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

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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

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answer is 1.68.

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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

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