You are Here: Home

# redox questions.. watch

1. 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. posted topic twice by accident...
3. (Original post by Freshman123)
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.
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 alos 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 yo have to either know, or be able to pconstruct the half equation for the manganate (VII) ion using hydrogen ions to remove the oxygen leaving Mn2+

half equation 1: MnO4[sup-[/sup] + 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[sup-[/sup] + 16H+ + 10e --> 2Mn2+ + 8H2O
10Cl- --> 5Cl2 + 10e
---------------------------------------------------------------
2MnO4[sup-[/sup] + 16H+ + 10Cl- --> 2Mn2+ + 8H2O + 5Cl2
4. (Original post by Freshman123)
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)
5. (Original post by charco)
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.
6. (Original post by Freshman123)
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.
The first stage is basically learning. You can't 'work out' what the results of a reduction or oxidation will be without prior knowledge.

The element iron has three common oxidation states (learned)
These are 0, 2 and 3 (learned)
0 represents the element (learned)
Therefore any redox processes usually occur between 2 and 3 (there is also a 6 oxidation state, but it is almost never seen)

Manganese has several oxidation states 0, 2, 3, 4, 5, 6, 7 (learned)
The common ones are 2,4,7 (learned)
Mn(VII) usually reduces to Mn(II) (learned)

Fundamentally you have to put in the legwork and get around the major oxidants and reductants to get used to the products of their reactions.

For example
-----------
When the thiosulphate ion behaves as a reducing agent it makes the tetrathionate ion. This would be impossible to work out, it must be learned.
2S2O32- --> S4O62- + 2e

sorry to be the harbinger of bad news...
7. Thanks very much for all the help Charco. I guess I'll just have to get around to learning those then.

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: February 17, 2010
The home of Results and Clearing

### 3,098

people online now

### 1,567,000

students helped last year
Today on TSR

### University open days

1. London Metropolitan University
Sat, 18 Aug '18
2. Edge Hill University
Sat, 18 Aug '18
3. Bournemouth University