Question

$Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O;$ $E{}^{\circ }=1.51V$ $Mn{O}_2+4H^{+}+2e^{-}\to M{n}^{2+}+2H_{2}O;$ $E{}^{\circ }=1.23V$

• A. 1.70 V
• B. 0.91 V
• C. 1.37 V
• D. 0.548 V

Answer 1

Answer: (A)

For the reaction, $Mn{O}_4^{-}+4H^{+}+3e^{-}\to Mn{O}_2+2H_{2}O$ $-E_3=\frac{-1.51\times 5+2\times 1.23}{3}$ $\therefore$ $E_3=1.70V$