Answer:
Anode (oxidation): Cr(s) ⇒ Cr³⁺(aq) + 3 e⁻
Cathode (reduction): Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
Explanation:
Let's consider the notation of a galvanic cell.
Cr(s) | Cr³⁺(aq) || Ag⁺(aq) | Ag(s)
On the left, it is represented the anode (oxidation) and on the right, it is represented the cathode (reduction).
The half-reactions are:
Anode (oxidation): Cr(s) ⇒ Cr³⁺(aq) + 3 e⁻
Cathode (reduction): Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
To have the global reaction, we have to multiply the reduction by 3 (so the number of electrons gained and lost are the same) and add both half-reactions.
Global reaction: Cr(s) + 3 Ag⁺(aq) ⇒ Cr³⁺(aq) + 3 Ag(s)