Answer:
Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s).
Explanation:
- Since E°Ag⁺ is more positive (+ 0.80 V) than that of E°Cu²⁺ (+ 0.34 V).
So, Ag will be the cathode that the reduction reaction is occurred on it and Cu will be the anode where the oxidation reaction is occurred on it.
Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s).
- Ag⁺ is reduced to Ag and Cu is oxidized to Cu²⁺.
