For the voltaic cell Cr(s)/Cr3+(aq)//Ni+2 (aq)/Ni(s), list the reactions at the cathode, anode, and the overall net reaction.

To determine the reactions at the cathode, anode, and overall net reaction for the voltaic cell Cr(s)/Cr3+(aq)//Ni+2(aq)/Ni(s), we need to understand the half-reactions occurring at each electrode.

1. Cathode (Reduction):
The cathode is where reduction takes place. In this case, Ni+2(aq) is being reduced to Ni(s). The half-reaction at the cathode is therefore:

Ni+2(aq) + 2e- → Ni(s)

2. Anode (Oxidation):
The anode is where oxidation occurs. In this case, Cr(s) is being oxidized to Cr3+(aq). To write the half-reaction at the anode, we must balance the atoms and charges. The balanced half-reaction is:

Cr(s) → Cr3+(aq) + 3e-

3. Overall Net Reaction:
To find the overall net reaction, we combine the two half-reactions and cancel out any common species. In this case, the electrons cancel out when the two half-reactions are added together. The overall net reaction is:

Cr(s) + Ni+2(aq) → Cr3+(aq) + Ni(s)

Therefore, the reactions at the cathode, anode, and the overall net reaction for the voltaic cell are as follows:

Cathode (Reduction): Ni+2(aq) + 2e- → Ni(s)
Anode (Oxidation): Cr(s) → Cr3+(aq) + 3e-
Overall Net Reaction: Cr(s) + Ni+2(aq) → Cr3+(aq) + Ni(s)