3.1.7 Oxidation reduction and redox equations

Question 1

What is oxidation & oxidising agent

Answer 1

process of the loss of electrons - electron acceptors

Question 2

What is reduction & reducing agent

Answer 2

process of electron gain - electron donors

Question 3

oxidation state

Answer 3

number assigned to element in chemical compound - represent number of electrons lost/gained by an atom of that element - compound

Question 4

oxidation state of uncombined elements , hydrogen , group 1 alkali metals ? oxidation state of group 2, aluminium, O2 , fluorine ? oxidation state of chlorine , sum of all oxidation states (neutral compound and complex ion )?

Answer 4

• 0
  +1 , except metal hydrides (NaH)
  +1

+2
+3
-2 , except peroxides/compounds with f
-1

-1 , except in compounds with F&O
0 (neutral)
Charge on ion (complex ion )

Question 5

How to write half equations

Answer 5

1. Identify reduced or oxidised elements , balance atoms of the element.

Example: in Cr2O7^2- -> Cr^3+, Cr - reduced, balancing Cr the equation becomes: Cr2O7^2- -> 2Cr^3+

2. Work out change in oxidation number & multiply it by the number of atoms of element. Add this many electrons to the left (reduction) or right (oxidation).

Example: Oxidation number of Cr changes from +6 to +3, a change of 3; there are two Cr atoms so 6 electrons need to be added; this is reduction so add the electrons to the left.
Cr2O7^2- + 6e- -> 2Cr^3+

3. If there are extra O atoms on one side, add twice as many H+ on that side and equal amount of H2O on the other side. If there are also extra H atoms, add equal amount of H+ on the other side.

Question 6

How to combine half-equations to give an overall redox equation?

Answer 6

1. Multiply one or both equations by integers so that the electrons cancel out.

Example:
Equation 1: Cr2O7^2- + 6e- + 14H+ -> 2Cr^3 + 7H2O

Equation 2: H2O2 -> 2H2 + O2 + 2e-

There are 6 electrons in equation 1 but 2 electrons in equation 2. So multiply equation 2 by 3 to give 6 electrons.
3H2O2 -> 6H2 + 3O2 + 6e-

2. Add the equations together and cancel the electrons.

Cr2O7^2- + 14H+ + 3H2O2 -> 2Cr^3 + 7H2O + 6H2 + 3O2
The 6e- disappears from both sides.

3. Sometimes H+ or H2O may also appear on both sides. These should be cancelled too.