Identify the oxidant and reductant of each reaction

Question

I need number 7, but if you could do 6 too than that would be great!

6. Balance the equations below assuming they occur in a basic solution.

(a) SO32−(𝑎𝑞)+Cu(OH)2(𝑠)⟶SO42−(𝑎𝑞)+Cu(OH)(𝑠)SO32−(aq)+Cu(OH)2(s)⟶SO42−(aq)+Cu(OH)(s)

(b) O2(𝑔)+Mn(OH)2(𝑠)⟶MnO2(𝑠)O2(g)+Mn(OH)2(s)⟶MnO2(s)

(c) NO3−(𝑎𝑞)+H2(𝑔)⟶NO(𝑔)NO3−(aq)+H2(g)⟶NO(g)

(d) Al(𝑠)+CrO42−(𝑎𝑞)⟶Al(OH)3(𝑠)+Cr(OH)4−(𝑎𝑞)Al(s)+CrO42−(aq)⟶Al(OH)3(s)+Cr(OH)4−(aq)

7. Identify the oxidant and reductant of each reaction of the previous exercise.

Answer 1

To identify the oxidant (oxidizing agent), you want to find the agent that is reduced. If it is reduced (gains electrons), it will be the oxidant.

For #7, we have...

(a) SO₃²⁻(𝑎𝑞)+Cu(OH)₂(𝑠) ⟶ SO₄²⁻(aq) + Cu(OH)(s) (unbalanced)

....4+...............2+.....................6+................1+............oxidation numbers for S and Cu

S went from 4+ to 6+ so it lost electrons and was oxidized

Cu went from 2+ to 1+ so it gained electrons and was reduced. It is the oxidant as in Cu(OH)2

One more so you get the idea, and the hopefully you can do the others

(b) O2(𝑔) + Mn(OH)2(𝑠) ⟶ MnO2(s) + 2H2O (unbalanced)

.....0..............2+........................4+...2- ..........oxidation numbers for oxygen and Mn

O goes from zero to 2- so it is reduced making it the oxidant. This would be in basic

Too many in # 6 to do, but will do one so you get the idea... After running through it, you'll see why there are too many for me to do,..In fact this answer is too long and it won't let me post all of it. I'll try adding the rest to the comments or another of your quesitons.