In the following reaction occurring in a voltaic cell, what is the Eo(cell)?
2 Ag+(aq) + S2−(aq) ⇒ 2 Ag(s) + S(s)
Ag+(aq) + e− ⇒ Ag(s) Eo = 0.800 V
S(s) + 2e− ⇒ S2−(aq) Eo = -0.476 V
The Eo(cell) can be calculated by adding the individual standard reduction potentials of the half-reactions involved in the voltaic cell.
Eo(cell) = Eo(reduction) + Eo(oxidation)
In this case, the reduction half-reaction is Ag+(aq) + e− ⇒ Ag(s) with Eo(reduction) = 0.800 V.
The oxidation half-reaction is S(s) + 2e− ⇒ S2−(aq) with Eo(oxidation) = -0.476 V.
Eo(cell) = 0.800 V + (-0.476 V) = 0.324 V
Therefore, the Eo(cell) for the given reaction in the voltaic cell is 0.324 V.