Question

Split up the following equation into two half equations, one for oxidation and one for reduction: Cu + 2Ag → Cu2+ + 2Ag

Answer 1

The given chemical equation can be split into oxidation and reduction half-reactions. Copper is oxidized at the anode forming the oxidation half-reaction: Cu(s) → Cu2+(aq) + 2e−. Silver is reduced at the cathode forming the reduction half-reaction: 2 × (Ag+(aq) + e− → Ag(s)).

Step-by-step explanation:

To split the given chemical equation into oxidation and reduction half-equations, we first identify which species is being oxidized and which is being reduced. In the reaction Cu + 2Ag → Cu2+ + 2Ag, copper undergoes oxidation and silver undergoes reduction.

The oxidation half-reaction at the anode is:
Cu(s) → Cu2+(aq) + 2e−

The reduction half-reaction at the cathode is:
2 × (Ag+(aq) + e− → Ag(s))

After writing the half-reactions, we can add them together to verify the overall reaction:

2 Ag+(aq) + Cu(s) → 2 Ag(s) + Cu2+(aq)