Balance the following redox equations by the ion-electron method: ? a) h2o2 + fe 2+ > fe 3+ + h2o (in the acidic solution) c) cn- + mno4- > cno- +mno2 (in basic solution) e) s2o2/3- + i2 > i- + s4o2/6- (in acidic solution)

We balance the given reactions above by following the rules in balancing redox reactions in acidic or basic solutions. Balance the atoms aside from the O and H atoms. Then we balance the Os and Hs by adding H2O or H+. Finally, we balance the total charge of the reactant and product by adding e-. We do as follows:

A) H2O2 + Fe 2+ ---> Fe 3+ + H2O (in the acidic solution)
   2H+ + H2O2 + Fe 2+ ---> Fe 3+ + 2H2O 
   e- + 2H+ + H2O2 + Fe 2+ ---> Fe 3+ + 2H2O 

C) CN- + MnO4- ---> CNO- +MnO2 (in basic solution)
     CN- + MnO4- ---> CNO- +MnO2 + H2O
     2H+ + CN- + MnO4- ---> CNO- +MnO2 + H2O
     2OH- + 2H+ + CN- + MnO4- ---> CNO- +MnO2 + H2O + 2OH-
     2H2O + CN- + MnO4- ---> CNO- +MnO2 + H2O + 2OH-
     e- + H2O + CN- + MnO4- ---> CNO- +MnO2 + 2OH-

E) S2O2/3- + I2 ---> I- + S4O2/6- (in acidic solution)
    2S2O2/3- + I2 ---> 2I- + S4O2/6- 
    4H+ + 2S2O2/3- + I2 ---> 2I- + S4O2/6- + 2H2O
    6e- + 4H+ + 2S2O2/3- + I2 ---> 2I- + S4O2/6- + 2H2O

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