Question

Helpppppppppp!!!!!!!

Answer 1

Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s).

Step-by-step 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.

• So, the choice will be:

Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s).

• Ag⁺ is reduced to Ag and Cu is oxidized to Cu²⁺.