---
title: "Oxidation and Reduction"
book: "School Chemistry — Grades 1 to 12"
subject: chemistry
language: en
chapter: 35
exercises: 15
source: https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction
license: CC-BY-NC-SA-4.0
credit: "One Chemistry Book, One Course (one-course.com)"
---

# Chapter 35 — Oxidation and Reduction

A zinc strip is dipped into a blue solution of copper sulfate. Within minutes its surface darkens under a spongy, reddish-brown coat; an hour later the blue of the solution has faded. Copper [metal](https://one-course.com/books/chemistry/1/en/chapter/20-the-periodic-table-a-first-look#def-g9-periodic-table-first-look-metal) has appeared where there was none, and zinc has gone into solution as invisible [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion). Nothing was heated, no gas was given off, no oxygen took part — and yet chemists call this an [oxidation](#def-g11-redox-oxidation). The word once meant “combining with oxygen”; it now means something more general and more useful: the loss of [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus). This chapter follows the [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus).

**You already know.**

An [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) is an [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) or a group of [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) that has gained or lost [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) ([Chapter 17](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#ch-g9-ions)). A [metal](https://one-course.com/books/chemistry/1/en/chapter/20-the-periodic-table-a-first-look#def-g9-periodic-table-first-look-metal) such as iron or zinc reacts with hydrochloric acid to give hydrogen and [metal](https://one-course.com/books/chemistry/1/en/chapter/20-the-periodic-table-a-first-look#def-g9-periodic-table-first-look-metal) [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) ([Chapter 19](https://one-course.com/books/chemistry/1/en/chapter/19-metals-reactions-with-acids-and-corrosion#ch-g9-metals-acids-corrosion)). An [amount of substance](https://one-course.com/books/chemistry/1/en/chapter/25-the-mole-and-molar-mass#def-g10-the-mole-amount) $n$, in [moles](https://one-course.com/books/chemistry/1/en/chapter/25-the-mole-and-molar-mass#def-g10-the-mole-amount), is computed from a mass $m$ by $n = m/M$ ([Chapter 25](https://one-course.com/books/chemistry/1/en/chapter/25-the-mole-and-molar-mass#ch-g10-the-mole)).

![A zinc strip in copper sulfate solution: copper is deposited on the zinc, and zinc ions take its place in the solution.](https://one-course.com/images/onecourse/chapters/chemistry-1/g11-redox/img-a004bd67229a.jpg)

*A zinc strip in copper sulfate solution: copper is deposited on the zinc, and zinc [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) take its place in the solution.*

## 35.1 Oxidants and reductants

**Definition 35.1 (Oxidant and reductant).**

An *oxidant* is a species able to gain one or more [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus). A *reductant* is a species able to give one or more [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) away.

**Definition 35.2 (Oxidation and reduction).**

An *oxidation* is a loss of [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus); a *reduction* is a gain of [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus). A [reductant](#def-g11-redox-oxidant) is oxidised; an [oxidant](#def-g11-redox-oxidant) is reduced.

**Remark 35.3 (A way to remember).**

In *[reduction](#def-g11-redox-oxidation)* the charge of the species is reduced: $\ce{Cu^{2+}}$ gaining two [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) becomes $\ce{Cu}$, charge $+2 \to 0$. In an [oxidation](#def-g11-redox-oxidation) the charge goes up: $\ce{Zn -> Zn^{2+} + 2e-}$, $0 \to +2$.

**Definition 35.4 (Redox couple and half-equation).**

An [oxidant](#def-g11-redox-oxidant) and the [reductant](#def-g11-redox-oxidant) it becomes by gaining [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) form a *redox couple*, written Ox/Red with the [oxidant](#def-g11-redox-oxidant) first: $\ce{Cu^{2+}}$/$\ce{Cu}$, $\ce{Zn^{2+}}$/$\ce{Zn}$. The exchange of [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) inside a couple is described by its *half-equation*,

$$
\text{Ox} + n\,\ce{e-} \ce{<=>} \text{Red},
$$

balanced in [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) and in charge, for instance $\ce{Cu^{2+}(aq) + 2e- <=> Cu(s)}$. The double arrow says that the exchange can go either way.

