A voltaic cell is based on the reduction of Ag^+(aq) to Ag(s) and the oxidation of Sn(s) to SN^2+(aq).

Write half-reactions for the cell's anode and cathode, and write a balanced cell reaction.

To determine the half-reactions for the anode and cathode, we need to identify the species being reduced (the cathode) and the species being oxidized (the anode) in the given voltaic cell.

Given:
Reduction of Ag^+(aq) to Ag(s) (cathode)
Oxidation of Sn(s) to Sn^2+(aq) (anode)

Half-reactions:
1. Reduction half-reaction (cathode):
Ag^+(aq) + e^- -> Ag(s)

Explanation:
In the reduction half-reaction, Ag^+(aq) gains an electron (e^-) to form Ag(s). This half-reaction shows the reduction process occurring at the cathode, where positive silver ions (Ag^+) are reduced and deposited as solid silver (Ag).

2. Oxidation half-reaction (anode):
Sn(s) -> Sn^2+(aq) + 2e^-

Explanation:
In the oxidation half-reaction, solid tin (Sn) loses two electrons (2e^-) to form Sn^2+(aq). This half-reaction shows the oxidation process occurring at the anode, where solid tin (Sn) loses electrons and forms aqueous tin ions (Sn^2+).

To combine these two half-reactions into a balanced cell reaction, we need to ensure that the number of electrons transferred is equal in both reactions.

Balanced cell reaction:
2Ag^+(aq) + Sn(s) -> 2Ag(s) + Sn^2+(aq)

Explanation:
By multiplying the reduction half-reaction (cathode) by 2 and adding it to the oxidation half-reaction (anode), we get the balanced cell reaction. The final balanced reaction equation shows that 2 silver ions (Ag^+) react with one tin (Sn) atom to produce 2 solid silver (Ag) and one tin ion (Sn^2+).

What part about this do you not understand?