what is the oxidation half reaction, reduction half reaction, and the complete balanced reaction of the following?

(a) Al(s) + MnO4‾(aq) → MnO2(s) + Al(OH)4‾(aq)
(b) Cl2(g) → Cl ‾(aq) + ClO ‾(aq)
(c) NO2‾(aq) + Al(s) → NH3(g) + AlO2‾(aq)
(d) MnO4‾(aq) + S2-(aq) → MnS(s) + S(s)
(e) CN ‾(aq) + MnO4‾(aq) → CNO‾(aq) + MnO2(s)
(f) Cu(s) + NO3‾(aq) → Cu2+(aq) + NO(g)
(g) Cr2O72-(aq) + Cl ‾(aq) → Cr3+(aq) + Cl2(g)
(h) Pb(s) + PbO2(s) + H2SO4(aq) → PbSO4(s)
(i) Mn2+(aq) + NaBiO3(s) → Na+(aq) + Bi3+(aq) + MnO4‾(aq)
(j) H3AsO4(aq) + Zn(s) → AsH3(g) + Zn2+(aq)
(k) As2O3(s) + NO3‾ (aq) → H3AsO4(aq) + NO(g)
(l) Br ‾(aq) + MnO4‾(aq) → Br2(l) + Mn2+(aq)
(m) CH3OH(aq) + Cr2O72-(aq) → CH2O(aq) + Cr3+(aq)

H2O2 (aq) +MnO4 - → Mn2+ (aq) + O2 (g) problem

(a) The oxidation half-reaction: Al(s) → Al(OH)4‾(aq)

The reduction half-reaction: 8H+(aq) + MnO4‾(aq) + 5e‾ → MnO2(s) + 4H2O(l)
The balanced reaction: 2Al(s) + 3MnO4‾(aq) + 10H2O(l) → 3MnO2(s) + 2Al(OH)4‾(aq)

(b) The oxidation half-reaction: Cl2(g) → 2Cl ‾(aq)
The reduction half-reaction: Cl2(g) + 2e‾ → 2ClO ‾(aq)
The balanced reaction: Cl2(g) + 2OH‾(aq) → Cl ‾(aq) + ClO ‾(aq) + H2O(l)

(c) The oxidation half-reaction: NO2‾(aq) → NH3(g)
The reduction half-reaction: 3H2O(l) + 3e‾ → AlO2‾(aq) + 6OH‾(aq)
The balanced reaction: 2NO2‾(aq) + 2Al(s) + 4H2O(l) → 2NH3(g) + AlO2‾(aq) + 4OH‾(aq)

(d) The oxidation half-reaction: S2-(aq) → S(s)
The reduction half-reaction: MnO4‾(aq) + 4H2O(l) + 5e‾ → MnS(s) + 8OH‾(aq)
The balanced reaction: 8OH‾(aq) + MnO4‾(aq) + 5S2-(aq) → MnS(s) + 4H2O(l) + 5S(s)

(e) The oxidation half-reaction: CN ‾(aq) → CNO‾(aq) + e‾
The reduction half-reaction: MnO4‾(aq) + 8H+(aq) + 5e‾ → MnO2(s) + 4H2O(l)
The balanced reaction: 16CN ‾(aq) + 2MnO4‾(aq) + 8H+(aq) → 5CNO‾(aq) + 2MnO2(s) + 4H2O(l)

(f) The oxidation half-reaction: Cu(s) → Cu2+(aq) + 2e‾
The reduction half-reaction: 2NO3‾(aq) + 10H+(aq) + 8e‾ → 2NO(g) + 5H2O(l)
The balanced reaction: 6Cu(s) + 8NO3‾(aq) + 18H+(aq) → 6Cu2+(aq) + 2NO(g) + 9H2O(l)

(g) The oxidation half-reaction: Cl ‾(aq) → Cl2(g) + 2e‾
The reduction half-reaction: 2Cr2O72-(aq) + 14H+(aq) + 6e‾ → 2Cr3+(aq) + 7H2O(l)
The balanced reaction: 14H+(aq) + 2Cr2O72-(aq) + 12Cl ‾(aq) → 2Cr3+(aq) + 7H2O(l) + 6Cl2(g)

