REDOX & OXIDATION NUMBERS

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

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

RULES All uncombined elements have an oxidation umber of zero e.g. Zn, Cl2, O2, Ar all have an oxidation number of zero The oxidation numbers of the elements in a compound add up to zero e.g. In NaCl Na = +1 Cl = -1 Sum = +1-1 = 0 The oxidation number of a monoatomic ion is equal to the charge on the ion e.g. Zn2+ = +2 Cl- = -1 In a polyatomic ion (CO32-) the sum of the individual oxidation numbers of the elements adds up to the charge on the ion e.g. in CO32- C = +4 and O = -2 Sum = +4 (3 x -2) = -2

Several elements have invariable oxidation numbers in their common compounds

Some examples – Give the oxidation number of: N in NO2 S in SO42- Fe in FeCl3 N in NO3- F in F2 S in SO3 Na in NaOH

QUESTIONS

Consider the following redox reaction: The half equation to show what is happening to Br2 is: Br2 + 2e-  2Br- 0 -2 Here Br2 has gained electrons since its gone from 0 to -2 So Br2 has been reduced The half equation to show what is happening to the I- ion is: 2I-  I2 + 2e- -2 0 Here I- has lost electrons since it has gone from -2 to 0 So I- has been oxidised to form I2

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 Work out oxidation numbers for element being oxidised/ reduced i.e.Zn  Zn2+ 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 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: Fe2+/Fe3+ F2/F- O2/O2- Al3+/Al Sn2+/Sn4+ Cu2+/Cu

Say you needed to show a half equation for the following: MnO4-/Mn2+ Then we would have: MnO4-  Mn2+ This equation is impossible to balance as we do not have oxygen on the RHS In these situations we MUST add H2O on the RHS and H+ on the LHS of the equation i.e. MnO4- + H+  Mn2+ + H2O 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 MnO4-  Mn2+ Add H2O and H+ on appropriate sides if needed MnO4- + H+  Mn2+ + H2O 2. Then balance the number of atoms of each element on both sides: MnO4- + 8H+  Mn2+ + 4H2O 3. Then balance the change in oxidation state with electrons: MnO4- + 8H+  Mn2+ + 4H2O -1 +8  +2 0 +7  +2

∴ 5e- must be added to the RHS to balance the charges i.e. MnO4- + 8H+ + 5e-  Mn2+ + 4H2O

The oxidation state of oxygen and hydrogen does not change MnO4- + 8H+ + 5e-  Mn2+ + 4H2O 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// The oxidation state of oxygen and hydrogen does not change Oxygen remains as -2 Hydrogen remains as +1

Example: Write the half equation for the change SO42-  SO2 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 H2O and H+ on appropriate sides to balance this equation 1. So, SO42- + H+  SO2 + H2O 2. Now balance the number of atoms on each side: SO42- + 4H+  SO2 + 2H2O

3. Now balance the charges on each side using electrons; SO42- + 4H+  SO2 + 2H2O -2 +4  0 0 +2 0 So, 2e- must be added to the LHS SO42- + 4H+ + 2e-  SO2 + 2H2O

SO42- + 4H+ + 2e-  SO2 + 2H2O The oxidation state of S in SO42- is +6 The oxidation state of S in SO2 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:   PbO2  Pb2+ Cl-  Cl2 S2O32-  S4O62- I2  I- IO3-  I2 ClO-  ClO3-, ClO-  Cl- H2SO4  SO2 Br-  Br2 H2SO4  S H2SO4  H2S SO32-  SO42-

COMBINING HALF EQUATIONS TO MAKE FULL IONIC EQUATIONS Write both half equations first (one which will show an oxidation and one which will show a reduction) 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: MnO4-/Mn2+ C2O42-/CO2 ANSWER MnO4- + 8H+ + 5e  Mn2+ + 4H2O C2O42-  2CO2 + 2e-

2MnO4- + 16H+ + 5C2O42- → 2Mn2+ + 10CO2 + 8H2O Now combine both half equations by eliminating the electrons: (REDUCTION) MnO4- + 8H+ + 5e-  Mn2+ + 4H2O (X2) (OXIDATION) C2O42-  2CO2 + 2e- (X5) 2MnO4- + 16H+ + 5C2O42- → 2Mn2+ + 10 CO2 + 8H2O 2MnO4- + 16H+ + 5C2O42- → 2Mn2+ + 10CO2 + 8H2O The oxidation state of Mn here is +7 The oxidation state of C here is +3 The oxidation state of Mn here is +2 The oxidation state of C here is +4 So this is a redox reaction as: Mn has been reduced from +7 to +2 And C has been oxidised from +3 to +4 So MnO4- is acting as an OXIDISING AGENT C2O42- is acting as a REDUCING AGENT

Write two half equations to show: SO42-/H2S I-/I2 ANSWER REDUCTION SO42- + 10H+ + 8e-  H2S+ 4H2O OXIDATION 2I- → I2 + 2e- Now combine both half equations by eliminating the electrons: SO42- + 10H+ + 8e-  H2S+ 4H2O (X1) 2I- → I2 + 2e- (X4) 8I- + SO42- + 10H+  H2S + 4I2 + 4H2O

8I- + SO42- + 10H+  H2S + 4I2 + 4H2O So in this redox reaction: Here oxidation state of I is -1 Here the oxidation state of S is +6 Here the oxidation state if S is -2 Here the oxidation state of I is zero So in this redox reaction: I is being oxidised from -1 to 0 S is being reduced from +6 to -2 I- is acting as a REDUCING AGENT SO42- 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) PbO2  Pb2+, Cl-  Cl2 b) S2O32-  S4O62-, I2  2I- c) IO3-  I2, I-  I2 d) ClO-  ClO3-, ClO-  Cl- e) H2SO4  SO2, Br-  Br2 f) H2SO4  S, I-  I2 g) H2SO4  H2S, I-  I2 h) ClO-  Cl-, I-  I2 i) PbO2  Pb2+, SO32-  SO42-