**Example 35.5 (Metals and their ions).**

Every [metal](https://one-course.com/books/chemistry/1/en/chapter/20-the-periodic-table-a-first-look#def-g9-periodic-table-first-look-metal) forms a couple with its [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion):

| couple | [half-equation](#def-g11-redox-couple) |
| --- | --- |
| $\ce{Ag+}$/$\ce{Ag}$ | $\ce{Ag+ + e- <=> Ag}$ |
| $\ce{Cu^{2+}}$/$\ce{Cu}$ | $\ce{Cu^{2+} + 2e- <=> Cu}$ |
| $\ce{Fe^{2+}}$/$\ce{Fe}$ | $\ce{Fe^{2+} + 2e- <=> Fe}$ |
| $\ce{Zn^{2+}}$/$\ce{Zn}$ | $\ce{Zn^{2+} + 2e- <=> Zn}$ |
| $\ce{Al^{3+}}$/$\ce{Al}$ | $\ce{Al^{3+} + 3e- <=> Al}$ |

[Ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) can form couples too: $\ce{Fe^{3+} + e- <=> Fe^{2+}}$ for the couple $\ce{Fe^{3+}}$/$\ce{Fe^{2+}}$, and $\ce{I2 + 2e- <=> 2I-}$ for $\ce{I2}$/$\ce{I-}$. So can hydrogen: $\ce{2H+ + 2e- <=> H2}$ for $\ce{H+}$/$\ce{H2}$.

**Remark 35.6 (One species, two roles).**

A species can be the [reductant](#def-g11-redox-oxidant) of one couple and the [oxidant](#def-g11-redox-oxidant) of another. The [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) $\ce{Fe^{2+}}$ is the [oxidant](#def-g11-redox-oxidant) of $\ce{Fe^{2+}}$/$\ce{Fe}$ and the [reductant](#def-g11-redox-oxidant) of $\ce{Fe^{3+}}$/$\ce{Fe^{2+}}$. Which role it plays depends on its partner in the reaction.

## 35.2 Redox reactions

**Definition 35.7 (Redox reaction).**

A *redox reaction* is a transfer of [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) from the [reductant](#def-g11-redox-oxidant) of one couple to the [oxidant](#def-g11-redox-oxidant) of another. The [reductant](#def-g11-redox-oxidant) is oxidised and the [oxidant](#def-g11-redox-oxidant) is reduced, at the same time.

**Proposition 35.8 (No free electrons).**

[Electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) do not exist free in a solution: every [electron](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) given by the [reductant](#def-g11-redox-oxidant) is taken by the [oxidant](#def-g11-redox-oxidant). The equation of a [redox reaction](#def-g11-redox-reaction) is therefore obtained by adding the two [half-equations](#def-g11-redox-couple), each multiplied by the number that makes the [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) given equal to the [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) taken; the [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) then cancel and do not appear in the equation.

**Example 35.9 (Zinc and copper ions).**

In the opening scene the zinc is oxidised and the copper [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) are reduced:

$$
\begin{array}{ll}
  \ce{Zn(s) -> Zn^{2+}(aq) + 2e-} & \text{(oxidation)}\\
  \ce{Cu^{2+}(aq) + 2e- -> Cu(s)} & \text{(reduction)}\\
  \hline
  \ce{Zn(s) + Cu^{2+}(aq) -> Zn^{2+}(aq) + Cu(s)} &
\end{array}
$$

Two [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) pass from each zinc [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) to one copper [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion). The blue colour of the solution is the colour of the $\ce{Cu^{2+}}$ [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion); it fades as they are used up, while the colourless $\ce{Zn^{2+}}$ [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) take their place.

![At the surface of the zinc. A zinc atom loses two electrons and leaves as a Zn2+ ion; the two electrons travel through the metal to a Cu2+ ion touching the surface, which becomes a copper atom and stays on the strip.](https://one-course.com/images/onecourse/chapters/chemistry-1/g11-redox/fig-9e277f49d25d.svg)

*At the surface of the zinc. A zinc [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) loses two [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) and leaves as a $\ce{Zn^{2+}}$ [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion); the two [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) travel through the [metal](https://one-course.com/books/chemistry/1/en/chapter/20-the-periodic-table-a-first-look#def-g9-periodic-table-first-look-metal) to a $\ce{Cu^{2+}}$ [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) touching the surface, which becomes a copper [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) and stays on the strip.*

![What can be seen: the deposit of copper on the immersed part of the strip, and the fading of the blue colour of the Cu2+ ions.](https://one-course.com/images/onecourse/chapters/chemistry-1/g11-redox/fig-6d6966293895.svg)

*What can be seen: the deposit of copper on the immersed part of the strip, and the fading of the blue colour of the $\ce{Cu^{2+}}$ [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion).*

**Example 35.10 (Metals in acid, revisited).**

When iron [dissolves](https://one-course.com/books/chemistry/1/en/chapter/2-mixing-and-dissolving#def-g2-mixing-and-dissolving-dissolve) in hydrochloric acid, the iron is oxidised and the $\ce{H+}$ [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) are reduced to hydrogen:

$$
\ce{Fe(s) + 2H+(aq) -> Fe^{2+}(aq) + H2(g)}.
$$

The couples are $\ce{Fe^{2+}}$/$\ce{Fe}$ and $\ce{H+}$/$\ce{H2}$; the chloride [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) take no part.

