Calculate the e.f.m of the cell `Cr|Cr^(3+)(0.1M)||Fe^(2+)(0.01M)|Fe`
[given that `E_(Cr^(3+)//Cr)^(@)=-0.75,E_(Fe^(2+)//Fe)^(@)=-0.45V]`

Explanation

To calculate the EMF (Electromotive Force) of the cell represented as `Cr|Cr^(3+)(0.1M)||Fe^(2+)(0.01M)|Fe`, we will follow these steps:

### Step 1: Identify the half-reactions
The half-reactions for the cell are:
- Anode (oxidation): \( \text{Cr} \rightarrow \text{Cr}^{3+} + 3e^- \) (oxidation potential)
- Cathode (reduction): \( \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \) (reduction potential)

### Step 2: Write the standard electrode potentials
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