A voltaic cell is set up at `25^(@)C`with the following half cells :
`Al^(3+)`(0.001 M) and `Ni^(2+)`(0.50 M)
Write the equation for the reaction when the cell generates the electric current. Also determine the cell potential (Given `E_(Ni^(2+)//Ni)^(@)=-0.25 V, E_(Al^(3+)//Al)^(@)=-1.66 V)`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the half-reactions and their standard reduction potentials.
We have two half-cells:
1. Aluminum: \( \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \) with \( E^\circ = -1.66 \, \text{V} \)
2. Nickel: \( \text{Ni}^{2+} + 2e^- \rightarrow \text{Ni} \) with \( E^\circ = -0.25 \, \text{V} \)

### Step 2: Determine which half-reaction will occur at the cathode and which at the anode.
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