电化学

The oxidation number 氧化数 (also called the oxidation state) shows how many electrons 电子 an atom has lost or gained compared with the free element. You work it out using simple rules:

• an uncombined element has an oxidation number of $0$.
• a simple ion has an oxidation number equal to its charge (so $\text{Mg}^{2+}$ is $+2$).
• Group 1 is always $+1$, Group 2 is always $+2$.
• hydrogen is $+1$ (but $-1$ in metal hydrides).
• oxygen is $-2$ (but $-1$ in peroxides).
• fluorine is always $-1$.
• the oxidation numbers in a neutral compound add up to $0$; in an ion they add up to the charge.

A Roman numeral shows the size of the oxidation number of an element, for example iron(II) means $+2$ and manganese(VII) in $\text{KMnO}_4$ means $+7$.

Worked example. Find the oxidation number of sulfur in the sulfate ion, $\text{SO}_4^{2-}$.

Each oxygen is $-2$, so the four oxygens give $4 \times (-2) = -8$. In an ion the numbers add up to the charge, here $-2$. If sulfur is $x$:

Worked example. Find the oxidation number of nitrogen in the nitrate ion, $\text{NO}_3^-$.

A redox 氧化还原 reaction is one where electrons move from one species to another.

• oxidation 氧化 is the loss of electrons. The oxidation number goes up.
• reduction 还原 is the gain of electrons. The oxidation number goes down.

A useful memory aid is OIL RIG: Oxidation Is Loss, Reduction Is Gain.

Oxidation and reduction always happen together, because the electrons lost by one species are gained by another. This electron transfer 电子转移 is why we call it a redox reaction.

Changes in oxidation number help you balance a redox equation:

1. find the element whose oxidation number rises (it is oxidised) and the one whose number falls (it is reduced).
2. the total rise must equal the total fall, because every electron lost is gained somewhere.
3. choose the ratio of the two species so the rise and fall match, then balance the rest of the equation.

Worked example. Balance the reaction of manganate(VII) with iron(II) in acid: $\text{MnO}_4^- + \text{Fe}^{2+} + \text{H}^+ \rightarrow \text{Mn}^{2+} + \text{Fe}^{3+} + \text{H}_2\text{O}$.

Manganese falls from $+7$ to $+2$ (a fall of $5$); iron rises from $+2$ to $+3$ (a rise of $1$). To make the total fall equal the total rise, take $5$ iron ions for every $1$ manganate ion. Then balance oxygen with water ($4\,\text{H}_2\text{O}$) and hydrogen with $\text{H}^+$ ($8\,\text{H}^+$):

Disproportionation 歧化 is a special redox reaction in which the same element is both oxidised and reduced at the same time. For example, when chlorine reacts with cold water, some chlorine atoms are reduced (to $\text{Cl}^-$) and others are oxidised (to $\text{ClO}^-$).

• an oxidising agent 氧化剂 takes electrons away from another species. In doing so, it is itself reduced.
• a reducing agent 还原剂 gives electrons to another species. In doing so, it is itself oxidised.

So in any redox reaction, the oxidising agent causes the oxidation of the other species, while the reducing agent causes the reduction.

Common mistakes.

• Saying the oxidising agent is oxidised. An oxidising agent takes electrons, so it is itself reduced; in the manganate(VII)/iron(II) reaction, $\text{MnO}_4^-$ is the oxidising agent and is reduced to $\text{Mn}^{2+}$.
• Writing an oxidation number as a charge, such as $2+$ for iron(II). Write the sign first: $+2$. Charges on ions are written number-first ($\text{Fe}^{2+}$).
• Giving oxygen $-2$ in hydrogen peroxide. In peroxides oxygen is $-1$; in $\text{OF}_2$ it is even $+2$, because fluorine always wins.
• Balancing the atoms but not the charge. Check that the total charge is the same on both sides: in the manganate(VII) equation, $-1 + 10 + 8 = +17$ on the left and $+2 + 15 = +17$ on the right.
• Calling every reaction with an element in two products disproportionation. The same element must start in one oxidation state and go both up and down.

• Apply the oxidation-number rules in order: O is $-2$ except in peroxides ($-1$); H is $+1$ except in metal hydrides ($-1$).
• In a redox equation the total increase equals the total decrease in oxidation number — use this to fix the ratio, then balance O with $\text{H}_2\text{O}$ and H with $\text{H}^+$.
• An oxidising agent is itself reduced; name both what is oxidised/reduced and the agent for full marks.
• Disproportionation is the same element both oxidised and reduced — show both oxidation-number changes.