Answer:
Anode: Cr(s) ⇒ Cr³⁺(aq) + 3 e⁻
Step-by-step explanation:
In the anode takes place the oxidation, in which the reducing agent loses electrons.
In the cathode takes place the reduction, in which the oxidizing agent gains electrons.
The corresponding half-reactions are:
Anode: Cr(s) ⇒ Cr³⁺(aq) + 3 e⁻
Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)