2 a) Balance the following Oxidation-reduction reactions that occur in acidic solution using half reaction method and identify the oxidizing agent, the reducing agent, the substance being oxidized, and the substance being reduced
Mn<2+>(aq) + NaBiO3(s) = Bi<3+> (aq) + MnO4<->(aq)
2 b) Same method:
Cu(s) + NO3<-> (aq) = Cu<2+> (aq) + NO(g)
2(a) Mn<2+>(aq) ---> MnO4<->(aq)
Balancing O, Mn<2+> + 4H2O ---> MnO4<->
Balancing H, Mn<2+> + 4H2O ---> MnO4<-> + 8H<+>
Balancing charges, Mn<2+> + 4H2O ---> MnO4<-> + 8H<+> + 5e<->
Similarly, BiO3<-> ---> Bi<3+>
Balancing O, BiO3<-> ---> Bi<3+> + 3H2O
Balancing H, BiO3<-> + 6H<+>---> Bi<3+> + 3H2O
Balancing charges, BiO3<-> + 6H<+> + 2e<-> ---> Bi<3+> + 3H2O ...
The solution is comprised of an explanation of how to balance a redox reaction in an acidic solution.
Cell Potentials and Balancing Redox Equations
Please see the attached file for the fully formatted problems.
In addition to mass balance, oxidation-reduction reactions must be balanced such that the number of electrons lost in the oxidation equals the number of electrons gained in the reduction. This balancing can be done by two methods: the half-reaction method or the oxidation number method. The half-reaction method balances the electrons lost in the oxidation half-reaction with the electrons gained in the reduction half-reaction. In either method , , and may be added to complete the mass balance. Which substances are used depends on the reaction conditions.
In acidic solution, the bromate ion can be used to react with a number of metal ions. One such reaction is
Since this reaction takes place in acidic solution, and will be involved in the reaction. Places for these species are indicated by the blanks in the following restatement of the equation:
BrO_3^- (aq)+ Sn^(2+) (aq)+ ____→Br^- (aq)+Sn^(4+)(aq) + ______
Calculate the standard cell potential at 25C for the following reaction:
Mg(s) +Fe2+(aq) -> Mg2+(aq) + Fe(s)
when H=-617 kJ and S= -301J/K.
E = ________VView Full Posting Details