.53 V.