**In the lab — The silver tree.**

A coil of clean copper wire is hung in a beaker of dilute silver nitrate solution, $\ce{Ag+(aq) + NO3^-(aq)}$. Within an hour, branching needles of shiny silver grow along the wire, and the colourless solution turns pale blue: $\ce{Cu(s) + 2Ag+(aq) -> Cu^{2+}(aq) + 2Ag(s)}$. The experiment is made by the teacher, with gloves and goggles, and the solution is collected afterwards as hazardous waste.

**Safety.**

![](https://one-course.com/images/onecourse/chapters/chemistry-1/g11-redox/fig-105f4dfb6e4f.svg)

![](https://one-course.com/images/onecourse/chapters/chemistry-1/g11-redox/fig-93c17238ab7b.svg)

![](https://one-course.com/images/onecourse/chapters/chemistry-1/g11-redox/fig-37c19258c5e2.svg)

![](https://one-course.com/images/onecourse/chapters/chemistry-1/g11-redox/fig-259f91be137f.svg)

Silver nitrate: an oxidiser that can feed a fire (GHS03); corrosive to skin and eyes, and it stains the skin black (GHS05); a health hazard (GHS08); very toxic to aquatic life, so it never goes down the sink (GHS09).

![Silver crystals growing on a copper wire in silver nitrate solution.](https://one-course.com/images/onecourse/chapters/chemistry-1/g11-redox/img-990674e45412.jpg)

*Silver crystals growing on a copper wire in silver nitrate solution.*

## 35.3 Balancing half-equations

Many important couples contain oxygen and are written for [acidic solutions](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-acidic), where $\ce{H+}$ [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) are present. Their [half-equations](#def-g11-redox-couple) need water [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) and $\ce{H+}$ [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) to balance.

**Method 35.11 (Balancing a half-equation in acidic solution).**

To write the [half-equation](#def-g11-redox-couple) of a couple Ox/Red in [acidic solution](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-acidic):

1. write Ox on the left and Red on the right, and balance the [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) that is neither O nor H;
2. balance the oxygen [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) with water [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) $\ce{H2O}$ ;
3. balance the hydrogen [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) with $\ce{H+}$ [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) ;
4. balance the charge with [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) $\ce{e-}$ , on the left;
5. check: same number of each [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) , same total charge on both sides.

**Example 35.12 (The permanganate ion).**

For the couple $\ce{MnO4^-}$/$\ce{Mn^{2+}}$ (deep purple to almost colourless): manganese is balanced; four O on the left give $\ce{4H2O}$ on the right; their eight H give $\ce{8H+}$ on the left; the charge on the left is $-1 + 8 = +7$, on the right $+2$, so five [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) are added on the left:

$$
\ce{MnO4^-(aq) + 8H+(aq) + 5e- <=> Mn^{2+}(aq) + 4H2O(l)}.
$$

**Example 35.13 (The dichromate ion).**

For $\ce{Cr2O7^{2-}}$/$\ce{Cr^{3+}}$ (orange to green): two chromium on the right; seven O give $\ce{7H2O}$; fourteen H give $\ce{14H+}$; charge $-2 + 14 = +12$ on the left against $2 \times 3 = +6$ on the right, so six [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus):

$$
\ce{Cr2O7^{2-}(aq) + 14H+(aq) + 6e- <=> 2Cr^{3+}(aq) + 7H2O(l)}.
$$

**Method 35.14 (Writing a redox equation).**

To write the equation of the reaction between the [oxidant](#def-g11-redox-oxidant) Ox$_1$ of one couple and the [reductant](#def-g11-redox-oxidant) Red$_2$ of another:

1. write both [half-equations](#def-g11-redox-couple) , the first in the direction of the [reduction](#def-g11-redox-oxidation) (Ox $_1 \to$ Red $_1$ ), the second in the direction of the [oxidation](#def-g11-redox-oxidation) (Red $_2 \to$ Ox $_2$ );
2. multiply them by the smallest numbers that make the [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) equal;
3. add them; the [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) cancel, and so may some $\ce{H+}$ and $\ce{H2O}$ that appear on both sides;
4. check [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) and charge once more.

**Example 35.15 (Permanganate and iron(II)).**

The permanganate [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) oxidises $\ce{Fe^{2+}}$ [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) to $\ce{Fe^{3+}}$:

$$
\begin{array}{ll}
  \ce{MnO4^- + 8H+ + 5e- -> Mn^{2+} + 4H2O} & \times 1\\
  \ce{Fe^{2+} -> Fe^{3+} + e-} & \times 5\\
  \hline
  \ce{MnO4^- + 8H+ + 5Fe^{2+} -> Mn^{2+} + 5Fe^{3+} + 4H2O} &
\end{array}
$$

Charge on the left: $-1 + 8 + 10 = +17$; on the right: $2 + 15 = +17$. This reaction is used in the next chapter to measure an amount of iron(II).

