Balance the following redox equation and identify the element oxidized, the element reduced, the oxidizing agent, and the reducing agent. show all of the work used to solve the problem. br- + mno2 yields br2 + mn2+

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

the right answer is c i just did the question.

c. electrical charges of ionic bonds.