7 Balance the following oxidationreduction reactions by meth

Solution

a) C2O42-(aq) + MnO4-(aq) ------> Mn2+(aq) + CO2(g) [acidic]

oxidation half : C2O42-(aq) -2e- -------> 2CO2(g) ]x5

reduction half : MnO4-(aq) +5e- ------> Mn2+(aq) ]x2

Net reaction : 5C2O42-(aq) + 2MnO4-(aq) +16H+ ------> 2Mn2+(aq) + 10CO2(g) + 8H2O

b) Mn2+(aq) + S2O82-(aq) ---------> MnO2(s) + SO42-(aq) [acidic]

oxidation half : Mn2+(aq) -2e --------> MnO2(s)

reduction half : S2O82-(aq) +2e- ------> 2SO42-(aq)

Net reaction : Mn2+(aq) + S2O82-(aq) + 4H+ ---------> MnO2(s) + 2SO42-(aq) +2H2O

c) CN-(aq) + MnO4-(aq) ---------> CNO-(aq) + MnO2 [basic]

oxidation half : CN-(aq) -2 e- ---------> CNO-(aq) ]x3

reduction half : MnO4-(aq) +3e- ------->MnO2(S) ]x2

Net reaction : 3CN-(aq) + 2MnO4-(aq) + H2O --------> 3CNO-(aq) + 2MnO2 + 2OH-