## 35.4 Oxidants and reductants around us

**Proposition 35.16 (Some common oxidants and reductants).**

The following species are met again and again in the laboratory and in everyday life:

| species | role | [half-equation](#def-g11-redox-couple) of its couple |
| --- | --- | --- |
| dioxygen $\ce{O2}$ | [oxidant](#def-g11-redox-oxidant) | $\ce{O2 + 4H+ + 4e- <=> 2H2O}$ |
| dichlorine $\ce{Cl2}$ | [oxidant](#def-g11-redox-oxidant) | $\ce{Cl2 + 2e- <=> 2Cl-}$ |
| hypochlorite $\ce{ClO-}$ (bleach) | [oxidant](#def-g11-redox-oxidant) | $\ce{2ClO- + 4H+ + 2e- <=> Cl2 + 2H2O}$ |
| hydrogen peroxide $\ce{H2O2}$ | [oxidant](#def-g11-redox-oxidant) | $\ce{H2O2 + 2H+ + 2e- <=> 2H2O}$ |
| permanganate $\ce{MnO4^-}$ | [oxidant](#def-g11-redox-oxidant) | $\ce{MnO4^- + 8H+ + 5e- <=> Mn^{2+} + 4H2O}$ |
| [metals](https://one-course.com/books/chemistry/1/en/chapter/20-the-periodic-table-a-first-look#def-g9-periodic-table-first-look-metal) $\ce{M}$ ($\ce{Fe}$, $\ce{Zn}$, $\ce{Al}$…) | [reductants](#def-g11-redox-oxidant) | $\ce{M^{n+} + $n$ e- <=> M}$ |
| iodide $\ce{I-}$ | [reductant](#def-g11-redox-oxidant) | $\ce{I2 + 2e- <=> 2I-}$ |
| thiosulfate $\ce{S2O3^{2-}}$ | [reductant](#def-g11-redox-oxidant) | $\ce{S4O6^{2-} + 2e- <=> 2S2O3^{2-}}$ |
| ascorbic acid $\ce{C6H8O6}$ (vitamin C) | [reductant](#def-g11-redox-oxidant) | $\ce{C6H6O6 + 2H+ + 2e- <=> C6H8O6}$ |

**Example 35.17 (Copper turns green).**

Copper roofs and statues, left in the open [air](https://one-course.com/books/chemistry/1/en/chapter/4-air-a-mixture-of-gases#def-g4-air-a-mixture-of-gases-air) for decades, slowly turn pale green. Oxygen from the [air](https://one-course.com/books/chemistry/1/en/chapter/4-air-a-mixture-of-gases#def-g4-air-a-mixture-of-gases-air) oxidises the copper to $\ce{Cu^{2+}}$ [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion), which, with water and carbon dioxide, form a thin green crust, the patina. Unlike [rust](https://one-course.com/books/chemistry/1/en/chapter/19-metals-reactions-with-acids-and-corrosion#def-g9-metals-acids-corrosion-corrosion) on iron, the patina sticks to the [metal](https://one-course.com/books/chemistry/1/en/chapter/20-the-periodic-table-a-first-look#def-g9-periodic-table-first-look-metal) and protects it from further attack.

![A copper roof after decades in the air: oxidised copper forms a green patina.](https://one-course.com/images/onecourse/chapters/chemistry-1/g11-redox/img-134e04d9c0be.jpg)

*A copper roof after decades in the [air](https://one-course.com/books/chemistry/1/en/chapter/4-air-a-mixture-of-gases#def-g4-air-a-mixture-of-gases-air): oxidised copper forms a green patina.*

**Remark 35.18 (Antioxidants).**

An “antioxidant” added to food is a [reductant](#def-g11-redox-oxidant) that is oxidised by the oxygen of the [air](https://one-course.com/books/chemistry/1/en/chapter/4-air-a-mixture-of-gases#def-g4-air-a-mixture-of-gases-air) before the food is. Vitamin C (ascorbic acid) is the commonest: a cut apple sprinkled with lemon juice browns much more slowly, because the vitamin C of the lemon is oxidised first.

## 35.5 Exercises

**Exercise 35.1 ★.**

Define an [oxidant](#def-g11-redox-oxidant), a [reductant](#def-g11-redox-oxidant), an [oxidation](#def-g11-redox-oxidation) and a [reduction](#def-g11-redox-oxidation).

**Solution of Exercise 35.1.**

[Oxidant](#def-g11-redox-oxidant): a species able to gain [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus). [Reductant](#def-g11-redox-oxidant): a species able to give [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) away. [Oxidation](#def-g11-redox-oxidation): a loss of [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus). [Reduction](#def-g11-redox-oxidation): a gain of [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus).

