Balancing redox reactions 2

Balance oxidation-reduction reactions using redox methods Include: oxidation number method, and half- reaction method Additional KEY Terms

Some redox reactions require an acidic or basic solution. Half-reaction method It depends on the reaction and substance needing to be oxidized – I will always tell you if it is in an acidic or basic solution Half-reaction method is used for balancing redox reactions in the presence of acid or base. Acid / base not oxidized or reduced in reaction Usually converted to water

Cr 2 O 7 2– (aq) + SO 3 2– (aq) → Cr 3+ (aq) + SO 4 2– (aq) Step 1: Assign O#s and write half-reactions Balancing in Acidic Solutions Oxidation:SO 3 2- → SO e – Reduction:Cr 2 O e – → Cr 3+ Aqueous ions cannot exist by themselves – the spectator ions must have already been removed

Step 2: Balance all elements except H and O Step 3: Balance oxygen atoms by adding H 2 O Oxidation:SO 3 2- → SO e – Reduction:Cr 2 O e – → Cr 3+ 2 Oxidation:SO H 2 O → SO e – Reduction:Cr 2 O e – → 2 Cr H 2 O Be sure to adjust the electrons to the number of atoms for each half-reaction Each Cr gains 3e - but there are 2 Cr atoms - so a total of 6e - gained 3 6

Step 4: Balance hydrogen atoms adding H + ions Step 5: Balance the number of electrons between half-reactions Oxidation: SO H 2 O → SO e – Reduction: Cr 2 O e – → 2 Cr H 2 O + 2 H + 14 H + + Oxidation: 3 x ( SO H 2 O → SO e – + 2 H + ) Reduction: 14 H + + Cr 2 O e – → 2 Cr H 2 O 3 SO H 2 O → 3 SO e – + 6 H + Multiply the whole equation by the common multiple needed to make the half-reaction electrons equal

6. Add the two half-reactions Oxidation: 3 SO H 2 O → 3 SO e – + 6 H + Reduction: 14 H + + Cr 2 O e – → 2 Cr H 2 O 8 H + + Cr 2 O SO 3 2- → 2 Cr SO H 2 O Take care here - cancel out what you can and combine the half- reactions into a single equation 8 4

MnO 4 – + I – → MnO 2 + I 2 8 H MnO 4 – + 6 I – → 2 MnO I H 2 O Balance the following reaction in a acidic solution Oxidation: I - → I 2 + e – Reduction: MnO e – → MnO H 2 O 4 H x () 2x ( ) Oxidation: 6 I - → 3 I 2 + 6e – Reduction: 8 H MnO e – → 2 MnO H 2 O No spectator ion on the reactant side – keep compound together – so you have to keep the product together too 1 2

Balancing in Basic Solutions Steps 1-4 are the same as in Acid solutions MnO 4 – + C 2 O 4 2– → CO 2 + MnO Oxidation: C 2 O 4 2- → CO 2 + e – Reduction: MnO e – → MnO H 2 O4 H Tip: Take your time and do each step on a new line to avoid any mis-steps

**5b. Eliminate H + / OH - by forming water Oxidation C 2 O 4 2- → 2 CO 2 + 2e – Reduction: 4 H + + MnO e – → MnO H 2 O + 4 OH - 4 OH - + **5a. Add the same number of OH - as H + to BOTH sides of the equation The point here is to cancel any water to simplify the half-reactions Oxidation C 2 O 4 2- → 2 CO 2 + 2e – Reduction: 4 OH H + + MnO e – → MnO H 2 O + 4 OH - 4 H 2 O 2

4 H 2 O + 2 MnO 4 – + 3 C 2 O 4 2– → 2 MnO CO OH – Oxidation 3 C 2 O 4 2- → 6 CO 2 + 6e – Reduction: 4 H 2 O + 2 MnO e – → 2 MnO OH - Step 7: Add the two half-reactions Oxidation C 2 O 4 2- → 2 CO 2 + 2e – Reduction: 2 H 2 O + MnO e – → MnO OH - 3 x () 2x ( ) Step 6: Balance the number of electrons between half-reactions

N 2 O + ClO – → NO 2 – + Cl – Balance the following reaction in a basic solution Oxidation: N 2 O → NO e – Reduction: ClO - + 2e – → Cl H 2 O + 2 H H H 2 O Steps

2 OH – + 2 ClO – + N 2 O → 2 Cl – + 2 NO 2 – + H 2 O Oxidation: N 2 O → NO e – Reduction: ClO - + 2e – → Cl - 23 H 2 O + 2 H H H 2 O + 6 OH - 2 OH H 2 O OH - 6 OH H 2 O 1 Reduction: ClO - + 2e – → Cl - 1 H 2 O OH - Oxidation: N 2 O → NO e – H 2 O 6 OH - + 2x ( ) 2 H 2 O + 2 ClO - + 4e – → 2 Cl OH - 12 Steps 5-7

CAN YOU / HAVE YOU? Balance oxidation-reduction reactions using redox methods Include: oxidation number method, and half- reaction method Additional KEY Terms