Question

Given below are half cell reactions:
$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O$
$E_{M{n}^{2+}/Mn{O}_4^{-}}^{\circ }=-1.510V$
$\frac{1}{2}{O}_2+2H^{+}+2e^{-}\to H_{2}O,$
$E_{O_2/H_{2}O}^{\circ }=+1.223V$
Will the permanganate ion, $Mn{O}_4^{-}$liberate $O_2$ from water in the presence of an acid ?

• A. No, because $E_{\text{cell }}^{\circ }=-2.733V$
• B. Yes, because $E_{\text{cell }}^{\circ }=+0.287V$
• C. No, because $E_{\text{cell }}^{\circ }=-0.287V$
• D. Yes, because $E_{\text{cell }}^{\circ }=+2.733V$

Answer 1

Answer: (B)