---
title: "Electrochemical Cells and Electrolysis"
book: "School Chemistry — Grades 1 to 12"
subject: chemistry
language: en
chapter: 47
exercises: 15
source: https://one-course.com/books/chemistry/1/en/chapter/47-electrochemical-cells-and-electrolysis
license: CC-BY-NC-SA-4.0
credit: "One Chemistry Book, One Course (one-course.com)"
---

# Chapter 47 — Electrochemical Cells and Electrolysis

A phone shows 1 % on a winter evening and dies. Plugged in overnight, it is full again by morning. Inside its battery the same [chemical reaction](https://one-course.com/books/chemistry/1/en/chapter/12-chemical-reactions-reactants-and-products#def-g7-chemical-reactions-reaction) runs one way during the day, giving out electrical energy, and is driven the other way at night by the charger. A [redox reaction](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-reaction) becomes a source of electricity when its [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) are made to travel through a wire; and electricity, pushed through a solution, can force a [redox reaction](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-reaction) that would never happen by itself.

**You already know.**

[Oxidants](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-oxidant) gain [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus), [reductants](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-oxidant) lose them; [half-equations](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-couple), [redox couples](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-couple) and [redox reactions](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-reaction) ([Chapter 35](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#ch-g11-redox)). A system evolves until $Q = K$ ([Chapter 43](https://one-course.com/books/chemistry/1/en/chapter/43-chemical-equilibrium-quotient-and-constant#ch-g12-equilibrium)). Solutions of [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) conduct ([Chapter 17](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#ch-g9-ions)); the [mole](https://one-course.com/books/chemistry/1/en/chapter/25-the-mole-and-molar-mass#def-g10-the-mole-amount) and the [Avogadro constant](https://one-course.com/books/chemistry/1/en/chapter/25-the-mole-and-molar-mass#def-g10-the-mole-avogadro) ([Chapter 25](https://one-course.com/books/chemistry/1/en/chapter/25-the-mole-and-molar-mass#ch-g10-the-mole)).

![Charging a phone: a reaction driven backwards overnight.](https://one-course.com/images/onecourse/chapters/chemistry-1/g12-cells-and-electrolysis/img-49517e36d302.jpg)

*Charging a phone: a reaction driven backwards overnight.*

## 47.1 A redox reaction split in two

When a zinc strip is dipped in copper sulfate solution, zinc is oxidised and copper [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) are reduced at the surface of the strip, $\ce{Zn + Cu^{2+} -> Zn^{2+} + Cu}$: the [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) pass directly from [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 [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion), and the energy is lost as a little heat. Separate the two halves, and join them by a wire: the [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) must now travel through the wire, and can light a lamp on the way.

**Definition 47.1 (Electrochemical cell, half-cell, salt bridge).**

An *electrochemical cell* produces electrical energy from a [redox reaction](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-reaction) whose [oxidation](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-oxidation) and [reduction](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-oxidation) take place in two separate compartments. Each compartment, 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) electrode dipped in a solution of its [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion), is a *half-cell*. A *salt bridge*, a tube or a paper strip soaked in a solution of [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion), joins the two solutions and closes the circuit while keeping them apart.

**Definition 47.2 (Anode, cathode).**

The electrode where an [oxidation](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-oxidation) takes place is the *anode*; the electrode where a [reduction](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-oxidation) takes place is the *cathode*.

![The Daniell cell. Electrons leave the zinc anode through the wire and enter the copper cathode; in the salt bridge, ions move to keep each solution neutral.](https://one-course.com/images/onecourse/chapters/chemistry-1/g12-cells-and-electrolysis/fig-82aed336061d.svg)

*The Daniell cell. [Electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) leave the zinc [anode](#def-g12-cells-and-electrolysis-anode) through the wire and enter the copper [cathode](#def-g12-cells-and-electrolysis-anode); in the [salt bridge](#def-g12-cells-and-electrolysis-cell), [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) move to keep each solution neutral.*

**Proposition 47.3 (Where the electrons go).**

In a cell delivering current, [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) leave the cell by the [anode](#def-g12-cells-and-electrolysis-anode), where they are released by the [oxidation](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-oxidation), travel through the external circuit, and enter by the [cathode](#def-g12-cells-and-electrolysis-anode), where they are taken by the [reduction](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-oxidation). The [anode](#def-g12-cells-and-electrolysis-anode) is the negative terminal, the [cathode](#def-g12-cells-and-electrolysis-anode) the positive one. Inside, the current is carried by [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion): [cations](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) move towards the [cathode](#def-g12-cells-and-electrolysis-anode), [anions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) towards the [anode](#def-g12-cells-and-electrolysis-anode).

