A redox reaction involves both oxidation and reduction. In 2Na + Cl₂ → 2NaCl, sodium (Na) is oxidized from 0 to +1, and chlorine (Cl) is reduced from 0 to -1. The other reactions do not involve a change in oxidation state for all elements.

In H₂S, sulfur has an oxidation state of -2, and in S, it is 0. Since the oxidation state of sulfur increases from -2 to 0, H₂S is oxidized.

In Cr₂O₇²⁻, oxygen has an oxidation state of -2. Let the oxidation state of Cr be x. So, 2x + 7(-2) = -2. Solving, 2x - 14 = -2, hence 2x = 12, x = +6.

KMnO₄ acts as an oxidizing agent because manganese in KMnO₄ is reduced from +7 to +2 in MnSO₄, indicating a gain of electrons.

BaCl₂ + H₂SO₄ → BaSO₄ + 2HCl is not a redox reaction as there is no change in the oxidation states of any element. It is a double displacement reaction. Other options involve changes in oxidation states.

In N₂O₅, oxygen has an oxidation state of -2. Let the oxidation state of N be x. So, 2x + 5(-2) = 0, 2x - 10 = 0, 2x = 10, x = +5.

A reducing agent is the species that loses electrons and gets oxidized. Here, Sn²⁺ loses electrons to become Sn⁴⁺ (oxidation state increases from +2 to +4), so Sn²⁺ is the reducing agent.

In Cu₂O and Cu₂S, copper has oxidation states of +1. In the product Cu, the oxidation state is 0. Since there is a decrease in oxidation state, copper is reduced.

In a disproportionation reaction, the same species is both oxidized and reduced. Here, Cu⁺ (oxidation state +1) is oxidized to Cu²⁺ (oxidation state +2) and reduced to Cu (oxidation state 0). Thus, Cu⁺ is oxidized to Cu²⁺.

The n-factor is the number of electrons gained or lost per mole of the substance. In acidic medium, Mn in KMnO₄ (oxidation state +7) is reduced to Mn²⁺ (oxidation state +2). The change in oxidation state is 7 - 2 = 5, so the n-factor is 5.