Balance each of the following redox reactions and classify it as a combination, decomposition or displacement reaction.

a) Mg(s)+H2O(g)->Mg(OH)2(s)+H2
b) Cr(NO3)(aq)+Al(s)->Al(NO3)(aq)+Cr(s)
c) PF3(g)+F2(g)->PF5(g)

a) Mg + 2H2O ==> Mg(OH)2 + 2H2

single replacement/displacement
What do you not understand about this. It seems simple enough. I shall be happy to explain anything you don't understand.

I know how to balance, I have no idea what the heck any of the classifications are

a) Mg(s) + H2O(g) -> Mg(OH)2(s) + H2

To balance this redox reaction, we need to assign oxidation numbers to the elements involved:

Oxidation number of Mg(s): 0 (because it is an uncombined element)
Oxidation number of H2O(g): +1 for H and -2 for O
Oxidation number of Mg(OH)2(s): +2 for Mg, -2 for O, and +1 for H
Oxidation number of H2: 0 (because it is an uncombined element)

Now we can write the balanced equation by following these steps:

1. Balance the atoms other than H and O:
Mg(s) + H2O(g) -> Mg(OH)2(s) + H2

2. Balance the oxygen atoms by adding H2O to the appropriate side:
Mg(s) + H2O(g) -> Mg(OH)2(s) + H2O(g)

3. Balance the hydrogen atoms by adding H+ to the appropriate side:
Mg(s) + 2H2O(g) -> Mg(OH)2(s) + H2O(g) + H2

4. Balance the charges by adding electrons to the appropriate side:
Mg(s) + 2H2O(g) -> Mg(OH)2(s) + 2H+(aq) + H2O(g) + H2 + 2e-

5. Balance the charges by adding electrons to the other side:
Mg(s) + 2H2O(g) + 2e- -> Mg(OH)2(s) + 2H+(aq) + H2O(g) + H2

Now the reaction is balanced, and it is a combination reaction.

b) Cr(NO3)(aq) + Al(s) -> Al(NO3)(aq) + Cr(s)

To balance this redox reaction, we need to assign oxidation numbers to the elements involved:

Oxidation number of Cr(NO3)(aq): +3 for Cr, -2 for O, and -1 for each NO3
Oxidation number of Al(s): 0 (because it is an uncombined element)

Now we can write the balanced equation by following these steps:

1. Balance the atoms other than H and O:
Cr(NO3)(aq) + Al(s) -> Al(NO3)(aq) + Cr(s)

2. Balance the charges by adding electrons to the appropriate side:
Cr(NO3)(aq) + Al(s) -> Al(NO3)(aq) + Cr(s) + 3e-

Now the reaction is balanced, and it is a displacement reaction.

c) PF3(g) + F2(g) -> PF5(g)

To balance this redox reaction, we need to assign oxidation numbers to the elements involved:

Oxidation number of PF3(g): +5 for P and -1 for each F
Oxidation number of F2(g): 0 (because it is an uncombined element)
Oxidation number of PF5(g): +5 for P and -1 for each F

Now we can write the balanced equation by following these steps:

1. Balance the atoms other than H and O:
PF3(g) + F2(g) -> PF5(g)

2. Balance the charges by adding electrons to the appropriate side:
PF3(g) + F2(g) -> PF5(g) + 3e-

Now the reaction is balanced, and it is a combination reaction.

To balance redox reactions, you need to assign oxidation numbers to each element to identify the changes in oxidation states. Once you have determined the changes in oxidation states, you can balance the equations by adding coefficients to the reactants and products.

a) Mg(s) + H2O(g) -> Mg(OH)2(s) + H2

Let's first assign oxidation numbers to each element:
Mg: 0 (unchanged)
H: +1 in H2O and 0 in H2
O: -2 in H2O and -2 in Mg(OH)2

We can see that hydrogen (H) is being reduced from +1 to 0, while oxygen (O) is being oxidized from -2 to 0.

Now let's balance the equation by adding coefficients:
Mg(s) + 2 H2O(g) -> Mg(OH)2(s) + H2

This reaction is a decomposition reaction, as a single reactant (Mg(s)) breaks down into multiple products.

b) Cr(NO3)(aq) + Al(s) -> Al(NO3)(aq) + Cr(s)

Assigning oxidation numbers:
Cr: +3 in Cr(NO3) and 0 in Cr
Al: 0 (unchanged)
N: +5 in NO3 and -3 in NO3
O: -2 in NO3

We can see that chromium (Cr) is being reduced from +3 to 0, while nitrogen (N) is being reduced from +5 to -3.

Balance the equation:
2 Cr(NO3)(aq) + 3 Al(s) -> 3 Al(NO3)(aq) + 2 Cr(s)

This reaction is a displacement reaction, as one element (Al) displaces another element (Cr) from its compound.

c) PF3(g) + F2(g) -> PF5(g)

Assigning oxidation numbers:
P: +3 in PF3 and +5 in PF5
F: -1 (unchanged)

We can see that phosphorus (P) is being oxidized from +3 to +5, while fluorine (F) is being reduced from 0 to -1.

Balance the equation:
4 PF3(g) + 5 F2(g) -> 4 PF5(g)

This reaction is a combination reaction, as two elements (PF3 and F2) combine to form a compound (PF5).