U4.9 · Oxidation-Reduction (Redox) ReactionsPractice · 氧化还原反应

Oxidation numbers in a compound sum to 0. Take H as +1, K as +1, and O as −2, then solve for Cl: $\ce{HClO}$ +1, $\ce{ClO2}$ +4, $\ce{KClO3}$ +5, $\ce{Cl2O7}$ +7. In $\ce{Cl2O7}$, for example, 2x + 7(−2) = 0, so x = +7, the highest oxidation number Cl can have.

The reducing agent is the reactant that is oxidized, losing electrons. C is +2 in CO and +4 in $\ce{CO2}$: it is oxidized, so CO is the reducing agent. Fe drops from +3 in $\ce{Fe2O3}$ to 0: it is reduced, so $\ce{Fe2O3}$ is the oxidizing agent. As written, 3 C atoms lose 6 electrons and 2 Fe atoms gain 6. Fe and $\ce{CO2}$ are products.

Fe goes from 0 to +3 in $\ce{Fe2O3}$, losing 3 electrons each, so 4 Fe lose 12. O goes from 0 to −2, gaining 2 each; 3 $\ce{O2}$ hold 6 O atoms, which gain 12. Lost equals gained, so 12 electrons are transferred (not 24: the electrons lost are the same ones gained).

Reduction is a gain of electrons, so the electrons appear on the left. $\ce{Fe^3+}$ gains 1 electron to become $\ce{Fe^2+}$: it is reduced. A is the oxidation half-reaction (Zn loses electrons); C puts the electron on the right, leaving charges of +3 and +1; D is the reduction of $\ce{Zn^2+}$, which doesn’t happen in this reaction.

O is −1 in $\ce{H2O2}$ and −2 in $\ce{H2O}$, so each $\ce{H2O2}$ gains 2 electrons (one per O), while each $\ce{Fe^2+}$ loses 1. They react 1 : 2, so n($\ce{Fe^2+}$) = 0.0150 × 2 = 0.0300 mol.

The solid gained 4.52 − 3.00 = 1.52 g. For every 2 mol of Ag⁺ that react, 2 × 107.87 g of silver deposits and 63.55 g of copper dissolves, a net gain of 152.19 g. So n(Cu) = 1.522 ÷ 152.19 = 0.0100 mol and n(Ag⁺) = 2 × 0.0100 = 0.0200 mol. Dividing by the molar mass of silver alone (0.0141) forgets that copper dissolves at the same time.