TS EAMCET · Chemistry · Electrochemistry
The reduction potential of a half cell consisting of a Pt electrode immersed in \(2.0 \mathrm{M} \mathrm{Fe}^{2+}\) and \(0.02 \mathrm{M} \mathrm{Fe}^{3+}\) solution (in V) is Given \(\left(\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059, \mathrm{E}_{\mathrm{Fe}^{3+} \mid \mathrm{Fe}^{2+}}^0=0.771 \mathrm{~V}\right)\)
- A 0.543
- B 0.653
- C 0.733
- D 0.822
Answer & Solution
Correct Answer
(B) 0.653
Step-by-step Solution
Detailed explanation
\(\mathrm{Fe}^{3+}+\mathrm{e}^{-} \rightarrow \mathrm{Fe}^{2+}\) Using Nernst equation,…
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