Oxidation and reduction questions Watch
3. Describe the following reactions, shown as half-equations, in words and use the OilRig mnemonic to work out what is happening in Redox terms.
a. 2K --> 2K+ + 2e- and 1/2O2 + 2e- --> O2-
b. Na --> Na+ + e- and 1/2Cl2 + e- --> Cl-
4. Convert these displacement reactions into their ionic form and use this to show where oxidation and reduction are taking place (ignoring the complex anions).
a. Mg(s) + FeSO4(aq) --> Fe(s) + MgSO4(aq)
b. Ni(s) + Pb(NO3)2(aq) --> Pb(s) + Ni(NO3)2(aq)
K is being oxidised. Why? Because it's oxidation state has changed from 0 to +1.
It's lost electrons and therefore become more positive.
There are 2 K's so the overall charge on the right side of the equation is 2+.
We therefore need 2e- on the right hand side to balance the charge out which makes the overall charge neutral. Just like the left side of 2K.
The other question, Oxygen is becoming reduced. It's gained electrons and become more negative.
The same principle is to balance out the overall charges on either side of the equation. Since the side on the right has overall charge of 2-, on the left side, you need half O2, which is 1 O, and 2e- which makes the oxidation state on the left 2-.
Basically if oxidation state goes up, it's been oxidised, if it goes down, it's been reduced.
Reductions always have electrons on the left side of the half equation, oxidation has it on the right side because after the reaction, it's lost electrons.
So the complex ions are the SO42- etc, so you make the equation use the Mg2+ ion and show what happens to it when it's in MgSO4.
To do this, you need to learn your oxidation number rules.
Attached to this post.