**Proof.** The [oxidation](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-oxidation) produces [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) at the [anode](#def-g12-cells-and-electrolysis-anode); the [reduction](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-oxidation) consumes them at the [cathode](#def-g12-cells-and-electrolysis-anode); the wire is the only path between the two. Each solution would otherwise become charged, which the [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) of the bridge prevent. ∎

**Method 47.4 (Describing a cell).**

1. Write the overall [redox reaction](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-reaction) that takes place spontaneously.
2. Split it into two [half-equations](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-couple) : the [oxidation](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-oxidation) ( [anode](#def-g12-cells-and-electrolysis-anode) ) and the [reduction](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-oxidation) ( [cathode](#def-g12-cells-and-electrolysis-anode) ).
3. Mark the polarity: [anode](#def-g12-cells-and-electrolysis-anode) $-$ , [cathode](#def-g12-cells-and-electrolysis-anode) $+$ ; the direction of the [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) in the wire ( [anode](#def-g12-cells-and-electrolysis-anode) to [cathode](#def-g12-cells-and-electrolysis-anode) ) and of the current ( [cathode](#def-g12-cells-and-electrolysis-anode) to [anode](#def-g12-cells-and-electrolysis-anode) ).
4. Describe the motion of the [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) in the bridge.

**History — Volta’s pile, 1800.**

In 1800 the physicist Alessandro Volta stacked discs of two different [metals](https://one-course.com/books/chemistry/1/en/chapter/20-the-periodic-table-a-first-look#def-g9-periodic-table-first-look-metal), zinc and copper or silver, separated by pieces of cloth soaked in brine, and obtained a steady electric current: the first battery. It gave physicists their first continuous source of electricity, and chemists a new tool: within a few years [electrolysis](#def-g12-cells-and-electrolysis-electrolysis) had isolated sodium and potassium for the first time.

![A voltaic pile in a museum. Photo GuidoB, CC BY-SA 3.0.](https://one-course.com/images/onecourse/chapters/chemistry-1/g12-cells-and-electrolysis/img-90c8f7e514de.jpg)

*A voltaic pile in a museum. Photo GuidoB, CC BY-SA 3.0.*

## 47.2 The voltage of a cell

**Definition 47.5 (Cell voltage).**

The *cell voltage* is the voltage measured between the terminals of a cell delivering no current (open circuit), with a voltmeter; it is positive when the voltmeter’s positive terminal is on the [cathode](#def-g12-cells-and-electrolysis-anode).

**Example 47.6 (The Daniell cell).**

With both solutions at $1\,\mathrm{mol}/\mathrm{L}$, the Daniell cell gives $1.10\,\mathrm{V}$. How such a voltage is predicted from tables of electrode potentials is shown in the Year 1 volume.

**Remark 47.7 (Why a cell runs down).**

The reaction of the cell, $\ce{Zn + Cu^{2+} -> Zn^{2+} + Cu}$, has a very large constant $K$. While the cell works, $Q = [\ce{Zn^{2+}}]/[\ce{Cu^{2+}}]$ grows, and the reaction keeps running as long as $Q < K$. When a [reactant](https://one-course.com/books/chemistry/1/en/chapter/12-chemical-reactions-reactants-and-products#def-g7-chemical-reactions-reactant) runs out (the zinc, or the copper [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion)), the current stops: the cell is “flat”.

## 47.3 Charge and capacity

**Definition 47.8 (Faraday constant, capacity).**

The *Faraday constant* $F$ is the electric charge of one [mole](https://one-course.com/books/chemistry/1/en/chapter/25-the-mole-and-molar-mass#def-g10-the-mole-amount) of [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus), $F = N_A e = 96\,485\,\mathrm{C}/\mathrm{mol}$. The *capacity* of a cell is the largest electric charge it can deliver, often given in ampere-hours: $1\,\mathrm{A}\,\mathrm{h} = 3600\,\mathrm{C}$.

