Answer:
C)
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.
