Discussion Forum



1)

$MnO_{4}^{-}+8H^{+}+5e^{-} \rightarrow  Mn^{2+}+4H_{2}O;$

$E^{0}=1.51V$

$MnO_{2}+4H^{+}+2e^{-} \rightarrow  Mn^{2+}+2H_{2}O$;

$E^{0}=1.23 VE^{0}_{MnO_{4}^{-}| MnO_{2}}$ is 


A) 1.70 V

B) 0.91 V

C) 1.37 V

D) 0.548V

Answer:

Option A

Explanation:

 On subtracting  eqn.(ii) from (i)  we get

 $MnO_{4}^{-}+4H^{+}+3e^{-} \rightarrow MnO_{2}+2H_{2}O$

 $\therefore$   $-E_{3}= \frac{-1.51 \times 5+2 \times 1.23}{3}$

                $\therefore$    $E_{3}=1.70V$