For the following cell, calculate the emf:
`Al//Al^(3+) (0.01 M)"||" Fe^(2+) (0.02M)"|"Fe`
Given: `E_(Al^(3+)//Al)^(Theta) = -1.66V, E_(Fe^(2+)//Fe)^(Theta)= -0.44V`

Explanation

To calculate the emf of the given electrochemical cell, we will follow these steps:

### Step 1: Identify the half-reactions and their standard electrode potentials.
- The half-reaction for aluminum is:
\[ \text{Al}^{3+} + 3e^- \rightarrow \text{Al} \quad E^\circ = -1.66 \, \text{V} \]
- The half-reaction for iron is:
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