Archive : January 2018

How to solve common redox questions

Mr Khemistry


Qn: Sulfur dioxide causes a colour change in acidified potassium dichromate. Given that the potassium dichromate and sulfur dioxide are present in the reacting ratio of 1:3, suggest a likely sulfur−containing product for the reaction.

This is considered quite a common redox question. 

Firstly, what do you need to know? And how to apply what you know to solve this question?

Diagram showing working step

First step

Basic knowledge of common oxidizing and reducing agents. Look up your lecture notes, there should be a table of common O.A. and R.A. together with their respective products. This will allow you to identify which is the oxidizing/reducing agent.

You will also need to calculating the reacting mole ratio between the O.A. and R.A. However, in this particular question, the reacting ratio is given.

diagram showing how to solve redox question

Second step to solve redox question

From the Cr half-equation (you will need to know how to balance half-equations in case they are not found in data booklet), you can determine the number of electrons gained for Cr since you know the product formed. (Note that you will not be able to do the same for sulfur dioxide as you do not know the product form)

Electron transfer is the main concept in Redox.

All electrons lost are equal to the electrons gained.

Diagram showing working steps for redox

Step 3

Once we know 6 electrons are gained by the two Chromium atoms, we can deduce that the three sulfur atoms lost 6 electrons.

Therefore, each sulfur atom would have lost 2 electrons and gained +2 oxidation state. 

Diagram showing solution for redox question

Final step

Since sulfur in sulfur dioxide has an oxidation state of +4, the sulfur in the new sulfur-containing product will have an oxidation state of +6, i.e. sulfur trioxide or sulfate ion.

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How do organic compounds achieve stability?

Mr Khemistry

Important Concept for Organic Chemistry

Organic Chemistry
Two main ways organic compounds achieve stability:








Electrons are delocalized to form different resonance structures, which lowers the energy level of the organic compound. This contributes to stability.

Dispersion of charge

Carbocation is electron deficient while R groups are electron-donating. Thus the more R groups are bonded to the carbocation, the more stable (less electron deficient) the carbocation will be. 

This concept also applies to negative ions, phenoxide ion is more stable as the negative charge can be dispersed into the benzene ring.

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A level Forum Talk @ Popular Bookfest 2017

Mr Khemistry

A level Forum Talk

Took part in the A level Forum Talk at the recently concluded Popular Bookfest. Quite an experience speaking at such a huge venue.

Mr Eric Kua speaking at Popular bookfest 2017

Group photo at the end of the talk

Subjects represented included Maths, Physics, Chemistry, Biology, Economics and GP. Looking forward to more talks, especially for those who have just finished O levels.

For more photos from the event, click here


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