**Proposition 47.9 (Charge and amount of electrons).**

A current $I$ flowing for a time $\Delta t$ carries the charge $Q = I\,\Delta t$, that is an amount of [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus)

$$
n(e^-) = \frac{Q}{F} = \frac{I\,\Delta t}{F} .
$$

**Proof.** Each [electron](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) carries the elementary charge $e$; $n$ [moles](https://one-course.com/books/chemistry/1/en/chapter/25-the-mole-and-molar-mass#def-g10-the-mole-amount) of [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) carry $n N_A e = n F$. ∎

**Example 47.10 (Capacity of a Daniell cell).**

The zinc electrode holds $6.5\,\mathrm{g}$ of zinc that can react, that is $6.5/65.4 = 0.099\,\mathrm{mol}$. Each zinc [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) gives two [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus): $n(e^-) = 0.199\,\mathrm{mol}$, so $Q = 0.199 \times 96485 = 1.9 \times 10^{4}\,\mathrm{C}$, about $1.9 \times 10^{4}\,/3600 = 5.3\,\mathrm{A}\,\mathrm{h}$, if the copper [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) do not run out first.

## 47.4 Electrolysis

**Definition 47.11 (Electrolysis, accumulator).**

An *electrolysis* is a [redox reaction](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-reaction) forced by an electrical generator, in the direction opposite to the one it would take by itself. An *accumulator* is a cell that can be recharged: during charging, the generator drives its reaction backwards, as an electrolysis.

**In the lab — Electrolysis of water.**

Two electrodes dip in water made conducting with a little sodium sulfate, each under an inverted test tube full of the solution. With a generator of a few volts, bubbles rise from both electrodes: at the [cathode](#def-g12-cells-and-electrolysis-anode) (connected to the $-$ terminal) dihydrogen, at the [anode](#def-g12-cells-and-electrolysis-anode) dioxygen, twice as much of the first as of the second. A lighted splint pops in the first gas; a glowing splint relights in the second.

![Electrolysis of water: twice as much dihydrogen as dioxygen collects above the electrodes.](https://one-course.com/images/onecourse/chapters/chemistry-1/g12-cells-and-electrolysis/fig-a5a843c0072b.svg)

*[Electrolysis](#def-g12-cells-and-electrolysis-electrolysis) of water: twice as much dihydrogen as dioxygen collects above the electrodes.*

**Example 47.12 (Water, split).**

At the [cathode](#def-g12-cells-and-electrolysis-anode): $\ce{2H2O + 2e- -> H2 + 2OH-}$; at the [anode](#def-g12-cells-and-electrolysis-anode): $\ce{6H2O -> O2 + 4H3O+ + 4e-}$. For four [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus), two [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) of dihydrogen and one of dioxygen, while the [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) $\ce{H3O+}$ and $\ce{OH-}$ formed at the two electrodes recombine into water: overall $\ce{2H2O -> 2H2 + O2}$, the reverse of the [combustion](https://one-course.com/books/chemistry/1/en/chapter/5-burning-what-a-fire-needs#def-g4-what-a-fire-needs-burning) of hydrogen, which needs energy to happen.

**Method 47.13 (Mass deposited by an electrolysis).**

1. Write the [half-equation](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-couple) at the electrode where 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) forms, for example $\ce{Cu^{2+} + 2e- -> Cu}$ .
2. Compute the charge $Q = I \Delta t$ and $n(e^-) = Q/F$ .
3. Divide by the number of [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) per [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) : $n(\ce{Cu}) =  n(e^-)/2$ ; then $m = n \times M$ .

**Example 47.14 (Refining copper).**

Impure copper from the smelter is made the [anode](#def-g12-cells-and-electrolysis-anode) of an [electrolysis](#def-g12-cells-and-electrolysis-electrolysis) cell in copper sulfate solution, and a thin sheet of pure copper the [cathode](#def-g12-cells-and-electrolysis-anode). At the [anode](#def-g12-cells-and-electrolysis-anode) copper [dissolves](https://one-course.com/books/chemistry/1/en/chapter/2-mixing-and-dissolving#def-g2-mixing-and-dissolving-dissolve), $\ce{Cu -> Cu^{2+} + 2e-}$; at the [cathode](#def-g12-cells-and-electrolysis-anode) pure copper is deposited, $\ce{Cu^{2+} + 2e- -> Cu}$; the impurities fall to the bottom or stay in solution. A current of $100\,\mathrm{A}$ for $24\,\mathrm{h}$ carries $Q = 100 \times 86400 = 8.64 \times 10^{6}\,\mathrm{C}$, that is $n(e^-) = 89.5\,\mathrm{mol}$, which deposits $89.5/2 = 44.8\,\mathrm{mol}$ of copper: $2.84\,\mathrm{kg}$.

