1) MnO−4+8H++5e−→Mn2++4H2O; E0=1.51V MnO2+4H++2e−→Mn2++2H2O; E0=1.23VE0MnO−4|MnO2 is A) 1.70 V B) 0.91 V C) 1.37 V D) 0.548V Answer: Option AExplanation: On subtracting eqn.(ii) from (i) we get MnO−4+4H++3e−→MnO2+2H2O ∴ −E3=−1.51×5+2×1.233 ∴ E3=1.70V