Chapter 2 · Electrochemistry

2.1 Galvanic Cells and Electrode Potentials

Keep the two halves of a redox reaction apart, join them with a wire and a salt bridge, and the electrons do useful work on the way. Standard electrode potentials tell you which way they flow and with what voltage: $E^{\circ}_{cell}=E^{\circ}_{R}-E^{\circ}_{L}$.

Drop a strip of zinc into blue copper(II) sulphate solution and, within minutes, the zinc turns reddish-brown with copper while the blue colour slowly fades. The beaker also warms up a little. Electrons have passed from zinc atoms to copper(II) ions, and the energy released has simply become heat. Now put the zinc and the copper(II) ions in separate beakers, join the metals with a wire and the solutions with a salt bridge. The same electrons now travel through the wire, and on the way they can light an LED, turn a motor or charge a phone. That is all a battery is: a spontaneous redox reaction whose two halves are kept apart so that the electron transfer happens through an external circuit.

In Class 11 you met the Daniell cell and the idea of an electrode potential. This section makes those ideas precise and quantitative. You will learn to draw and name any galvanic cell, to read the standard cell notation, to measure every electrode against one agreed reference (the standard hydrogen electrode) and to use a short table of standard electrode potentials to predict the voltage of a cell and the direction of a redox reaction. The next section adds concentration (the Nernst equation) and energy ($\Delta_{r}G^{\circ}=-nFE^{\circ}_{cell}$).

Electrochemical Cells and the Daniell Cell

Electrochemistry is the study of how chemical energy and electrical energy are converted into each other. The conversion happens in a device with two electrodes dipping into one or two electrolyte solutions.

📘 Definition 2.1 — Electrochemical cells: galvanic and electrolytic

An electrochemical cell is a device in which a redox reaction is linked to a flow of electrons through an external circuit. There are two kinds:

  • A galvanic cell (also called a voltaic cell) converts the chemical energy of a spontaneous redox reaction into electrical energy. Its reaction has $\Delta_{r}G\lt0$, and it supplies a current.
  • An electrolytic cell uses electrical energy from an external source to drive a non-spontaneous redox reaction. This is electrolysis, the subject of Section 2.4.

In both kinds, the electrode at which oxidation occurs is called the anode and the electrode at which reduction occurs is called the cathode. (Memory aid: anode and oxidation both begin with a vowel; cathode and reduction both begin with a consonant.)

The classic galvanic cell is the Daniell cell. A zinc rod dips into zinc sulphate solution, and a copper rod dips into copper(II) sulphate solution. Each metal with its own solution is a half-cell, also called an electrode or a redox couple ($\mathrm{Zn^{2+}/Zn}$, $\mathrm{Cu^{2+}/Cu}$). The two rods are joined by a wire through a voltmeter, and the two solutions are joined by a salt bridge: a U-tube filled with a jelly (agar-agar or gelatin) containing an inert electrolyte such as $\mathrm{KCl}$, $\mathrm{KNO_{3}}$ or $\mathrm{NH_{4}NO_{3}}$. The reactions are

$$\text{anode (oxidation):}\quad \mathrm{Zn(s)\rightarrow Zn^{2+}(aq)+2e^{-}}$$ $$\text{cathode (reduction):}\quad \mathrm{Cu^{2+}(aq)+2e^{-}\rightarrow Cu(s)}$$ $$\text{cell reaction:}\quad \mathrm{Zn(s)+Cu^{2+}(aq)\rightarrow Zn^{2+}(aq)+Cu(s)}$$

With both solutions at $1\ \mathrm{mol\,L^{-1}}$ and $298\ \mathrm{K}$ the voltmeter reads $1.10\ \mathrm{V}$, with the copper rod positive.

📘 Definition 2.2 — How a galvanic cell works: electrodes, electron flow and the salt bridge
  • At the anode the metal is oxidised and releases electrons into the rod. Electrons pile up there, so in a galvanic cell the anode is the negative electrode.
  • At the cathode ions from the solution take electrons from the rod and are reduced. The rod is left electron-poor, so in a galvanic cell the cathode is the positive electrode.
  • Electrons flow through the external wire from anode to cathode. Conventional current flows the other way, from cathode ($+$) to anode ($-$) in the wire.
  • The salt bridge (i) completes the circuit by letting ions, not electrons, carry charge between the two solutions, and (ii) keeps each solution electrically neutral. Without it, the anode solution would gain excess positive charge ($\mathrm{Zn^{2+}}$) and the cathode solution excess negative charge ($\mathrm{SO_{4}^{2-}}$ left behind), and the current would stop almost at once. Anions of the bridge move towards the anode compartment and cations towards the cathode compartment. It also keeps the two solutions from mixing.
📐 Rules 2.1 — Cell notation (the IUPAC convention)
  1. Write the anode on the left and the cathode on the right.
  2. A single vertical line $|$ marks a boundary between phases (for example metal | solution, or gas | solution). A double line $\|$ marks a salt bridge.
  3. Inside each half-cell, write the species in the order in which electrons meet them. Give concentrations or pressures in brackets when they matter, e.g. $\mathrm{Zn^{2+}(aq,\ 1\ M)}$, $\mathrm{H_{2}(g,\ 1\ bar)}$.
  4. If a couple has no solid metal (a gas, or two ions such as $\mathrm{Fe^{3+}/Fe^{2+}}$), an inert electrode such as platinum or graphite is written at the outer end. Two ions in the same solution are separated by a comma.
$$\mathrm{Zn(s)\,|\,Zn^{2+}(aq)\,\|\,Cu^{2+}(aq)\,|\,Cu(s)}$$