![Copper refining by electrolysis: copper passes from the impure anode to the pure cathode through the solution.](https://one-course.com/images/onecourse/chapters/chemistry-1/g12-cells-and-electrolysis/fig-997828af9042.svg)

*Copper refining by [electrolysis](#def-g12-cells-and-electrolysis-electrolysis): copper passes from the impure [anode](#def-g12-cells-and-electrolysis-anode) to the pure [cathode](#def-g12-cells-and-electrolysis-anode) through the solution.*

## 47.5 Batteries, accumulators and fuel cells

Every battery is a cell, every rechargeable battery an [accumulator](#def-g12-cells-and-electrolysis-electrolysis). A car’s lead–acid [accumulator](#def-g12-cells-and-electrolysis-electrolysis) works with lead, lead oxide and sulfuric acid; a phone’s lithium-ion [accumulator](#def-g12-cells-and-electrolysis-electrolysis) moves lithium [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) between two electrodes, one of graphite and one of 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) oxide, through a liquid or polymer electrolyte, while the [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) go round the outside circuit. 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 is a cell whose [reactants](https://one-course.com/books/chemistry/1/en/chapter/12-chemical-reactions-reactants-and-products#def-g7-chemical-reactions-reactant) are fed in continuously: a hydrogen [fuel](https://one-course.com/books/chemistry/1/en/chapter/5-burning-what-a-fire-needs#def-g4-what-a-fire-needs-fuel) cell runs the reaction $\ce{2H2 + O2 -> 2H2O}$, the reverse of the [electrolysis](#def-g12-cells-and-electrolysis-electrolysis) of water, and gives electricity and water. Dihydrogen made by [electrolysis](#def-g12-cells-and-electrolysis-electrolysis) with electricity from wind or sun, then used in [fuel](https://one-course.com/books/chemistry/1/en/chapter/5-burning-what-a-fire-needs#def-g4-what-a-fire-needs-fuel) cells, is one way of storing that electricity.

![An electrolyser hall: water is split into dihydrogen and dioxygen.](https://one-course.com/images/onecourse/chapters/chemistry-1/g12-cells-and-electrolysis/img-2386bb811d27.jpg)

*An electrolyser hall: water is split into dihydrogen and dioxygen.*

## 47.6 Exercises

**Exercise 47.1 ★.**

In the Daniell cell, which electrode is the [anode](#def-g12-cells-and-electrolysis-anode)? The [cathode](#def-g12-cells-and-electrolysis-anode)? Write the [half-equation](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-couple) at each.

**Solution of Exercise 47.1.**

The zinc electrode is the [anode](#def-g12-cells-and-electrolysis-anode), $\ce{Zn -> Zn^{2+} + 2e-}$; the copper electrode is the [cathode](#def-g12-cells-and-electrolysis-anode), $\ce{Cu^{2+} + 2e- -> Cu}$.

**Exercise 47.2 ★.**

A current of $0.20\,\mathrm{A}$ flows for $30\,\mathrm{min}$. Compute the charge carried and the amount of [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus).

**Solution of Exercise 47.2.**

$Q = 0.20 \times 1800 = 360\,\mathrm{C}$; $n(e^-) = 360/96485 =
3.7 \times 10^{-3}\,\mathrm{mol}$.

**Exercise 47.3 ★.**

What is the role of the [salt bridge](#def-g12-cells-and-electrolysis-cell)? What happens if it is removed?

**Solution of Exercise 47.3.**

It closes the circuit and keeps each solution electrically neutral, its [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) moving into the compartments. Without it the circuit is open: no current flows, the voltmeter reads zero and a lamp stays dark.