**Exercise 35.2 ★.**

For each pair, write the [redox couple](#def-g11-redox-couple) in the form Ox/Red: $\ce{Fe}$ and $\ce{Fe^{2+}}$; $\ce{Ag}$ and $\ce{Ag+}$; $\ce{I-}$ and $\ce{I2}$; $\ce{Fe^{2+}}$ and $\ce{Fe^{3+}}$.

**Solution of Exercise 35.2.**

$\ce{Fe^{2+}}$/$\ce{Fe}$; $\ce{Ag+}$/$\ce{Ag}$; $\ce{I2}$/$\ce{I-}$; $\ce{Fe^{3+}}$/$\ce{Fe^{2+}}$.

**Exercise 35.3 ★.**

Write the [half-equations](#def-g11-redox-couple) of the couples $\ce{Ag+}$/$\ce{Ag}$, $\ce{Al^{3+}}$/$\ce{Al}$, $\ce{Fe^{3+}}$/$\ce{Fe^{2+}}$ and $\ce{Cl2}$/$\ce{Cl-}$.

**Solution of Exercise 35.3.**

$\ce{Ag+ + e- <=> Ag}$; $\ce{Al^{3+} + 3e- <=> Al}$; $\ce{Fe^{3+} + e- <=> Fe^{2+}}$; $\ce{Cl2 + 2e- <=> 2Cl-}$.

**Exercise 35.4 ★.**

Is the transformation $\ce{Fe -> Fe^{2+} + 2e-}$ an [oxidation](#def-g11-redox-oxidation) or a [reduction](#def-g11-redox-oxidation)? Is iron acting as an [oxidant](#def-g11-redox-oxidant) or a [reductant](#def-g11-redox-oxidant)?

**Solution of Exercise 35.4.**

An [oxidation](#def-g11-redox-oxidation) ([electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) are lost); iron acts as a [reductant](#def-g11-redox-oxidant).

**Exercise 35.5 ★.**

In the reaction $\ce{Zn + Cu^{2+} -> Zn^{2+} + Cu}$, name the species oxidised, the species reduced, the [oxidant](#def-g11-redox-oxidant) and the [reductant](#def-g11-redox-oxidant), and give the number of [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) transferred for each zinc [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom).

**Solution of Exercise 35.5.**

Zinc is oxidised and is the [reductant](#def-g11-redox-oxidant); $\ce{Cu^{2+}}$ is reduced and is the [oxidant](#def-g11-redox-oxidant). Two [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) per zinc [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom).

**Exercise 35.6 ★★.**

Balance, in [acidic solution](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-acidic), the [half-equation](#def-g11-redox-couple) of the couple $\ce{H2O2}$/$\ce{H2O}$, following [Method 35.11](#met-g11-redox-balance) step by step.

**Solution of Exercise 35.6.**

Start from $\ce{H2O2 -> H2O}$: oxygen: two O on the left, so $\ce{2H2O}$ on the right; hydrogen: 2 on the left, 4 on the right, so $\ce{2H+}$ on the left; charge: $+2$ on the left, 0 on the right, so two [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) on the left: $\ce{H2O2 + 2H+ + 2e- <=> 2H2O}$.

**Exercise 35.7 ★★.**

Write the equation of the reaction between iodine $\ce{I2}$ and the thiosulfate [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) $\ce{S2O3^{2-}}$, which gives iodide [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) and tetrathionate [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) $\ce{S4O6^{2-}}$.

**Solution of Exercise 35.7.**

$\ce{I2 + 2e- -> 2I-}$ and $\ce{2S2O3^{2-} -> S4O6^{2-} + 2e-}$; two [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) each, so $\ce{I2 + 2S2O3^{2-} -> 2I- + S4O6^{2-}}$.

**Exercise 35.8 ★★.**

Write the equation of the silver tree reaction from its two [half-equations](#def-g11-redox-couple), and explain why the solution turns blue.

**Solution of Exercise 35.8.**

$\ce{Cu -> Cu^{2+} + 2e-}$ and $2\times$ ($\ce{Ag+ + e- -> Ag}$): $\ce{Cu + 2Ag+ -> Cu^{2+} + 2Ag}$. The $\ce{Cu^{2+}}$ [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) formed are blue.

**Exercise 35.9 ★★.**

Vitamin C, $\ce{C6H8O6}$, reacts with iodine $\ce{I2}$. Write the equation of the reaction, and say which species is the [reductant](#def-g11-redox-oxidant).

