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KCET · Chemistry · Electrochemistry

Given: \( E_{M n^{+7}}^{0} \mid M n^{+2}=1.5 \mathrm{~V} \) and \( E_{M n^{+4}}^{0} \mid M n^{+2}=1.2 \mathrm{~V} \), then \( E_{M n^{+7}}^{0} \mid M n^{+4} \) is

  1. A \( 1.7 \mathrm{~V} \)
  2. B \( 2.1 \mathrm{~V} \)
  3. C \( 0.3 \mathrm{~V} \)
  4. D \( 0.1 \mathrm{~V} \)
Verified Solution

Answer & Solution

Correct Answer

(A) \( 1.7 \mathrm{~V} \)

Step-by-step Solution

Detailed explanation

(A)
\(M n^{+7}+5 e^{-} \rightarrow M n^{2+}\)
\(\Delta G^{0}=-n \mathrm{FE}^{0} \Rightarrow \Delta \mathrm{G}^{0}=-5 \mathrm{~F} 1.5\)
\(M n^{+4}+2 e^{-} \rightarrow M n^{2+}\)
\(\Delta \mathrm{G}^{0}=-2 \mathrm{~F} 1.2\)
Required equation
\(M n^{+7}+3 e^{-} \rightarrow M n^{4+}\)
\(\Delta \mathrm{G}^{0}=-3 \mathrm{FE}^{0}\)
\(\therefore-3 \mathrm{FE}^{0}=-7.5 \mathrm{~F}+2.4 \mathrm{~F}\)
\(\therefore \mathrm{E}^{0}=1.7 \mathrm{~V}\)