**Exercise 47.4 ★.**

Convert a [capacity](#def-g12-cells-and-electrolysis-capacity) of $2.5\,\mathrm{A}\,\mathrm{h}$ into coulombs.

**Solution of Exercise 47.4.**

$2.5 \times 3600 = 9000\,\mathrm{C}$.

**Exercise 47.5 ★.**

In an [electrolysis](#def-g12-cells-and-electrolysis-electrolysis), is the reaction spontaneous? What supplies the energy?

**Solution of Exercise 47.5.**

No: the reaction goes in the direction opposite to the spontaneous one. The generator supplies the energy.

**Exercise 47.6 ★★.**

A cell is made of a silver electrode in silver nitrate and a copper electrode in copper sulfate. The voltmeter shows that [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) leave by the copper. Write the [half-equations](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-couple), name the [anode](#def-g12-cells-and-electrolysis-anode) and the [cathode](#def-g12-cells-and-electrolysis-anode), and give the overall reaction.

**Solution of Exercise 47.6.**

[Anode](#def-g12-cells-and-electrolysis-anode) (copper, $-$): $\ce{Cu -> Cu^{2+} + 2e-}$. [Cathode](#def-g12-cells-and-electrolysis-anode) (silver, $+$): $\ce{Ag+ + e- -> Ag}$. Overall: $\ce{Cu + 2Ag+ -> Cu^{2+} + 2Ag}$.

**Exercise 47.7 ★★.**

A watch battery delivers $10\,\text{µ}\mathrm{A}$ continuously for three years. What is its [capacity](#def-g12-cells-and-electrolysis-capacity) in $\mathrm{mA}\,\mathrm{h}$?

**Solution of Exercise 47.7.**

Three years are $3 \times 365 \times 24 = 26\,280\,\mathrm{h}$; $0.010 \times 26280 = 263\,\mathrm{mA}\,\mathrm{h}$.

**Exercise 47.8 ★★.**

A spoon is silver-plated with a current of $0.50\,\mathrm{A}$ for $40\,\mathrm{min}$, $\ce{Ag+ + e- -> Ag}$. What mass of silver is deposited?

**Solution of Exercise 47.8.**

$Q = 0.50 \times 2400 = 1200\,\mathrm{C}$; $n(e^-) = n(\ce{Ag}) =
1200/96485 = 0.0124\,\mathrm{mol}$; $m = 0.0124 \times 107.9 =
1.34\,\mathrm{g}$.

**Exercise 47.9 ★★.**

On the figure of the Daniell cell, give the direction of the current in the wire, and the direction in which the $\ce{K+}$ and $\ce{NO3-}$ [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) of the bridge move.

**Solution of Exercise 47.9.**

The current goes in the wire from the copper ($+$) to the zinc ($-$), opposite to the [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus). $\ce{K+}$ [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) move towards the copper compartment, which loses $\ce{Cu^{2+}}$; $\ce{NO3-}$ [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) towards the zinc compartment, which gains $\ce{Zn^{2+}}$.

**Exercise 47.10 ★★.**

Compare a cell and an [electrolysis](#def-g12-cells-and-electrolysis-electrolysis): the sign of the [anode](#def-g12-cells-and-electrolysis-anode), the direction of the reaction, the conversion of energy.

**Solution of Exercise 47.10.**

In both the [anode](#def-g12-cells-and-electrolysis-anode) is where the [oxidation](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-oxidation) happens. In a cell the [anode](#def-g12-cells-and-electrolysis-anode) is $-$, the reaction is spontaneous and chemical energy becomes electrical energy; in an [electrolysis](#def-g12-cells-and-electrolysis-electrolysis) the [anode](#def-g12-cells-and-electrolysis-anode) is joined to the $+$ of the generator, the reaction is forced and electrical energy becomes chemical energy.

**Exercise 47.11 ★★.**

In the [electrolysis](#def-g12-cells-and-electrolysis-electrolysis) of water, $24\,\mathrm{mL}$ of dihydrogen are collected. What volume of dioxygen is collected at the same time? Why?

**Solution of Exercise 47.11.**

$12\,\mathrm{mL}$: the equation $\ce{2H2O -> 2H2 + O2}$ gives two [molecules](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-molecule) of dihydrogen for one of dioxygen, and equal amounts of gas take equal volumes.

