REDOX OXIDATION NUMBERS REDOX REACTIONS OILRIG Oxidation Is

REDOX & OXIDATION NUMBERS

REDOX REACTIONS OILRIG – Oxidation Is Loss of electrons Reduction Is gain of electrons This is really important Think of January sales!

OXIDATION NUMBER RULES Ø Oxidation numbers which are assigned to ions (of each element) in a compound help us to write half equations and full ionic equations. Ø This is very important when deciding which species in an ionic equation is being oxidised and which is being reduced

RULES 1. All uncombined elements have an oxidation umber of zero e. g. Zn, Cl 2, O 2, Ar all have an oxidation number of zero 2. The oxidation numbers of the elements in a compound add up to zero e. g. In Na. Cl Na = +1 Cl = -1 Sum = +1 -1 = 0 3. The oxidation number of a monoatomic ion is equal to the charge on the ion e. g. Zn 2+ = +2 Cl- = -1 4. In a polyatomic ion (CO 32 -) the sum of the individual oxidation numbers of the elements adds up to the charge on the ion e. g. in CO 32 - C = +4 and O = -2 Sum = +4 (3 x -2) = -2

5. Several elements have invariable oxidation numbers in their common compounds

Some examples – Give the oxidation number of: 1. N in NO 2 2. S in SO 423. Fe in Fe. Cl 3 4. N in NO 35. F in F 2 6. S in SO 3 7. Na in Na. OH

QUESTIONS

Consider the following redox reaction: The half equation to show what is happening to Br 2 is: The half equation to show what is happening to the I- ion is: Br 2 + 2 e- 2 Br- 2 I- I 2 + 2 e- 0 -2 Here Br 2 has gained electrons since its gone from 0 to -2 So Br 2 has been reduced -2 0 Here I- has lost electrons since it has gone from -2 to 0 So I- has been oxidised to form I 2

NOTE// An oxidising agent (or oxidant) is the species that causes another element to oxidise. It is itself reduced in the reaction A reducing agent (or reductant) is the species that causes another element reduce. It is itself oxidised in the reaction NOTE ALSO// When naming oxidising and reducing agents always refer to full name of substance and not just name of element

WRITING HALF EQUATIONS RULES 1. Work out oxidation numbers for element being oxidised/ reduced i. e. Zn 2+ Zn changes from 0 to +2 2. Add electrons equal to the change in oxidation number For reduction add e’s to reactants For oxidation add e’s to products 3. Check to see that the sum of the charges on the reactant side equals the sum of the charges on the product side

QUESTIONS Write half equations for the following reactions and state whether the half equation is showing a reduction or oxidation: a) Fe 2+/Fe 3+ b)F 2/Fc) O 2/O 2 d)Al 3+/Al e) Sn 2+/Sn 4+ f) Cu 2+/Cu

Say you needed to show a half equation for the following: Mn. O 4 -/Mn 2+ Then we would have: Mn. O 4 - Mn 2+ This equation is impossible to balance as we do not have oxygen on the RHS In these situations we MUST add H 2 O on the RHS and H+ on the LHS of the equation i. e. Mn. O 4 - + H+ Mn 2+ + H 2 O Now we have all the atoms of each element on each side to enable us to balance this half equation.

Example: Write the half equation for the change Mn. O 4 - Mn 2+ 1. Mn. O 4 + H+ - Mn 2+ + H 2 O Add H 2 O and H+ on appropriate sides if needed 2. Then balance the number of atoms of each element on both sides: Mn. O 4 - + 8 H+ Mn 2+ + 4 H 2 O 3. Then balance the change in oxidation state with electrons: Mn. O 4 - + 8 H+ Mn 2+ + 4 H 2 O -1 +8 +2 0 +7 +2

Mn. O 4 - + 8 H+ + 5 e- Mn 2+ + 4 H 2 O The oxidation state of Mn here is +7 The oxidation state of Mn here is +2 So this half equation is showing that Mn is being reduced from +7 to +2 NOTE// ØThe oxidation state of oxygen and hydrogen does not change ØOxygen remains as -2 ØHydrogen remains as +1

Example: Write the half equation for the change SO 42 - SO 2 Again it is impossible to balance this equation despite the fact that we have the same atoms of each element on both sides, so again we must turn to using H 2 O and H+ on appropriate sides to balance this equation 1. So, SO 42 - + H+ SO 2 + H 2 O 2. Now balance the number of atoms on each side: SO 42 - + 4 H+ SO 2 + 2 H 2 O