**Solution of Exercise 35.9.**

$\ce{C6H8O6 -> C6H6O6 + 2H+ + 2e-}$ and $\ce{I2 + 2e- -> 2I-}$: $\ce{C6H8O6 + I2 -> C6H6O6 + 2I- + 2H+}$. Vitamin C is the [reductant](#def-g11-redox-oxidant).

**Exercise 35.10 ★★.**

Aluminium reacts with copper(II) [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion). Write the equation, and explain why the [half-equations](#def-g11-redox-couple) must be multiplied by 2 and by 3.

**Solution of Exercise 35.10.**

$\ce{Al -> Al^{3+} + 3e-}$ gives 3 [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus), $\ce{Cu^{2+} + 2e- -> Cu}$ takes 2; the smallest common number is 6, so the first is multiplied by 2 and the second by 3: $\ce{2Al + 3Cu^{2+} -> 2Al^{3+} + 3Cu}$.

**Exercise 35.11 ★★.**

Hydrogen peroxide oxidises iron(II) [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) to iron(III) [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) in [acidic solution](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-acidic). Write the equation.

**Solution of Exercise 35.11.**

$\ce{H2O2 + 2H+ + 2e- -> 2H2O}$ and $2\times$ ($\ce{Fe^{2+} -> Fe^{3+} + e-}$): $\ce{H2O2 + 2H+ + 2Fe^{2+} -> 2H2O + 2Fe^{3+}}$.

**Exercise 35.12 ★★★.**

A zinc strip loses $1.30\,\mathrm{g}$ in a solution containing an excess of copper(II) [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion). What mass of copper is deposited? Take $M(\ce{Zn}) = 65.4\,\mathrm{g}/\mathrm{mol}$ and $M(\ce{Cu}) = 63.5\,\mathrm{g}/\mathrm{mol}$.

**Solution of Exercise 35.12.**

$n(\ce{Zn}) = 1.30/65.4 = 1.99 \times 10^{-2}\,\mathrm{mol}$. One copper [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) is deposited per zinc [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) oxidised, so $n(\ce{Cu}) = 1.99 \times 10^{-2}\,\mathrm{mol}$ and $m(\ce{Cu}) = 1.99\times 10^{-2} \times 63.5 = 1.26\,\mathrm{g}$.

**Exercise 35.13 ★★★.**

Write the equation of the [oxidation](#def-g11-redox-oxidation) of iron(II) [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) by dichromate [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) in [acidic solution](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-acidic). What amount of $\ce{Fe^{2+}}$ is oxidised by $0.010\,\mathrm{mol}$ of dichromate [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion)?

**Solution of Exercise 35.13.**

$\ce{Cr2O7^{2-} + 14H+ + 6e- -> 2Cr^{3+} + 7H2O}$ and $6\times$ ($\ce{Fe^{2+} -> Fe^{3+} + e-}$): $\ce{Cr2O7^{2-} + 14H+ + 6Fe^{2+} -> 2Cr^{3+} + 6Fe^{3+} + 7H2O}$. Six $\ce{Fe^{2+}}$ per dichromate [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion): $6 \times 0.010 = 0.060\,\mathrm{mol}$.

**Exercise 35.14 ★★★.**

Bleach contains hypochlorite [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) $\ce{ClO-}$ and chloride [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) $\ce{Cl-}$. Using the couples $\ce{ClO-}$/$\ce{Cl2}$ and $\ce{Cl2}$/$\ce{Cl-}$, write the equation of the reaction that takes place when bleach meets an acid, and explain why the labels of bleach and of acidic descaling products both warn never to [mix](https://one-course.com/books/chemistry/1/en/chapter/2-mixing-and-dissolving#def-g2-mixing-and-dissolving-mix) them.

**Solution of Exercise 35.14.**

$\ce{2ClO- + 4H+ + 2e- -> Cl2 + 2H2O}$ and $\ce{2Cl- -> Cl2 + 2e-}$; adding and dividing by 2: $\ce{ClO- + Cl- + 2H+ -> Cl2 + H2O}$. An acid supplies the $\ce{H+}$ [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion), and the reaction releases dichlorine, a toxic gas: hence the warning never to [mix](https://one-course.com/books/chemistry/1/en/chapter/2-mixing-and-dissolving#def-g2-mixing-and-dissolving-mix) the two products.

**Exercise 35.15 ★★★.**

In the figure of the zinc surface, the [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) travel through the [metal](https://one-course.com/books/chemistry/1/en/chapter/20-the-periodic-table-a-first-look#def-g9-periodic-table-first-look-metal), not through the solution. Using [Proposition 35.8](#prop-g11-redox-no-free-electrons), explain why the copper is deposited on the zinc strip itself and not elsewhere in the beaker.