Reading a cell notation backwards is just as useful: the left electrode is oxidised, the right one is reduced, and adding the two half-reactions (with equal numbers of electrons) gives the cell reaction.

✏️ Example 1 — Following the electrons and ions in a Daniell cell
A Daniell cell with a $\mathrm{KCl}$ salt bridge delivers a small current until $0.0200\ \mathrm{mol}$ of electrons has passed through the wire. (a) Name the anode and the cathode and give their signs. (b) In which direction do electrons move in the wire, and $\mathrm{K^{+}}$ and $\mathrm{Cl^{-}}$ ions in the salt bridge? (c) By how much does the mass of each rod change? ($\mathrm{Zn}=65.4$, $\mathrm{Cu}=63.5\ \mathrm{g\,mol^{-1}}$.)
(a) Electrodes Zinc is oxidised, so the zinc rod is the anode (negative). Copper(II) ions are reduced on the copper rod, so the copper rod is the cathode (positive).
(b) Flow Electrons move through the wire from zinc to copper. In the salt bridge, $\mathrm{Cl^{-}}$ ions move into the zinc compartment (which is gaining $\mathrm{Zn^{2+}}$ ions) and $\mathrm{K^{+}}$ ions move into the copper compartment (which is losing $\mathrm{Cu^{2+}}$ ions). Each solution stays neutral.
(c) Masses Each zinc atom gives $2$ electrons, so $0.0200\ \mathrm{mol}\ e^{-}$ removes $0.0100\ \mathrm{mol}$ of Zn: the zinc rod loses $0.0100\times65.4=0.654\ \mathrm{g}$. The same electrons deposit $0.0100\ \mathrm{mol}$ of Cu: the copper rod gains $0.0100\times63.5=0.635\ \mathrm{g}$.
Check The amounts in moles are equal because both half-reactions involve $2e^{-}$. The masses differ only because the molar masses differ. Electrons are neither made nor destroyed in the wire.
✏️ Example 2 — From reaction to cell notation, and back
(a) Write the cell notation for a galvanic cell based on $\mathrm{Mg(s)+2Ag^{+}(aq)\rightarrow Mg^{2+}(aq)+2Ag(s)}$.
(b) Write the cell notation for a cell in which hydrogen gas at $1\ \mathrm{bar}$ reduces $\mathrm{Cu^{2+}(aq)}$ to copper.
(c) Write the half-reactions and the cell reaction for $\mathrm{Zn(s)\,|\,Zn^{2+}(aq)\,\|\,Fe^{3+}(aq),\,Fe^{2+}(aq)\,|\,Pt(s)}$.
(a) Mg is oxidised (anode, left); $\mathrm{Ag^{+}}$ is reduced (cathode, right): $$\mathrm{Mg(s)\,|\,Mg^{2+}(aq)\,\|\,Ag^{+}(aq)\,|\,Ag(s)}$$ The coefficient $2$ in the equation does not appear in the notation.
(b) $\mathrm{H_{2}}$ is oxidised to $\mathrm{H^{+}}$, so the hydrogen electrode is the anode. It needs an inert platinum electrode, written at the outer end: $$\mathrm{Pt(s)\,|\,H_{2}(g,\ 1\ bar)\,|\,H^{+}(aq)\,\|\,Cu^{2+}(aq)\,|\,Cu(s)}$$
(c) Half-reactions Left (anode): $\mathrm{Zn(s)\rightarrow Zn^{2+}(aq)+2e^{-}}$. Right (cathode, on Pt): $\mathrm{Fe^{3+}(aq)+e^{-}\rightarrow Fe^{2+}(aq)}$.
(c) Cell reaction Multiply the cathode reaction by $2$ to balance electrons, then add: $$\mathrm{Zn(s)+2Fe^{3+}(aq)\rightarrow Zn^{2+}(aq)+2Fe^{2+}(aq)}$$ Check: atoms balance, and the charge is $+6$ on both sides.
Interactive 1 — Galvanic Cell Builder

