For the reaction MnO4−+8H++5e−→Mn2++4H2O,E0=1.50V MnO2+4H++2e−→Mn2++2H2O,E0=1.23V Thus , for the reaction MnO4−+4H++3e−→MnO2+2H2O,E0(in V) is

Electrons involved are different , hence Enet0≠E10+E20 But ΔGnet0=ΔG10+ΔG20 Signs are taken as per required reaction

Book Online Demo

Check Your IQ

Try Test

Courses

Offline Centres

For the reaction
          $M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O, E^{0} = 1.50 V$
          $M n O_{2} + 4 H^{+} + 2 e^{-} \to M n^{2 +} + 2 H_{2} O, E^{0} = 1.23 V$
         Thus , for the reaction
          $M n O_{4}^{-} + 4 H^{+} + 3 e^{-} \to M n O_{2} + 2 H_{2} O, E^{0}$ (in V) is

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 1.68.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Electrons involved are different , hence $E_{n e t}^{0} \neq E_{1}^{0} + E_{2}^{0}$

But $Δ G_{n e t}^{0} = Δ G_{1}^{0} + Δ G_{2}^{0}$

Signs are taken as per required reaction

Question Image

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE