Question

(A) H2SeO4(aq) + 2 Cl-(aq) + 2 H+(aq) → H2SeO3(aq) + Cl2(g) + H2O(l) (B) S8(s) + 8 O2(g) → 8 SO2(g) (C) 3 Br2(aq) + 6 OH-(aq) → 5 Br-(aq) + BrO3-(aq) + 3 H2O(l) (D) Ca2+(aq) + SO42-(aq) → CaSO4(s) (E) PtCl4(s) + 2 Cl-(aq) → PtCl62-(aq) A reaction in which the same reactant undergoes both oxidation and reduction

Answer 1

C)

Step-by-step explanation:

Oxidation occurs when the Nox number increases and a reduction occurs when the Nox number decreases. So, let's verify in which reaction the same reactant has elements that both occur after the reaction.

A) H₂SeO₄: H = +1, O = -2 (fixed)

Se: x +2*1 + 4*(-2) = 0

x + 2 - 8= 0

x = +6

Cl⁻: -1 (ions have Nox equal to its charge)

H⁺: +1

H₂SeO₃: H = + 1, O = -2

Se: x +2*1 + 3*(-2) = 0

x + 2 -6 = 0

x = + 4 (Se reduces)

Cl₂: Cl = 0 (simple molecule, so Cl oxides)

H₂O: H = +1, O = -2

B) S₈: S = 0

O₂: O = 0

SO₂: O = -2, S: x + 2*(-2) = 0 -> x = +4 (S oxides and O reduces)

C) Br₂: Br = 0

OH⁻ : O = -2, H = +1

Br⁻ = -1 (Br reduces)

BrO₃⁻: O = -2, Br: x +3x(-2) = -1 -> x = +5 (Br oxides)

H₂O: H = +1, O = -2

D) Ca²⁺: Ca = +2

SO₄²⁻: O = -2, S: x +4*(-2) = -2 -> x = +6

CaSO₄: O = -2, S = +6, Ca = +2

E) PtCl₄: Cl = -1 (fixed), Pt: x + 4*(-1) = 0 -> x = +4

Cl⁻: Cl = -1

PtCl₆²⁻: Cl = -1, Pt = x + 6*(-1) = -2 -> x = +4

So, only in letter C Br₂ is undergoes both oxidation and reduction.