redox questions..

I need help on how to do these sort of questions, will be much appreciated;

Q) By combining half - reaction equations, write balanced equations for the following reactions:

a) Fe3+ (aq) + I- (aq) -->

c) MnO4 - (aq) +Cl- (aq) +H+ (aq) -->

There's nothing after the arrows.... those are 2 questions on the sheet with about 20 of that type, if you could show me how to do these 2 I should be able to do the rest..
thanks.

I need help on how to do these sort of questions, will be much appreciated;

Q) By combining half - reaction equations, write balanced equations for the following reactions:

a) Fe3+ (aq) + I- (aq) -->

c) MnO4 - (aq) +Cl- (aq) +H+ (aq) -->

There's nothing after the arrows.... those are 2 questions on the sheet with about 20 of that type, if you could show me how to do these 2 I should be able to do the rest..
thanks.

²⁺
₂

Stage 1: construct the balanced HALF equations for each component in the equation using electrons to equalise the charges.

Fe³⁺ (aq) + I^- (aq) -->

The Fe3+ gets reduced to Fe²⁺ so it is:

Fe³⁺ + 1e --> Fe²⁺

Iodide gets oxidised to iodine:

2I^- --> I₂ + 2e

Stage 2: Manipulate the equations mathematically to make the electrons equal.

In this case you have to multiply the iron equation by 2. This gives the two half equations, which you can add together directly and the electrons cancel out:

2Fe³⁺ + 2e --> 2Fe²⁺
2I^- --> I₂ + 2e
--------------------------------------
2Fe³⁺ + 2I^- --> 2Fe²⁺ + I₂

As a checksum make sure that the sum of the charges on both sides are equal and also that the sum of the particles of each type are equal (of course).

-------------------------------------------------------------

MnO₄^- (aq) + Cl^- (aq) + H⁺ (aq) -->

2nd example: This is slightly harder as you have to either know, or be able to construct the half equation for the manganate (VII) ion using hydrogen ions to remove the oxygen leaving Mn2+

half equation 1: MnO₄^- + 8H⁺ + 5e --> Mn²⁺ + 4H₂O

half equation 2: 2Cl^- --> Cl₂ + 2e

Now you must make the electrons in both the equations equal. This means multiplying half equation 1 by 2 and half equation 2 by 5. They can then be added together.

2MnO₄^- + 16H⁺ + 10e --> 2Mn²⁺ + 8H₂O
10Cl^- --> 5Cl₂ + 10e
---------------------------------------------------------------
2MnO₄^- + 16H⁺ + 10Cl^- --> 2Mn²⁺ + 8H₂O + 5Cl₂

Double post to correct formatting errors (PC won't let me edit for some reason)

Thanks for explaining the 2nd stage well, it seems simple enough.
But with the first stage, how do you know what each one oxidises or gets reduced to? Such as the iron being reduced to Fe²⁺ and Mn to Mn²⁺. Is there a rule for it?
I probably should've completed all the basics first, although I understand why you added 8H20 which was to equal the oxygens you had on the left side.