**Exercise 47.12 ★★★.**

A Daniell cell contains $5.0\,\mathrm{g}$ of zinc and $100\,\mathrm{mL}$ of copper sulfate at $0.50\,\mathrm{mol}/\mathrm{L}$. Which [reactant](https://one-course.com/books/chemistry/1/en/chapter/12-chemical-reactions-reactants-and-products#def-g7-chemical-reactions-reactant) runs out first? What is the [capacity](#def-g12-cells-and-electrolysis-capacity) of the cell? How long can it deliver $50\,\mathrm{mA}$?

**Solution of Exercise 47.12.**

$n(\ce{Zn}) = 5.0/65.4 = 0.076\,\mathrm{mol}$; $n(\ce{Cu^{2+}}) = 0.100 \times 0.50 = 0.050\,\mathrm{mol}$; they react one to one, so the copper [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) run out first. $n(e^-) = 2 \times 0.050
= 0.100\,\mathrm{mol}$, $Q = 9.65 \times 10^{3}\,\mathrm{C} = 2.68\,\mathrm{A}\,\mathrm{h}$. At $50\,\mathrm{mA}$: $2.68/0.050 = 54\,\mathrm{h}$.

**Exercise 47.13 ★★★.**

A battery of $1.5\,\mathrm{V}$ delivers $2.0\,\mathrm{A}\,\mathrm{h}$. Compute the energy it stores ($E = Q U$), in joules and in watt-hours.

**Solution of Exercise 47.13.**

$Q = 2.0 \times 3600 = 7200\,\mathrm{C}$; $E = 7200 \times 1.5 =
1.08 \times 10^{4}\,\mathrm{J}$, that is $1.5 \times 2.0 = 3.0\,\mathrm{W}\,\mathrm{h}$.

**Exercise 47.14 ★★★.**

Water is electrolysed with $2.0\,\mathrm{A}$ for $1.0\,\mathrm{h}$. Compute the amounts, then the volumes at $20\,{}^{\circ}\mathrm{C}$ ($V_m = 24.1\,\mathrm{L}/\mathrm{mol}$), of dihydrogen and dioxygen produced.

**Solution of Exercise 47.14.**

$Q = 7200\,\mathrm{C}$, $n(e^-) = 0.0746\,\mathrm{mol}$. Two [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) per $\ce{H2}$: $n(\ce{H2}) = 0.0373\,\mathrm{mol}$, $V = 0.90\,\mathrm{L}$. Four per $\ce{O2}$: $n(\ce{O2}) = 0.0187\,\mathrm{mol}$, $V = 0.45\,\mathrm{L}$.

**Exercise 47.15 ★★★.**

In copper refining, the [anode](#def-g12-cells-and-electrolysis-anode) loses $3.00\,\mathrm{kg}$ of material while the [cathode](#def-g12-cells-and-electrolysis-anode) gains $2.84\,\mathrm{kg}$ of copper. What is the percentage of copper in the impure [anode](#def-g12-cells-and-electrolysis-anode), assuming that only copper is oxidised at the [anode](#def-g12-cells-and-electrolysis-anode) and the impurities simply fall off?

**Solution of Exercise 47.15.**

The copper dissolved at the [anode](#def-g12-cells-and-electrolysis-anode) equals the copper deposited, the same [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) passing at both: $2.84/3.00 = 94.7\,\%$ copper.

## 47.7 Problem: The Phone Battery

**Problem 47.1.**

Weekend problem — what mass of lithium shuttles between the electrodes of a $4000\,\mathrm{mA}\,\mathrm{h}$ phone battery?

A phone battery is labelled “$3.7\,\mathrm{V}$, $4000\,\mathrm{mA}\,\mathrm{h}$”. In a simplified picture, during discharge each lithium [atom](https://one-course.com/books/chemistry/1/en/chapter/11-atoms-and-molecules#def-g7-atoms-and-molecules-atom) stored in the graphite electrode gives one [electron](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) and leaves as a lithium [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion), $\ce{Li -> Li+ + e-}$; the [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) crosses the electrolyte and is taken up by 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) oxide electrode, which gains an [electron](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) at the same time.