(h) The oxidation half-reaction: Pb(s) → Pb2+(aq) + 2e‾
The reduction half-reaction: PbO2(s) + 4H+(aq) + 2e‾ → PbSO4(s) + 2H2O(l)
The balanced reaction: Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)

(i) The oxidation half-reaction: Mn2+(aq) → MnO4‾(aq) + 5e‾
The reduction half-reaction: NaBiO3(s) + 3H2O(l) + 9e‾ → Bi3+(aq) + 3OH‾(aq) + 3Na+(aq)
The balanced reaction: 3Mn2+(aq) + 3NaBiO3(s) + 9H2O(l) → 3Na+(aq) + 3Bi3+(aq) + 3MnO4‾(aq) + 9OH‾(aq)

(j) The oxidation half-reaction: Zn(s) → Zn2+(aq) + 2e‾
The reduction half-reaction: H3AsO4(aq) + 9H+(aq) + 3e‾ → AsH3(g) + 4H2O(l)
The balanced reaction: 2H3AsO4(aq) + 6Zn(s) + 9H+(aq) → 3AsH3(g) + 6Zn2+(aq) + 6H2O(l)

(k) The oxidation half-reaction: As2O3(s) + 6H2O(l) + 12e‾ → 4H3AsO4(aq) + 10OH‾(aq)
The reduction half-reaction: 3NO3‾(aq) + 12H+(aq) + 10e‾ → 3NO(g) + 6H2O(l)
The balanced reaction: 4As2O3(s) + 3NO3‾ (aq) + 18H+(aq) → 8H3AsO4(aq) + 3NO(g) + 10H2O(l)

(l) The oxidation half-reaction: 5Br ‾(aq) → Br2(l) + 2e‾
The reduction half-reaction: 8H+(aq) + MnO4‾(aq) + 5e‾ → Mn2+(aq) + 4H2O(l)
The balanced reaction: 2MnO4‾(aq) + 16H+(aq) + 10Br ‾(aq) → 2Mn2+(aq) + 5Br2(l) + 8H2O(l)

(m) The oxidation half-reaction: CH3OH(aq) → CH2O(aq) + 2H+(aq) + 2e‾
The reduction half-reaction: Cr2O72-(aq) + 14H+(aq) + 6e‾ → 2Cr3+(aq) + 7H2O(l)
The balanced reaction: 3CH3OH(aq) + 2Cr2O72-(aq) + 16H+(aq) → 2CH2O(aq) + 4Cr3+(aq) + 14H2O(l)

To determine the oxidation half-reaction, reduction half-reaction, and the complete balanced reaction, we need to follow these steps:

Step 1: Assign oxidation numbers.
Write down the oxidation numbers for each element in the reactants and products. The oxidation number is a representation of the charge an atom would have if the compound or ion were entirely ionic.

Step 2: Identify the element undergoing oxidation and reduction.
The element with a decreasing oxidation number is undergoing reduction, while the element with an increasing oxidation number is undergoing oxidation.

Step 3: Write the half-reactions.
For the oxidation half-reaction, write the reactant that is losing electrons. For the reduction half-reaction, write the reactant that is gaining electrons. Balance the number of atoms and charges on both sides of each half-reaction.

Step 4: Balance the charges in each half-reaction.
Add electrons (e-) to balance the charges on both sides of each half-reaction.

Step 5: Balance the number of atoms in each half-reaction.
Adjust the number of atoms on each side of each half-reaction to balance the number of atoms.

Step 6: Combine the half-reactions.
Multiply the balanced half-reactions by the appropriate values to ensure that the number of electrons transferred in the oxidation half-reaction is equal to the number of electrons gained in the reduction half-reaction. Then combine the two half-reactions to form the overall balanced equation.

Now let's apply these steps to the given reactions:

(a) Al(s) + MnO4‾(aq) → MnO2(s) + Al(OH)4‾(aq)
Step 1: Assign oxidation numbers:
Al: 0 Mn: +7 O: -2 H: +1

Step 2: Identify the element undergoing oxidation and reduction:
Al is undergoing oxidation (from 0 to +3) and Mn is undergoing reduction (from +7 to +4).

