How do I balance these oxidation reduction equations using the half cell method?

[amesalot57]

I do not understand how to balance these oxidation reduction equations.
These equations seem really complicated, I don't understand them. The question that i have to answer is
Balance the following equations by the half cell method. Show both half cell reactions and identify them as oxidation or reduction.
a) SO3^2- + MnO4-+ H+ (arrows) Mn2+ + SO4^2- + H2O
b) Cl2 + OH- (arrows) Cl- + ClO3- + H2O
c) SO4^2- + I- + H+ (arrows) S2- + I2 + H2O

 

[chemisttree]

You need to start by writing out the half reactions. For example the half reaction of iron +3 being reduced to iron +2 would be:

Fe+3 + e- ---> Fe+2

Try it from there.

 

[ravenprp]

Firstly, it's important to understand the basics of oxidation-reduction (redox) reactions. These reactions involve the transfer of electrons between reactants, with one species losing electrons (oxidation) and the other gaining electrons (reduction). In order to balance these equations, we need to ensure that the number of electrons lost by one species is equal to the number of electrons gained by the other species.

The half cell method is a systematic approach to balancing redox equations, which involves breaking the reaction into two half-cell reactions - one for the oxidation half and one for the reduction half. Each half-cell reaction involves the transfer of electrons between two species, and the overall redox reaction can be balanced by combining the two half-cell reactions.

To balance the given equations using the half cell method, follow these steps:

1. Write the two half-cell reactions for each equation, keeping in mind that oxidation occurs on the left side of the arrow and reduction occurs on the right side.

a) Oxidation half: SO3^2- (aq) (arrows) SO4^2- (aq) + 2e-
Reduction half: MnO4- (aq) + 8H+ (aq) + 5e- (arrows) Mn2+ (aq) + 4H2O (l)

b) Oxidation half: Cl2 (g) (arrows) 2Cl- (aq) + 2e-
Reduction half: OH- (aq) + 2e- (arrows) ClO3- (aq) + H2O (l)

c) Oxidation half: SO4^2- (aq) (arrows) S2- (aq) + 2e-
Reduction half: I- (aq) (arrows) I2 (aq) + 2e- + 2H2O (l)

2. Balance the number of electrons in each half-cell reaction by multiplying one or both reactions by a suitable number. In the given equations, the number of electrons is already balanced in each half-cell reaction.

3. Combine the two half-cell reactions, cancelling out any species that appear on both sides of the equation. In the given equations, the electrons cancel out and we are left with the following balanced overall reactions:

a) 2SO3^2- (aq) + MnO4- (aq) +