help please?

Q1
What species has been oxidised in this equation? __sulfur (S) __
4NO₃⁻ + 3S + 4H⁺ → 3SO₂ + 4NO + 2H₂O
(This is because the oxidation number of S increases from o to +2.)


Question 2
What species has been reduced in this equation? __nitrate ion (NO₃⁻) __
4NO₃⁻ + 3S + 4H⁺ → 3SO₂ + 4NO + 2H₂O
(This is because the oxidation number of N decreases from +5 to +2.)


Question 3
Has Aluminium been oxidised or reduced in this equation? __oxidised __
2Al(s) + 6H⁺(aq) → 2Al³⁺(aq) + 3H₂(g)
(Thisn is because the oxidation number of Al increases from 0 to +3.)


Question 4
Which of the following is the correct balanced chemical equation.
a. S₂O₃²⁻(aq) + I₂(aq) → S₄O₆²⁻(aq) + I⁻(aq)
b. S₂O₃²⁻(aq) + 2I₂(aq) → S₄O₆²⁻(aq) + I⁻(aq)
c. S₂O₃²⁻(aq) + I₂(aq) → S₄O₆²⁻(aq) + 2I⁻(aq)
d. 2S₂O₃²⁻(aq) + I₂(aq) → S₄O₆²⁻(aq) + 2I⁻(aq)
e. S₂O₃²⁻(aq) + I₂(aq) + SO₃ → S₄O₆²⁻(aq) + 2I⁻(aq)

The answer is: d. 2S₂O₃²⁻(aq) + I₂(aq) → S₄O₆²⁻(aq) + 2I⁻(aq)

(a. is false. There are 2S/3O/2I/2⁻ on the left, but 4S/6O/1I/3⁻ on the right.)
(b. is false. There are 2S/3O/4I/2⁻ on the left, but 4S/6O/1I/3⁻ on the right.)
(c. is false. There are 2S/3O/2I/2⁻ on the left, but 4S/6O/2I/4⁻ on the right.)
(d. is true. There are 4S/6O/2I/4⁻ on the left, and also 4S/6O/2I/4⁻ on the right.)
(e. is false. There are 3S/6O/2I/2⁻ on the left, but 4S/6O/2I/4⁻ on the right.)