Answer:
H2O + 2MnO4- + 3CN- ----> 2MnO2 + 2OH- + 3CNO-
Reducing agent: CN-
Oxidizing agent: MnO4-
6OH- + 3Br2 ---> 5Br- + BrO3- + 3H2O
Reducing agent & oxidizing agent: Br2 (Occurs at the same time)
2ClO2 + 2OH- + H2O2 ----> O2 + 2H2O + 2ClO2-
Reducing agent: H2O2
Oxidizing agent: ClO2
Explanation:
- Set the oxidation number of all compounds to identify which is reduced and which is oxidized
- Establish semireactions identifying how many electrons are gained or lost
- Look at the amount of oxygen and add H2O where there are more oxygen on the side of the reaction, while on the opposite side you will add the amount of OH- depending on the amount of H that was added from H2O
-As electrons are almost never balanced, the two half reactions are multiplied by a common number, so you will can simplify the electrons in the final reaction
- If stoichiometric coefficients are divisible by the same number, it should be simplified so that they are with the smallest possible number.
Br2 ------> BrO3-1 + Br-1 (You have all the steps on this)
At the same time bromine is oxidized to bromate and reduced to bromide
Br2 ----> BrO3- (oxidation)
Br2 ----> Br- (reduction)
12OH- + Br2 ----> 2BrO3- + 6H2O + 10e-
(As I have one Bromine and 2, I have to set 2 BrO3-, so instead of 3 H2O, I have to set 6H2O .- now, as I know that Br2 has 0 on oxidation number, the value in BrO3- is +5, so it won 5 electrons but as we put 2 BrO3-, finally it has won 10 electrons. In the end I have to set 12 OH- on the reactive side)
Charges and elements are balanced
2e- + Br2 ---> 2Br-
(As I have one Bromine and 2, I have to set 2 Br-, no oxygen, no H2O, and the Br2 has won 2e- to be reduced to bromide)
(12OH- + Br2 ----> 2BrO3- + 6H2O + 10e-) 2
(2e- + Br2 ---> 2Br-) 10
I have the e- disbalanced so I have to multiply in the first, by 2, in the second by 10
24OH- + 2Br2 + 20 e- + 10Br2 ----> 4BrO3- + 12H2O + 20e- + 20Br-
20 e- on both sides are cancellated
On reactive side 2Br2 + 10Br2 = 12Br2
24OH- + 12Br2 ----> 4BrO3- + 12H2O + 20Br-
The stoichiometric coefficients are multiple of 4, so
24/4OH- + 12/4Br2 ----> 4/4BrO3- + 12/4H2O + 20/4Br-
6OH- + 3Br2 ---> BrO3- + 3H2O + 5Br-
