For the reaction below identify the oxidizing agent and the reducing agent. Use the oxidation method. Show all work!

Question

For the reaction below identify the oxidizing agent and the reducing agent. Use the oxidation method. Show all work!

Answer 1

For this problem, you need to determine the oxidation state of all atoms on both sides of the reaction. Remember the steps for determining oxidation state:

1) The oxidation state of an atom in its elemental form (uncombined element such at C, Cl₂, P₄, etc.) is zero.

2) The charge of an ion is equal to the oxidation state of that ion (i.e. KCl is ionic, made up of K⁺¹ and Cl^-1, therefore the oxidation state of K = +1 and the oxidation state of Cl = -1)

3) The sum of the oxidation states of all atoms and/or ions will be equal to zero for a neutral compound or the charge for a polyatomic ion.

4) For compounds, we will use the following guidelines:

Hydrogen will usually be +1 (unless in an ionic compound)

Oxygen will usually be -2 (unless in a peroxide)

Fluorine will always be -1

Chloride will usually be -2

Note: the more electronegative element in a compound is assigned a negative oxidation state.

After assigning oxidation states to all atoms on both sides of the reaction, you will see which atoms are changing oxidation states. For instance, chlorine (in KCl) has a -1 oxidation state on the reactant side of the reaction (see step 2) and a zero oxidation state on the product side (Cl₂, see step 1).

You can then determine which of the atoms from above are gaining electrons (reduction), and which one is losing electrons (oxidation). For Cl^-1 → Cl⁰, chlorine is becoming more positive because it is losing electrons. Therefore, chlorine is oxidized.

After determining which atom is being oxidized, and which is reduced, you can now locate the reducing agent (the reactant that causes the reduction of the other reactant by giving the reactant electrons - it contains the atom being oxidized) are the oxidizing agent (the reactant that causes the oxidation of the other reactant by making the reactant lose electrons - it contains the atom being reduced). Chlorine is oxidized, and the reactant containing chlorine is KCl, so KCl is the reducing agent.

We can write out the half-reaction to show this with oxidation numbers:

-1 0

2 KCl → Cl₂ + 2e^-