# Redox: Oxidation and Reduction

I don't get the difference between the gain or loss of electrons in a reaction and changes in oxidation state. Based on my notes and some research:

1. Gain or loss of electrons

• A substance is oxidised if it loses electrons in a reaction.
• A substance is reduced if it gains electrons in a reaction.
2. Changes in oxidation state

• A substance is oxidised if the oxidation state of the element increases.
• A substance is reduced if the oxidation state of the element decreases.

Whenever I see a chemical equation, I am confused as I do not know if it is a gain or loss of electrons or changes in oxidation state. I have an assignment to do which involves redox reaction, but it is the holidays and I can't contact my teacher. Could someone explain so that I could understand? For example,

$$\ce{2 Ca (s) + O2 (g) → 2CaO (s)}$$

I thought there is a change in oxidation state as on the reactants side, the oxidation state of both $$\ce{Ca}$$ and $$\ce{O2}$$ are $$0$$. As for the product, the oxidation state for $$\ce{Ca}$$ is $$+2$$ and $$\ce{O}$$ is $$-2$$. Therefore, $$\ce{Ca}$$ is oxidised and $$\ce{O2}$$ is reduced.

But one of the questions on my worksheet asked about electron transfer for this chemical equation, and I do not understand why it is a electron transfer.

Look this as:

$$\ce{2Ca + O2 -> 2CaO}\qquad\text{(complete reaction)}$$

then we can this split into two sub-reactions:

$$\ce{Ca - 2 e- -> \overset{+2}{Ca}}\tag{1}$$

Here "$$+2$$" means that $$\ce{Ca}$$ lost $$2$$ electrons (or two negative units) and therefore it was oxidized. Since it lost something negative it has a higher number of protons (positive particles) remained. In other words the atoms is now positively charged. The initial state was $$0$$ (the same numbers of protons and electrons, or positive and negative particles/units) and it was increased by loosing two electrons (the number increased).

$$\ce{O + 2 e- -> \overset{-2}{O}}\tag{2}$$

The oxygen got two electrons or two negative particles. Now oxygen has two negative particles more than positive particles (protons) and therefore it is $$-2$$. In other words, the number (oxydation state) decreased ($$0 \to -2$$).

You have an electron transfer because on order for the reaction to take place must exist an exchange of electrons from one atom to the other. Basically, this means that electrons are shared between both atoms to form a bond. In your case you'll have that $$\ce{Ca}$$ loses two $$\ce{e-}$$ and oxygen gains two $$\ce{e-}$$ because it's more electronegative than $$\ce{Ca}$$. This can be represented as follows:

\begin{align} \ce{Ca &-> Ca^2+ + 2 e-}\\ \ce{O2 + 4 e- &-> 2 O^2-} \end{align}

This is not the real process, I mean is all more complex than that, but it will give you and idea about what I mean.