A common type of question in A-level Chemistry Preliminary Exams is redox calculation questions.

This type of question usually appears in Paper 1 (MCQ) which allows only 1.5 minutes to solve.

It sometimes appears in Papers 2 and 3 as well. A look at a few of these questions reveal a pattern in which the questions are asked:


1. CJC 2012 Paper 1

The mineral tellurite, TeO2 ( Mr= 160.0) is often used in the manufacture of optic fibres. It was found that 1.01g of TeO2 in an ore sample required exactly 60 cm3 of 0.035 moldm-3acidified K2Cr2O7 for complete reaction. In this reaction, Cr2O72- is converted into Cr3+.

What is the oxidation state of Te in the final product?

A.  +2 B.  +3 C.  +5 D.  +6

2. HCI 2012 Paper 1

When iron is reacted with aqueous iron (III) oxide, iron (II) ions are formed.
Assuming the reaction goes to completion, how many moles of Fe and Fe3+ (aq) would result in a mixture containing twice the number of moles of Fe2+(aq) compared to Fe3+(aq) once the reaction had taken place?

Moles of Fe Moles of Fe3+ (aq)
 A 2 4
 B 2 5
 C 2 6
 D 2 7


3. VJC, 2006, Paper 3

The reaction of an aqueous solution of xenon difluoride, XeF2, with potassium iodide produces an iodine-containing anion with iodine having a high oxidation number. XeF2 is reduced to Xe in the process. A student proposed that the anion could be either iodate (V),IO3, or iodate (VII), IO4.

When 0.530 g of potassium iodide was completely reacted with an aqueous solution of XeF2, all the iodine is converted into the iodate anion. Subsequently, when the resultant solutionis acidified, and added to excess potassium iodide solution, iodine is formed.

It is found that 28.40 cm3 of 0.900 moldm-3 of sodium thiosulphate is required to discharge its colour.

Deduce the identity of the iodate anion, showing your working clearly.

Solutions to CJC Paper 1

Tips: Most MCQ questions are set in the same manner as this question.

Of the 2 Redox half-equations, only one of them would we know the oxidation states (or identities) of the reactant and the product. For the other half equation, we only know the oxidation state (or identity) of the reactant OR the product.

How do we solve this in the shortest possible time?

The solution is very trivial:

Step 1

Write the reactant and product of the half-equation known and balance ALL atoms EXCEPT O and H:

Cr2O72- → 2Cr3+

Step 2

Calculate the oxidation states of the atom undergoing either reduction or oxidation.
Balance the charge by adding an appropriate number of electrons.

6e+ Cr2O72- 2 Cr3+
Cr: 2x(+6) Cr: 2x(+3)
=+12 =+6

Note: You can balance this half equation if you want to. But this is entirely unnecessary in MCQ. You should balance it if it appears in Paper 2 or 3.

Incidentally, this balanced equation is found in the Data Booklet, so you may wish to copy it if you would like to.

Step 3

Write the other half equation. Do not worry if the reactant or product is not known. We shall represent the unknown (either reactant or product) by X.

We balance the charge by adding n x electrons.

Note : Since Cr is reduced from +12 to +6 in one of the half equations, Te has to be oxidised. Thus electrons should appear on the right hand side of the half equation.

TeO2 → X + ne

Step 4

From the 2 half-equations written, derive the stoichiometric mole ratio of the redox reaction by making sure that both electrons lost = electrons gain.

(6e + Cr2O72- → 2 Cr3+) x n

(TeO2 → X + ne) x 6

Stoichiometric mole ratio of

Cr2O72- : TeO2
n : 6


Step 5

Determine the actual mole ratios of the 2 reactants in the redox reaction (the oxidising and reducing agents)

Moles of Cr2O72- = 60/1000 dm3 x 0.0350/ moldm-3 = 0.0021
Moles of TeO2 = 1.01 /(127.6+32) = 0.00633

Actual mole ratio of

Cr2O72- : TeO2
0.0021 : 0.00633
1 : 3


Step 6

Equate the stoichiometric and actual mole ratios (since there are no limiting and excess reactants in the question). Determine the value of n.

Stoichiometric mole ratio = actual mole ratio used

 n  =  2


Step 7

Calculate the final oxidation state of Te using the value of n
Initial oxidation state of Te in TeO2 = +4

Since it has lost 2 electrons in the reaction, final oxidation state of Te = +6



Point of Ponder:

What happens if there is excess reactant? The question will usually state a subsequent reaction between the excess reactant and another reactant, in which the amount of reactant in excess can be found.

From the initial amount of the excess reactant and its excess amount, the amount of this excess reactant that actually reacted in the redox reaction can be calculated.

The rest of the procedure continues from step 5.


Solutions to HCI Paper 1

Comments : This question is slightly different from the previous. We are to determine the mole ratio in which the reactants react, instead of the initial or final oxidation state of one of the reactants. The technique, however, is almost the same.

Step 1
Write a balanced equation for the reaction using half equation method

Fe + Fe2O3 + 6H+ → 3Fe2+ +3H2O

Step 2
Set up a mole ratio table for the reaction

In mol Fe + Fe2O3 + 6H+ 3Fe2+ + 3H2O
Initial X Y 0
Change -X -X +3X
Final 0 Y-X 3X

Note : Question did not mention anything about Fe in the final mixture, that means the final mixture does not contain Fe

Step 3
Since the question stated that the final mixture contains twice as much Fe2+ compared to Fe3+

3X  = 2(Y-X)
5X  = 2Y
—        =


How do we deal with multiple redox reactions like the one in VJC Paper 3 where the products of some reactions are not known? Get A levels chemistry top school papers and revision help here now!