In the following reaction occurring in a voltaic cell, what is the Eo(cell)?

Question

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

Answer 1

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.