Q14 / 180 · KCET 2019 · Chemistry

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

  • Correct answer
    A
    \( 1.7 \mathrm{~V} \)
  • B
    \( 2.1 \mathrm{~V} \)
  • C
    \( 0.3 \mathrm{~V} \)
  • D
    \( 0.1 \mathrm{~V} \)

Solution

(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 $M n^{+7}+3 e^{-} \rightarrow M n^{4+}$ $\Delta \mathrm{G}^{0}=-3 \mathrm{FE}^{0}$ $\therefore-3 \mathrm{FE}^{0}=-7.5 \mathrm{~F}+2.4 \mathrm{~F}$ $\therefore \mathrm{E}^{0}=1.7 \mathrm{~V}$

Hence, the correct answer is option: 'A'.

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