Step 3: Write the half-reactions:
Oxidation half-reaction: Al(s) → Al3+(aq) + 3e-
Reduction half-reaction: MnO4‾(aq) + 4H+(aq) + 3e- → MnO2(s) + 2H2O(l)

Step 4: Balance the charges in each half-reaction:
Oxidation: Al(s) → Al3+(aq) + 3e-
Reduction: MnO4‾(aq) + 8H+(aq) + 5e- → MnO2(s) + 4H2O(l)

Step 5: Balance the number of atoms in each half-reaction:
Oxidation: 2Al(s) → 2Al3+(aq) + 6e-
Reduction: 5MnO4‾(aq) + 16H+(aq) + 10e- → 5MnO2(s) + 8H2O(l)

Step 6: Combine the half-reactions:
Multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2 to balance the electrons:
10Al(s) + 30OH-(aq) → 10Al(OH)4‾(aq)
5MnO4‾(aq) + 16H+(aq) + 10e- → 5MnO2(s) + 8H2O(l)

Now combine the two reactions and eliminate the spectator ions (OH-):
10Al(s) + 5MnO4‾(aq) + 16H+(aq) → 5MnO2(s) + 8H2O(l) + 10Al(OH)4‾(aq)

(b) Cl2(g) → Cl ‾(aq) + ClO ‾(aq)
Step 1: Assign oxidation numbers:
Cl: 0 O: -2

Step 2: Identify the element undergoing oxidation and reduction:
Cl is undergoing reduction (from 0 to -1) and O is undergoing oxidation (from -1 to 0).

Step 3: Write the half-reactions:
Oxidation half-reaction: 2Cl-(aq) → Cl2(g) + 2e-
Reduction half-reaction: Cl2(g) + 2e- → 2ClO-(aq)

Step 4: Balance the charges in each half-reaction:
Oxidation: 2Cl-(aq) → Cl2(g) + 2e-
Reduction: Cl2(g) + 2e- → 2ClO-(aq)

Step 5: Balance the number of atoms in each half-reaction:
Oxidation: 2Cl-(aq) → Cl2(g) + 2e-
Reduction: Cl2(g) + 2e- → 2ClO-(aq)

Step 6: Combine the half-reactions:
1Cl2(g) + 2Cl-(aq) → 2ClO-(aq)

(c) NO2‾(aq) + Al(s) → NH3(g) + AlO2‾(aq)
Step 1: Assign oxidation numbers:
N: +3 O: -2 Al: 0 H: +1

Step 2: Identify the element undergoing oxidation and reduction:
N is undergoing reduction (from +3 to -3) and Al is undergoing oxidation (from 0 to +3).

Step 3: Write the half-reactions:
Oxidation half-reaction: Al(s) → Al3+(aq) + 3e-
Reduction half-reaction: NO2‾(aq) + 8H+(aq) + 3e- → NH3(g) + 2H2O(l)

Step 4: Balance the charges in each half-reaction:
Oxidation: Al(s) → Al3+(aq) + 3e-
Reduction: NO2‾(aq) + 8H+(aq) + 3e- → NH3(g) + 2H2O(l)

Step 5: Balance the number of atoms in each half-reaction:
Oxidation: 2Al(s) → 2Al3+(aq) + 6e-
Reduction: 3NO2‾(aq) + 8H+(aq) + 6e- → 3NH3(g) + 6H2O(l)

Step 6: Combine the half-reactions:
Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2 to balance the electrons:
6Al(s) + 6NO2‾(aq) + 16H+(aq) → 6Al3+(aq) + 6NH3(g) + 12H2O(l)

Now combine the two reactions and eliminate the spectator ions (H+):
6Al(s) + 6NO2‾(aq) → 6Al3+(aq) + 6NH3(g) + 12H2O(l)

Continue the steps for each reaction to find the oxidation half-reaction, reduction half-reaction, and the complete balanced reaction.