**Part I — The cell.**

1. During discharge, is the graphite electrode the [anode](#def-g12-cells-and-electrolysis-anode) or the [cathode](#def-g12-cells-and-electrolysis-anode) ? Why?
2. Which is the negative terminal?
3. In which direction do the [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) travel through the phone? And the lithium [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) inside the battery?
4. Why is this cell an [accumulator](#def-g12-cells-and-electrolysis-electrolysis) ?

**Part II — The charge.**

5. Convert $4000\,\mathrm{mA}\,\mathrm{h}$ into ampere-hours, then into coulombs.
6. Compute the amount of [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) this charge represents.
7. How many [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) is that?
8. The phone draws on average $0.25\,\mathrm{A}$ . How long does a full battery last?
9. The screen shows 20 %. What charge, in coulombs, is left?

**Part III — The energy.**

10. Compute the energy stored, $E = Q U$ , in joules.
11. Convert it into watt-hours.
12. Compare with the energy released by [burning](https://one-course.com/books/chemistry/1/en/chapter/5-burning-what-a-fire-needs#def-g4-what-a-fire-needs-burning) $1\,\mathrm{g}$ of methane ( $55.7\,\mathrm{kJ}$ ).
13. How many full charges does $1\,\mathrm{kW}\,\mathrm{h}$ of electricity give, if no energy were lost?

**Part IV — Recharging.**

14. During charging, which reaction takes place at the graphite electrode? Is it an [oxidation](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-oxidation) or a [reduction](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-oxidation) ?
15. Why is recharging an [electrolysis](#def-g12-cells-and-electrolysis-electrolysis) ?
16. The charger delivers $2.0\,\mathrm{A}$ . How long does a full recharge take at least?
17. What amount of lithium [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) crosses the electrolyte during one full charge?
18. Compute their mass.
19. State the final answer: what mass of lithium shuttles between the electrodes in one full charge?

**Solution of Problem 47.1.**

**1.** The [anode](#def-g12-cells-and-electrolysis-anode): lithium is oxidised there.

**2.** The graphite electrode, the [anode](#def-g12-cells-and-electrolysis-anode).

**3.** [Electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus) go through the phone from the graphite electrode to the oxide electrode; lithium [ions](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) cross the electrolyte in the same direction, from graphite to oxide.

**4.** Its reaction can be driven backwards by a charger.

**5.** $4.000\,\mathrm{A}\,\mathrm{h}$, that is $4.000 \times 3600 =
14\,400\,\mathrm{C}$.

**6.** $14400/96485 = 0.1492\,\mathrm{mol}$.

**7.** $0.1492 \times 6.022 \times 10^{23} = 8.99 \times 10^{22}$ [electrons](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus).

**8.** $4.000/0.25 = 16\,\mathrm{h}$.

**9.** $0.20 \times 14400 = 2880\,\mathrm{C}$.

**10.** $E = 14400 \times 3.7 = 5.3 \times 10^{4}\,\mathrm{J}$, about $53\,\mathrm{kJ}$.

**11.** $53280/3600 = 14.8\,\mathrm{W}\,\mathrm{h}$.

**12.** [Burning](https://one-course.com/books/chemistry/1/en/chapter/5-burning-what-a-fire-needs#def-g4-what-a-fire-needs-burning) $1\,\mathrm{g}$ of methane releases a little more energy ($55.7\,\mathrm{kJ}$) than a full phone battery stores.

**13.** $1000/14.8 = 67.6$: 67 full charges.

**14.** $\ce{Li+ + e- -> Li}$: a [reduction](https://one-course.com/books/chemistry/1/en/chapter/35-oxidation-and-reduction#def-g11-redox-oxidation).

**15.** The charger forces the reaction in the direction opposite to the spontaneous one.

**16.** $14400/2.0 = 7200\,\mathrm{s}$, $2.0\,\mathrm{h}$.

**17.** One lithium [ion](https://one-course.com/books/chemistry/1/en/chapter/17-ions-and-ionic-solutions#def-g9-ions-ion) per [electron](https://one-course.com/books/chemistry/1/en/chapter/16-inside-the-atom#def-g9-inside-the-atom-nucleus): $0.1492\,\mathrm{mol}$.

**18.** $0.1492 \times 6.9 = 1.03\,\mathrm{g}$.

**19.** About $1.03\,\mathrm{g}$ of lithium shuttles between the electrodes in one full charge.