Redox Reactions Questions and Answers

Answer: A

Magnesium changes from 0 to +2.

Free magnesium metal, \(\mathrm{Mg_{(s)}}\), is an element in its standard state, so its oxidation state is 0. In magnesium nitride, \(\mathrm{Mg_{3}N_{2}}\), nitrogen is the more electronegative element and takes -3:

\[ \begin{aligned} 3x + 2(-3) &= 0 \\ x &= +2 \end{aligned} \]

So each magnesium atom loses two electrons and is oxidised, while nitrogen goes from 0 to -3 and is reduced. +3, +4 and +6 are not oxidation states of magnesium, a group 2 metal that only ever shows +2 in compounds, and a negative value would mean magnesium had gained electrons.

Answer: C

The INCORRECT statement is Decreases in number of positive charges.

Oxidation is the loss of electrons. Losing an electron makes a species more positive: Fe²⁺ → Fe³⁺ + e^- increases the positive charge, and Cl^- → Cl decreases the negative charge. A decrease in positive charge is the opposite, a reduction.

The other statements describe oxidation in its classical and electronic senses: addition of oxygen, removal of hydrogen, removal of an electron, and the decrease in negative charge that follows from losing electrons.

Answer: E

The incorrect statement is Removal of an electron.

Reduction is the gain of electrons; removal (loss) of an electron is oxidation. The other four are all valid descriptions of reduction:

• removal of oxygen, as in CuO + H₂ giving Cu,
• addition of hydrogen, as in ethene giving ethane,
• addition of an electron, as in Fe³⁺ + e^- giving Fe²⁺,
• decrease in oxidation number, which is what electron gain does to the atom.

Answer: B

The reaction is \(\mathrm{2HgCl_{2} + SnCl_{2} \rightarrow Hg_{2}Cl_{2} + SnCl_{4}}\).

In HgCl₂ mercury is +2. In Hg₂Cl₂ (calomel, with an Hg-Hg bond) each mercury is +1. So the +2 state is reduced to +1, while tin is oxidised from +2 to +4.

The others involve different changes: in Mg + S, Mg goes 0 to +2; in the combustion of methane carbon goes -4 to +4; in the ferrocyanide reaction iron goes +2 to +3, an oxidation; and in H₂S + O₂ sulphur goes -2 to 0. None of them reduces +2 to +1.

Answer: D

Phosphorus changes 0 to -3 and 0 to +1.

This is a disproportionation: elemental phosphorus, P₄, is in the 0 state.

• In PH₃: \(x + 3(+1) = 0\), so P is -3.
• In H₂PO₂^-: \(2(+1) + x + 2(-2) = -1\), so P is +1.

So part of the phosphorus is reduced to -3 and part oxidised to +1. The option 0 to +3 would need phosphite, H₂PO₃^-, and 0 to +5 would need phosphate; neither is formed here.

Answer: C

Oxygen is in the -1 state in hydrogen peroxide.

In H2O2 the two oxygen atoms are bonded to each other (a peroxide link, -O-O-). Each hydrogen is +1, so the two oxygens share a total of -2, which is -1 each.

In water, manganese dioxide, carbon dioxide and potassium nitrate oxygen has its usual oxidation state of -2. The exceptions to -2 are peroxides (-1), superoxides (-1/2) and compounds with fluorine, such as OF2 (+2).

Answer: C

The oxidation numbers are -4 and +1.

In CH₄, hydrogen bonded to a non-metal has oxidation number +1. The molecule is neutral, so the oxidation numbers must add to zero:

\[ \begin{aligned} x + 4(+1) &= 0 \\ x &= -4 \end{aligned} \]

Carbon is more electronegative than hydrogen, which is why it carries the negative value. Any option giving carbon +4 or hydrogen 0 or -1 breaks this rule; hydrogen is -1 only in metal hydrides such as NaH.

Answer: C

The oxidation numbers are 0, +1, -1/2.

In \(\mathrm{O_2}\) oxygen is an element, so 0. In \(\mathrm{O_2F_2}\) fluorine, the most electronegative element, is always -1, so the two oxygens share +2: each is +1. In \(\mathrm{RbO_2}\), a superoxide, Rb is +1, so the two oxygens share -1: each is -1/2.

\[ \begin{aligned} \mathrm{O_2F_2}: 2x + 2(-1) &= 0 \Rightarrow x = +1 \\ \mathrm{RbO_2}: +1 + 2x &= 0 \Rightarrow x = -\tfrac{1}{2} \end{aligned} \]

The option with +2 uses \(\mathrm{OF_2}\) instead of \(\mathrm{O_2F_2}\); -1 is the value in peroxides, not superoxides.

Answer: A

The matching is (i)-(b), (ii)-(a), (iii)-(d), (iv)-(c), taking the compounds in the order listed.

With oxygen at -2: MnO has Mn(II), so manganese(II) oxide (b); MnO₂ has Mn(IV), manganese(IV) oxide (a); Fe₂O₃ has 2x = +6, so Fe(III), iron(III) oxide (d); Tl₂O has 2x = +2, so Tl(I), thallium(I) oxide (c).

Stock notation writes this oxidation number in Roman numerals after the metal's name. The other options swap Fe₂O₃ and MnO₂, or give Tl₂O a higher state, which the formula does not allow.

Answer: D

Chlorine is +7 in HClO₄.

In perchloric acid, hydrogen is +1 and each of the four oxygens is -2: 1 + x + 4(-2) = 0, so x = +7, chlorine's highest oxidation state.

In NaCl, KCl, HCl and SiCl₄ chlorine is bonded to a less electronegative element and has the oxidation number -1. In general chlorine's positive states appear only with oxygen or fluorine: +1 in HClO, +3 in HClO₂, +5 in HClO₃ and +7 in HClO₄.