ES / F332 / GLU 11

Oxidation States & REDOX reactions

1] Give the oxidation state of the underlined element in each of the following: [9marks]

a) NH2 - b) AlO2

- c) VO2+ d) POCl5 e) ClF3

f) S4O6 – g) P2O3 h) Cl 2O3 i) IO3

-

2]

Give the oxidation states of each element in the following: [8marks]

a) Copper (II) Chloride ? b) Sodium Sulphate (VI) c) Iron (III) Nitrate (V)

3] What is oxidised and what is reduced in the following reactions. [12marks]

(show the change in oxidation states to back up your answers)

a) BrO3- + 6I- + 6H+ Br- + 3I2 + 3H2O

b) 2VO3- + 4H+ + 3C2O4

2- 6CO2 + 2VO22- + 2H2O

c) Cr2O72-

(aq) + 14H+ (aq) + 6e- 2Cr3+ (aq) + 7H2O (l)

d) N2H6O (aq) + IO3- (aq) + 2H+ (aq) + Cl- (aq) N2 (g) + ICl (aq) + 4H2O (l)

e) Na2H10S2O8 (aq) + 4Br2 (aq) 2H2SO4 (aq) + 2NaBr (aq) + 6HBr (aq) f) 5As2O3 (s) + 4MnO4

- (aq) + 12H+ (aq) 5As2O5 (s) + 4Mn2+ (aq) + 6H2O (l)

4] Give the formula of the Oxidising agent and of the Reducing agent in each reaction in Q3 [6marks]

5] For the following balanced equation:

C + 2H2SO4 CO2 + 2SO2 + 2H2O which of the following statements is / are TRUE ? [2marks]

i) Oxygen is the ONLY element that does not change oxidation number during the reaction.

ii) Sulphuric acid is the oxidising agent.

iii) Carbon is the oxidising agent.

iv) The oxidation number of sulphur is +6 in H2SO4 .

Sometimes with oxidation states, the clue is in the name. For example, in “Iron (III) Chloride solution” the roman numerals tell you that iron has an oxidation state of (+3). These roman numerals always refer to positive oxidation states of the element that they are closest to. For more on this, check out page 206 Chemical Ideas.

Example: 2FeCl2 + Cl2 2FeCl3

An OXIDISING AGENT is a substance that brings about the oxidation of another substance and in the process is reduced itself.

Oxidising agents ACCEPT electrons.

�Cl has an ox state of (0) in Cl2 Cl (-1) in FeCl3 [Cl is reduced itself and is an Oxidising agent]

�You can show this further by doing the ionic half equations: Cl2 + 2e- 2Cl - [Gain of electrons]

A REDUCING AGENT is (basically the opposite of oxidising agent) a substance that brings about the reduction of another

substance, and in the process is oxidised itself. Reducing agents LOSE electrons.

�Fe (+2) in FeCl2 Fe (+3) in FeCl3 [Fe is Oxidised itself, and is therefore a Reducing agent]

�Half equation : Fe2+ Fe3+ + e- [Loss of electrons]

ES / F332 / GLU 11

6] Look at the following ionic half equations and for each decide whether it shows OXIDATION or

REDUCTION ? [4marks]

a) I2 + 2e- 2I- c) 2S2O32- S4O6

2- + 2e- b) Fe2+ Fe3+ + e- d) MnO4

- + 8H+ + 5e- Mn2+ + 4H2O

7] Decide which element has been oxidised and which reduced in each of the following [4marks]

(back up your answer with a reason / supporting comment):

a) Pure zinc metal is placed in hydrochloric acid. There is violent fizzing as hydrogen gas

is given off and zinc chloride is formed.

b) Chlorine gas is bubbled through potassium bromide solution. The liquid quickly changes

from colourless to a reddy-brown colour as the chlorine displaces bromine.

8] Balance these symbol equations and decide what has been oxidised and what has been reduced: [14marks]

a) IO3- + H+ + I- I2 + H2O

b) Cl2 + OH- Cl- + ClO- + H2O

c) MnO4- + 6H+ + HCOOH Mn2+ + 8H2O + CO2

d) Cr2O72- + 6Fe2+ + H+ Cr3+ + Fe3+ + H2O

e) NH3 + O2 N2 + H2O

f) SO2 + H2O + Br2 H+ + SO42- + Br-

g) NH3 + O2 NO + H2O

9] Combine the following half equations to form one equation in each case (without electrons): [3marks]

a) I2 + 2e- 2I- and 2S2O3

2- S4O6 2- + 2e-

b) Fe2+ Fe3+ + e- and MnO4- + 8H+ + 5e- Mn2+ + 4H2O

c) Cr Cr3+ + 3e- and Cd2+ + 2e- Cd

Ionic half-equations (like the ones in Q6) can be combined to make FULL symbol equations. The only tricky part is getting rid of the electrons on either side of the equation ! To do this, you have to multiply both half equations by appropriate whole numbers so that both equations show the same number of “free electrons”, then add the 2 equations together and the electrons should cancel out.

eg/ Q: Combine these 2 half equations: A A2+ + 2e- and B3+ + 3e- B

ANS:

STEP 1 (multiply both half equations whole number to get same number of electrons. In this case, both 3 and 2 go into the number 6 !)

3x [ A A2+ + 2e-] = 3A 3A2+ + 6e-

2x [ B3+ + 3e- B ] = 2B3+ + 6e- 2B STEP 2 ( Combine them. The electrons cancel out as you have 6 “reactant” electrons and 6 “product” electrons)

3A + 2B3+ 3A2+ + 2B