**Solution of Exercise 35.15.**

[Electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) cannot travel freely through the solution ([Proposition 35.8](#prop-g11-redox-no-free-electrons)): those lost by zinc [atoms](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) can only be taken by a $\ce{Cu^{2+}}$ [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) that touches the [metal](https://one-course.com/books/chemistry/1/en/chapter/20-the-periodic-table-a-first-look#def-g9-periodic-table-first-look-metal). The copper [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) is therefore formed at the surface of the strip, and stays there.

## 35.6 Problem: The Breath Test

**Problem 35.1.**

Weekend problem — the chemistry of a breath-alcohol tube: two couples, an orange-to-green colour change, the mass of ethanol a tube can detect, and the fuel cell that replaced it

The first breath-alcohol tests used a glass tube filled with grains of silica coated with orange potassium dichromate, $\ce{K2Cr2O7}$, in acid. Ethanol, $\ce{CH3CH2OH}$, in the breath blown through the tube reduces the dichromate [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) to green $\ce{Cr^{3+}}$ [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) and is oxidised to ethanoic acid, $\ce{CH3COOH}$; the length of the green zone measures the ethanol. Suppose that a tube contains $2.0\,\mathrm{mg}$ of potassium dichromate. Use $M(\ce{K}) = 39.1$, $M(\ce{Cr}) = 52.0$, $M(\ce{O}) = 16.0$, $M(\ce{C}) = 12.0$, $M(\ce{H}) = 1.0\,\mathrm{g}/\mathrm{mol}$, $N_A =
6.02 \times 10^{23}\,\mathrm{mol}^{-1}$ and $e = 1.60 \times 10^{-19}\,\mathrm{C}$.

**Part I — Two couples.**

1. Name the two [redox couples](#def-g11-redox-couple) involved, [oxidant](#def-g11-redox-oxidant) first.
2. Is ethanol oxidised or reduced? Is it the [oxidant](#def-g11-redox-oxidant) or the [reductant](#def-g11-redox-oxidant) ?
3. Write the [half-equation](#def-g11-redox-couple) of the couple $\ce{Cr2O7^{2-}}$ / $\ce{Cr^{3+}}$ in [acidic solution](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-acidic) .
4. Write the [half-equation](#def-g11-redox-couple) of the couple $\ce{CH3COOH}$ / $\ce{CH3CH2OH}$ in [acidic solution](https://one-course.com/books/chemistry/1/en/chapter/18-acids-bases-and-ph#def-g9-acids-bases-ph-acidic) .
5. Deduce the equation of the reaction in the tube.

**Part II — What one tube can measure.**

6. Explain the colour change of the grains.
7. Compute the [molar mass](https://one-course.com/books/chemistry/1/en/chapter/25-the-mole-and-molar-mass#def-g10-the-mole-molar-mass) of potassium dichromate.
8. Compute the amount of dichromate [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) in the tube.
9. What amount of ethanol is needed to turn the whole tube green?
10. Deduce the largest mass of ethanol one tube can measure.

**Part III — From breath to blood.** A person blows $1.0\,\mathrm{L}$ of breath through the tube; the green zone grows in proportion to the mass of ethanol that reaches it.

11. The breath contains $0.25\,\mathrm{mg}$ of ethanol per litre. What fraction of the dichromate is used?
12. The tube is $10\,\mathrm{cm}$ long. How long is the green zone?
13. Breath analysers assume that $2100\,\mathrm{L}$ of breath carry as much ethanol as $1\,\mathrm{L}$ of blood. Estimate the mass of ethanol per litre of blood, in grams.
14. Why does the green zone start at the inlet of the tube and grow towards the outlet, instead of the whole tube turning slightly green?
15. Why can a tube be used only once?

**Part IV — The [fuel](https://one-course.com/books/chemistry/1/en/chapter/5-burning-what-a-fire-needs#def-g4-what-a-fire-needs-fuel) cell.** Potassium dichromate carries the pictograms ![](https://one-course.com/images/onecourse/chapters/chemistry-1/g11-redox/fig-b92e5d8c18fd.svg) ![](https://one-course.com/images/onecourse/chapters/chemistry-1/g11-redox/fig-16d319f07c40.svg) ![](https://one-course.com/images/onecourse/chapters/chemistry-1/g11-redox/fig-8e4c7d47ff36.svg) ![](https://one-course.com/images/onecourse/chapters/chemistry-1/g11-redox/fig-c662a9a86cea.svg) ![](https://one-course.com/images/onecourse/chapters/chemistry-1/g11-redox/fig-62fdf82d203a.svg) ![](https://one-course.com/images/onecourse/chapters/chemistry-1/g11-redox/fig-831fae79d6bd.svg). Modern devices use instead a [fuel](https://one-course.com/books/chemistry/1/en/chapter/5-burning-what-a-fire-needs#def-g4-what-a-fire-needs-fuel) cell, in which the ethanol of the breath is oxidised by dioxygen on an electrode, and the [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) released flow through a wire.