3. Now balance the charges on each side using electrons; SO 42 - + 4 H+ SO 2 + 2 H 2 O -2 +4 0 0 +2 0 So, 2 e- must be added to the LHS SO 42 - + 4 H+ + 2 e- SO 2 + 2 H 2 O

SO 42 - + 4 H+ + 2 e- SO 2 + 2 H 2 O The oxidation state of S in SO 42 - is +6 The oxidation state of S in SO 2 is +4 So here S is being reduced from +6 to +4 Again the oxidation states of both oxygen and hydrogen do not change

QUESTIONS Write half equations for the following reactions and using oxidation numbers explain which element is being oxidised/reduced: a) Pb. O 2 Pb 2+ b) Cl- Cl 2 c) S 2 O 32 - S 4 O 62 d) I 2 Ie) IO 3 - I 2 f) Cl. O- Cl. O 3 -, g) Cl. O- Clh) H 2 SO 4 SO 2 i) Br- Br 2 j) H 2 SO 4 S k) H 2 SO 4 H 2 S l) Pb. O 2 Pb 2+ m) SO 32 - SO 42

COMBINING HALF EQUATIONS TO MAKE FULL IONIC EQUATIONS Ø Write both half equations first (one which will show an oxidation and one which will show a reduction) Ø To combine two half equations there must be equal numbers of electrons in the two half equations so that the electrons cancel out Write two half equations to show: Mn. O 4 -/Mn 2+ C 2 O 42 -/CO 2 ANSWER Mn. O 4 - + 8 H+ + 5 e Mn 2+ + 4 H 2 O C 2 O 42 - 2 CO 2 + 2 e-

Now combine both half equations by eliminating the electrons: (REDUCTION) Mn. O 4 - + 8 H+ + 5 e- Mn 2+ + 4 H 2 O (OXIDATION) C 2 O 42 - 2 CO 2 + 2 e- (X 2) (X 5) 2 Mn. O 4 - + 16 H+ + 5 C 2 O 42 - → 2 Mn 2+ + 10 CO 2 + 8 H 2 O 2 Mn. O 4 - + 16 H+ + 5 C 2 O 42 - → 2 Mn 2+ + 10 CO 2 + 8 H 2 O The oxidation state of Mn here is +7 The oxidation state of C here is +3 So this is a redox reaction as: And Mn has been reduced from +7 to +2 C has been oxidised from +3 to +4 The oxidation state of Mn here is +2 The oxidation state of C here is +4 So Mn. O 4 - is acting as an OXIDISING AGENT C 2 O 42 - is acting as a REDUCING AGENT

Write two half equations to show: SO 42 -/H 2 S I-/I 2 ANSWER REDUCTION OXIDATION SO 42 - + 10 H+ + 8 e- H 2 S+ 4 H 2 O 2 I- → I 2 + 2 e- Now combine both half equations by eliminating the electrons: SO 42 - + 10 H+ + 8 e- H 2 S+ 4 H 2 O 2 I- → I 2 + 2 e- (X 1) (X 4) 8 I- + SO 42 - + 10 H+ H 2 S + 4 I 2 + 4 H 2 O

8 I- + SO 42 - + 10 H+ H 2 S + 4 I 2 + 4 H 2 O Here oxidation state of I is -1 Here the oxidation state of S is +6 So in this redox reaction: I is being oxidised from -1 to 0 S is being reduced from +6 to -2 Here the oxidation state if S is -2 Here the oxidation state of I is zero I- is acting as a REDUCING AGENT SO 42 - is acting as an OXIDISING AGENT

QUESTIONS Write two half equations for each question and then combine the half equations. State which species has been oxidised and which has been reduced. Give the name of the oxidising agent and reducing agent in each full ionic equation. a) Pb. O 2 Pb 2+, Cl- Cl 2 b) S 2 O 32 - S 4 O 62 -, I 2 2 I c) IO 3 - I 2, I- I 2 d) Cl. O- Cl. O 3 -, Cl. O- Cl e) H 2 SO 4 SO 2, Br- Br 2 f) H 2 SO 4 S, I- I 2 g) H 2 SO 4 H 2 S, I- I 2 h) Cl. O- Cl-, I- I 2 i) Pb. O 2 Pb 2+, SO 32 - SO 42 -