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MnO4- + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O Show how Reaction (1) results...

MnO4- + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O

Show how Reaction (1) results from the two half reactions involving Fe3+/Fe2+ and MnO4-/Mn2+.

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Answer #1

Fe+2 oxides to Fe+3
and MnO4- reduces to Mn+2

at anode,
Fe+2 --> Fe+3
for balancing charge add e- to product side
Fe+2 --> Fe+3 + e-

at catode,
MnO4- --> Mn+2
for balancing O add 4 H2O to product side
MnO4- --> Mn+2 + 4H2O

for balancing H add 8H+ to reactant side
MnO4- + 8H+ --> Mn+2 + 4H2O

for balancing charge add 5e- to reactant side
MnO4- + 8H+ + 5e- --> Mn+2 + 4H2O

now multiply anode reaction by 5 and to cathode reaction
we will get required equation

so, the givenequation is sum of two half cell reaction

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