The charge of Br changed from –1 to 0, therefore it is the
element which is oxidized. Since it is oxidized then Br is also the reducing
agent.
The charge of Mn changed from +4 to +2 therefore it is the
element which is reduced. Since Mn is reduced, then MnO2 is the oxidizing
agent.
The half –reactions are:
Br: 2Br --> Br2 + 2e-
Mn: MnO2 --> Mn2+
First balance oxygen by adding H2O:
MnO2 --> Mn2+ + 2H2O
Then balance hydrogen by adding H+ ions:
4H+ + MnO2 --> Mn2 + 2H2O
Then the appropriate electrons:
4e- + 4H+ + MnO2 --> Mn2 + 2H2O
Multiply the half-reaction of Br by 2 because the half-reaction
of Mn has 4 electrons.
4Br --> 2Br2 + 4e-
Combine the two half reactions and cancel common factors:
4Br- + 4H+ + MnO2 --> 2Br2 + Mn2 + 2H2O
Heat= mass * change in temperature* specific heat
specific heat=409 J/kg K
