2.13 · Oxidation and ReductionPractice · 氧化还原

K is a group 1 metal, +1; O is −2. The oxidation numbers add up to 0: 2(+1) + x + 4(−2) = 0, so x = +6. Iron is usually +2 or +3; iron in the +6 state is a strong oxidizer, which is why ferrate kills microbes.

Using the rules (H is +1, O is −2, Na is +1, Cl⁻ is −1): $\ce{N2H4}$ −2, $\ce{NH4Cl}$ −3, $\ce{N2O}$ +1, $\ce{NaNO2}$ +3. The N in $\ce{NH4+}$ of $\ce{NH4Cl}$ is −3, nitrogen’s lowest oxidation number.

Mg goes from 0 to +2: it loses electrons, is oxidized, and is the reducing agent, with MgO the product of oxidation. The C in CO₂ goes from +4 to 0: it gains electrons, is reduced, so CO₂ is the oxidizing agent and C the product of reduction. As written, 2 Mg lose 4 electrons and 1 C gains 4, so each C atom formed takes 4 electrons. That’s why a magnesium fire can’t be put out with a CO₂ extinguisher.

In KClO₃, Cl is +5 and O is −2; in the products Cl is −1 and O is 0. Electrons move between two elements in the same compound: O loses them and Cl gains them. As written, 6 O atoms each lose 2, 12 in all, and 2 Cl atoms each gain 6, 12 in all, so the totals match. Making 3 O₂ transfers 12 electrons: 12 ÷ 3 = 4 per O₂.

Balance electrons first: Cu goes from 0 to +2 (loses 2), and each N goes from +5 to +4 (gains 1), so 1 Cu matches 2 NO₂. Then N: Cu(NO₃)₂ holds 2 more N that don’t change, so 4 HNO₃ in all; finally H gives the water: 4 H make 2 H₂O. $\ce{Cu(s) +}$ $\ce{4HNO3(aq) ->}$ $\ce{Cu(NO3)2(aq) +}$ $\ce{2NO2(g) +}$ $\ce{2H2O(l)}$. Only some of the HNO₃ acts as the oxidizing agent; the rest supplies NO₃⁻.

Cu loses 2 electrons to become $\ce{Cu^2+}$, and each $\ce{Fe^3+}$ gains just 1 to become $\ce{Fe^2+}$, so one Cu needs two $\ce{Fe^3+}$: B. A balances atoms but not charge (+3 on the left, +4 on the right). C reduces $\ce{Fe^3+}$ all the way to Fe, but copper is less active than iron and can’t do that, and the question says FeCl₂ forms. D is balanced but has the wrong product: the copper ends up as CuCl₂, that is, $\ce{Cu^2+}$.

Zinc is more active than silver, so zinc is the anode, where oxidation occurs (Zn loses 2 electrons to become $\ce{Zn^2+}$); silver is the cathode, where $\ce{Ag+}$ gains electrons and plates out as silver, so the silver gains mass. Electrons flow through the wire from the anode (zinc) to the cathode (silver). The zinc side builds up positive ions, so anions ($\ce{NO3-}$) from the bridge move toward the zinc side and $\ce{K+}$ moves toward the silver side. KCl isn’t used in this bridge because $\ce{Cl-}$ would precipitate $\ce{Ag+}$ as AgCl.

Overall, $\ce{Zn + 2Ag+ -> Zn^2+ + 2Ag}$: the 2 electrons lost by each Zn plate out 2 Ag. n(Zn) = 1.30 ÷ 65.38 = 0.01988 mol, so n(Ag) = 2 × 0.01988 = 0.03976 mol and m(Ag) = 0.03976 × 107.87 = 4.29 g.

The more active a metal, the more easily it loses electrons and the less readily its ion takes them back. Activity: Mg > Zn > Fe > Cu, so oxidizing strength of the ions runs the other way: $\ce{Cu^2+}$ > $\ce{Fe^2+}$ > $\ce{Zn^2+}$ > $\ce{Mg^2+}$. Copper is least active, so $\ce{Cu^2+}$ is the strongest oxidizing agent, which is why iron can displace copper from $\ce{CuSO4}$ solution.

S is −2 in H₂S and +4 in SO₂, and 0 in the product. The S from 2 H₂S rises to 0 and is oxidized (losing 2 × 2 = 4 electrons); the S from 1 SO₂ falls to 0 and is reduced (gaining 4). Both are sulfur, so the mass ratio equals the atom ratio, 2 : 1.