Respuesta :

superman1987

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.

  • So, the choice will be:

Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s).

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

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