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Given E(Cr^(3+)//Cr)^(@)= 0.72V, E(Fe^(2...

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 :

A

0.26V

B

0.399V

C

-0.339

D

-0.26V

Text Solution

AI Generated Solution

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 ...
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