16. Give two reasons, read from the pictograms, why dichromate tubes were abandoned.
17. Write the equation of the [oxidation](#def-g11-redox-oxidation) of ethanol by dioxygen, using the couple $\ce{O2}$ / $\ce{H2O}$ .
18. How many [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) does one ethanol [molecule](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) give up when it is oxidised to ethanoic acid?
19. For $0.25\,\mathrm{mg}$ of ethanol, compute the amount of ethanol, then the number of [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) released.
20. Deduce the electric charge that flows through the [fuel](https://one-course.com/books/chemistry/1/en/chapter/5-burning-what-a-fire-needs#def-g4-what-a-fire-needs-fuel) cell. Why is this charge a good measure of the ethanol in the breath?

**Solution of Problem 35.1.**

**1.** $\ce{Cr2O7^{2-}}$/$\ce{Cr^{3+}}$ and $\ce{CH3COOH}$/$\ce{CH3CH2OH}$.

**2.** Ethanol is oxidised: it is the [reductant](#def-g11-redox-oxidant).

**3.** $\ce{Cr2O7^{2-} + 14H+ + 6e- <=> 2Cr^{3+} + 7H2O}$.

**4.** Carbon is balanced (two on each side). Oxygen: two on the left, one on the right, so $\ce{H2O}$ on the right. Hydrogen: four on the left, $6 + 2 = 8$ on the right, so $\ce{4H+}$ on the left. Charge: $+4$ on the left, 0 on the right, so four [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) on the left: $\ce{CH3COOH + 4H+ + 4e- <=> CH3CH2OH + H2O}$.

**5.** Twelve [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus): $2\times$ the first, $3\times$ the second reversed: $\ce{2Cr2O7^{2-} + 3CH3CH2OH + 16H+ -> 4Cr^{3+} + 3CH3COOH + 11H2O}$.

**6.** Orange $\ce{Cr2O7^{2-}}$ [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) are reduced to green $\ce{Cr^{3+}}$ [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) wherever ethanol reaches them.

**7.** $M = 2\times 39.1 + 2\times 52.0 + 7\times 16.0 =
294.2\,\mathrm{g}/\mathrm{mol}$.

**8.** $n = 2.0\times 10^{-3}/294.2 = 6.8 \times 10^{-6}\,\mathrm{mol}$.

**9.** Three ethanol [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) per two dichromate [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion): $n = \tfrac{3}{2}\times 6.8\times 10^{-6} = 1.0 \times 10^{-5}\,\mathrm{mol}$.

**10.** $M(\ce{C2H6O}) = 46.0\,\mathrm{g}/\mathrm{mol}$, so $m = 1.02\times 10^{-5}\times 46.0 = 4.7 \times 10^{-4}\,\mathrm{g}$, about $0.47\,\mathrm{mg}$.

**11.** $0.25/0.47 \approx 0.53$: about $53\,\%$ of the dichromate.

**12.** $0.53 \times 10 \approx 5.3\,\mathrm{cm}$.

**13.** $0.25\,\mathrm{mg} \times 2100 = 525\,\mathrm{mg}$ per litre of blood, about $0.53\,\mathrm{g}/\mathrm{L}$.

**14.** The breath meets the grains at the inlet first; the dichromate there is in excess and removes all the ethanol, so the reaction is complete in a first stretch of the tube, which turns fully green, before any ethanol reaches further.

**15.** The reaction is total: the green $\ce{Cr^{3+}}$ [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) do not go back to dichromate, so a used tube cannot measure again.

**16.** It is acutely toxic (GHS06) and a serious health hazard (GHS08: its hazard statements include “may cause cancer”), as well as corrosive and toxic to aquatic life; a device used by the public must not contain it.

**17.** $\ce{O2 + 4H+ + 4e- -> 2H2O}$ and $\ce{CH3CH2OH + H2O -> CH3COOH + 4H+ + 4e-}$: $\ce{CH3CH2OH + O2 -> CH3COOH + H2O}$.

**18.** Four [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus).

**19.** $n = 0.25\times 10^{-3}/46.0 = 5.4 \times 10^{-6}\,\mathrm{mol}$ of ethanol; [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) $4n = 2.2 \times 10^{-5}\,\mathrm{mol}$, that is $2.2\times 10^{-5}\times 6.02\times 10^{23} = 1.3 \times 10^{19}$ [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus).

**20.** $Q = 1.3\times 10^{19}\times 1.60\times 10^{-19} \approx
2.1\,\mathrm{C}$. Each ethanol [molecule](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) sends exactly four [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) through the wire, so the charge is proportional to the amount of ethanol: measuring it measures the ethanol.