Topic 7 in Physical chem AQA A Level - redox

I understand the whole idea that oh oxidation is the loss of electrons reduction is the gain of electrons, blah blah blah and oxidagtion state rules, but how do I form ionic equations and half equations from full equations is i just don;t understand hownto do any of this so pls fgive a full detailed breakdown

Reply 1

TypicalNerd

Original post

by Jimmy Butler IV

You should always be told what your important products/reactants are and under what conditions the reaction is carried out, so you can follow the typical method, set up unbalanced equations and work them up from there.

For example, if I told you that in acidic conditions, dichromate ions can react with iodide ions to form iodine and chromium(III) ions:

(1. Let’s start with the easy bit. We can see iodide (I^-) is converted to iodine (I2), so why not begin with this half-equation. Putting the reactant and product on the correct sides of the equation:

I^-(aq) —> I2(aq)

(2. We can see that the numbers of iodine atoms and the charges are not balanced. To make both sides match, let’s scale up the LHS so there are a pair of iodide ions on either side:

2I^-(aq) —> I2(aq)

We can also see that since there must be 2 electrons on the RHS to balance the overall charge:

2I^-(aq) —> I2(aq) + 2e^-(aq)

And since electrons are liberated, iodide ions must be acting as the reducing agent (i.e. they are oxidised).

(3. We can start similarly for the reduction half-equation that Cr2O7^2- becomes Cr^3+, hence we have

Cr2O7^2-(aq) —> Cr^3+(aq)

We can already see we have a little bit of a problem here - how do we deal with the oxygens and the charges? If you have read and understood the problem, you will have spotted that the reaction is carried out in acidic conditions (e.g. you have H^+ ions floating about and these can react with the oxygens to give H2O). So we can now add these to their respective sides of the equation:

Cr2O7^2-(aq) + H^+(aq) —> Cr^3+(aq) + H2O(l)

(4. Now using the same principle as in (2., we have to balance the charges and number of atoms. Of course, it’s best to scale the number of chromium(III) ions up by 2 to match the LHS and the number of H2O molecules up by 7, since there are 7 oxygens on the LHS:

Cr2O7^2-(aq) + H^+(aq) —> 2Cr^3+(aq) + 7H2O(l)

We can now see that you need 14H^+ ions on the LHS, so we scale up accordingly:

Cr2O7^2-(aq) + 14H^+(aq) —> 2Cr^3+(aq) + 7H2O(l)

Checking the charges:

Total on LHS: -2 (from Cr2O7^2-) + 14 x +1 (from 14H^+) = +12

Total on RHS: 2 x +3 = +6

We can see that the difference in total charges is 6, so we add 6 electrons to the LHS to ensure the numbers all work:

Cr2O7^2-(aq) + 14H^+(aq) + 6e^- —> 2Cr^3+(aq) + 7H2O(l)

This makes sense. After all, the oxidation half-equation produced electrons, so this reduction half equation would need to consume them.

(5. Now to combine the equations. We want to do this in such a way that the final result contains no electrons on either side. As we can see the oxidation half equation has 2 electrons on the RHS and the reduction half equation has 6 electrons on the LHS.

These don’t match, so we need to scale one or both of these equations. In this case, it’s simple - we multiply the oxidation half-equation by 3 (i.e. 6I^-(aq) —> 3I2(aq) + 6e^-) and don’t scale the reduction half equation. Then when we add them together, the 6 electrons on either side cancel out:

Cr2O7^2-(aq) + 14H^+(aq) + 6e^- + 6I^-(aq) —> 2Cr^3+(aq) + 7H2O(l) + 3I2(aq) + 6e^-

The underlined electrons match and cancel:

Cr2O7^2-(aq) + 14H^+(aq) + 6I^-(aq) —> 2Cr^3+(aq) + 7H2O(l) + 3I2(aq)

And here’s the final answer. This was the standard method, which is rather long-winded. There is a much quicker method, however that I refer to as the “change in oxidation states method”. I’ll outline that in the next post.

Also, since I explained you use H^+ and H2O to balance out oxygens being lost in acidic conditions, it’s worth noting that you use OH^- and H2O to balance out oxygens being lost in basic conditions.

Reply 2

TypicalNerd

The change in oxidation states method is so much faster. Let’s do it for the same reaction as in my first post.

(1. We know that the reactants are Cr2O7^2- and I^- and that the products will be Cr^3+ and I2, so let’s work out the changes in oxidation states for the elements of interest.

Cr: +6 (in Cr2O7^2-) - +3 (in Cr^3+) = +3
I: -1 (in I^-) - 0 (in I2) = -1

We now can work out the reacting ratio of Cr to I by ignoring the signs (e.g. we treat both changes in oxidation states as positive numbers) and flipping them round. This makes the Cr to I ratio 1:3. This makes setting up the equation much quicker and we can skip having to deal with cancelling electrons entirely.

Because Cr2O7^2- contains two chromiums, to maintain the Cr to I ratio, you need 6 I^- ions to react with it and not just 3. This is a common mistake in this particular example worth avoiding. Hence, we have:

Cr2O7^2-(aq) + 6I^-(aq) —> Cr^3+(aq) + I2(aq)

(2. Much like with the earlier method, the solution is acidic and hence we can use H^+ and H2O to balance out the oxygens on the LHS:

Cr2O7^2-(aq) + H^+(aq) + 6I^-(aq) —> Cr^3+(aq) + H2O(l) + I2(aq)

Now it’s a matter of just making the numbers on each side match. To avoid repeating myself, I’ll just put the correct answer below and leave working it out as an exercise to the reader if they so wish:

Cr2O7^2-(aq) + 14H^+(aq) + 6I^-(aq) —> 2Cr^3+(aq) + 7H2O(l) + 3I2(aq)

And thus we are done.

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