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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 find the potential for 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 chromium and iron are: - For chromium: \( \text{Cr}^{3+} + 3e^- \rightarrow \text{Cr} \) with \( E^\circ = +0.72 \, V \) - For iron: \( \text{Fe}^{2+} + 2e^- \rightarrow \text{Fe} \) with \( E^\circ = -0.42 \, V \) ### Step 2: Determine the anode and cathode ...
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