Write the redox reaction between Sn2+ and MnO4- in acidic conditions. what is reduced/oxidized?
oxidation half reaction
Sn2+ (aq) ----> Sn4+ (aq) + 2e- -----1
reduction half reaction
MnO4- (aq) + 8H+ (aq) + 5e- -----> Mn2+ (aq) + 4H2O (l) ----2
in order to balace the electrons
multiply the equation1 with 5 and
equation 2 with 2
5Sn2+ (aq) ----> 5Sn4+ (aq) + 10e-
2MnO4- (aq) + 16H+ (aq) + 10e- -----> 2Mn2+ (aq) + 8H2O (l) add both equations
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2MnO4- (aq) + 5Sn2+ (aq) + 16H+ (aq) -------.. 2Mn2+ (aq) + 5Sn4+ (aq) +8H2O (l) this is the balanced redox reaction
MnO4- get reduced
Sn2+ get oxidised
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