Given `E_(Cr^(3+)//Cr)^(@)= 0.72V`, `E_(Fe^(2+)//Fe)^(@)=-0.42V`. The potential for the
`cell Cr|Cr^(3+)(0.1M)||Fe^(2+) (0.01M)`| Fe is :

Options

A. 0.26V
B. 0.399V
C. -0.339
D. -0.26V

Explanation

To solve the problem, we need to calculate the cell potential (E_cell) for the electrochemical cell given the standard reduction potentials for chromium and iron. Here’s a step-by-step breakdown of the solution:

### Step 1: Identify the half-reactions and their standard potentials
We have the following standard reduction potentials:
- For chromium: \( E^\circ_{Cr^{3+}/Cr} = 0.72 \, V \)
- For iron: \( E^\circ_{Fe^{2+}/Fe} = -0.42 \, V \)

### Step 2: Determine the anode and cathode
...