Pick any two half-cells from the table of standard electrode potentials. The builder puts the couple with the lower $E^{\circ}$ on the left as the anode. It then draws the cell, writes the cell notation and the cell reaction, and finds $E^{\circ}_{cell}=E^{\circ}_{R}-E^{\circ}_{L}$. Press Play to watch electrons run through the wire from anode to cathode while the salt bridge sends cations towards the cathode and anions towards the anode. The value of $\Delta_{r}G^{\circ}=-nFE^{\circ}_{cell}$ (Section 2.2) is shown too.

cell notation—
anode (oxidation)—
cathode (reduction)—
cell reaction—
$E^{\circ}_{cell}$ · $n$—
$\Delta_{r}G^{\circ}$—

Electrode Potential and the Standard Hydrogen Electrode

Why does the Daniell cell give $1.10\ \mathrm{V}$ and not some other value? When a metal rod dips into a solution of its own ions, two opposite processes begin at the surface: metal atoms leave as ions, leaving electrons behind in the rod, and ions from the solution take electrons from the rod and deposit as atoms. Very quickly the two rates become equal and a potential difference builds up between the rod and the solution. Zinc has a strong tendency to lose electrons, so its rod becomes negative relative to its solution. Copper(II) ions have a strong tendency to gain electrons, so the copper rod becomes positive relative to its solution. The cell voltage is the difference between these two tendencies.

📘 Definition 2.3 — Electrode potential and standard electrode potential

The electrode potential of a half-cell is the potential difference that develops between the electrode and its electrolyte. By IUPAC convention it is always quoted as a reduction potential: it measures the tendency of the couple to gain electrons, for the half-reaction written as $$\text{oxidised form}+n\,e^{-}\rightarrow\text{reduced form}.$$

When every species in the half-cell is in its standard state (each solute at unit concentration, $1\ \mathrm{mol\,L^{-1}}$; each gas at $1\ \mathrm{bar}$; pure solids and liquids), the value is the standard electrode potential $E^{\circ}$, for example $E^{\circ}_{\mathrm{Cu^{2+}/Cu}}=+0.34\ \mathrm{V}$. Values are usually given at $298\ \mathrm{K}$.

The potential difference between the two electrodes of a galvanic cell, measured when no current is drawn, is the cell potential or electromotive force (EMF) of the cell, $E_{cell}$. With all species in their standard states it is the standard EMF, $E^{\circ}_{cell}$.

There is a catch. The potential of a single electrode cannot be measured: any voltmeter needs two connections, so you always measure the difference between two electrodes. The way out is to agree on one reference electrode and give it the value zero. Every other electrode is then measured against it.

📘 Definition 2.4 — The standard hydrogen electrode (SHE)

The standard hydrogen electrode is a platinum electrode coated with finely divided platinum (platinum black), dipping into an acid solution in which the concentration of $\mathrm{H^{+}}$ ions is $1\ \mathrm{mol\,L^{-1}}$, with pure hydrogen gas at $1\ \mathrm{bar}$ bubbled over it. The platinum does not react; it only provides a surface and an electrical contact. The couple is

$$\mathrm{2H^{+}(aq)+2e^{-}\rightarrow H_{2}(g)},\qquad E^{\circ}_{\mathrm{H^{+}/H_{2}}}=0.00\ \mathrm{V}\ \text{(by definition, at all temperatures)}.$$

Its notation is $\mathrm{Pt(s)\,|\,H_{2}(g,\ 1\ bar)\,|\,H^{+}(aq,\ 1\ M)}$. To find the standard potential of any other electrode, build a cell with the SHE on the left (as the anode) and the test electrode on the right. Then $E^{\circ}_{cell}=E^{\circ}_{R}-0=E^{\circ}_{R}$.

Two measurements. For $\mathrm{Pt(s)\,|\,H_{2}(g,\ 1\ bar)\,|\,H^{+}(aq,\ 1\ M)\,\|}$ $\mathrm{Cu^{2+}(aq,\ 1\ M)\,|\,Cu(s)}$ the copper electrode is positive and the EMF is $0.34\ \mathrm{V}$, so $E^{\circ}_{\mathrm{Cu^{2+}/Cu}}=+0.34\ \mathrm{V}$: $\mathrm{Cu^{2+}}$ ions are reduced more easily than $\mathrm{H^{+}}$ ions, and hydrogen gas reduces them. For the corresponding cell with zinc, the zinc electrode is negative and the meter reads $0.76\ \mathrm{V}$, so $E^{\circ}_{\mathrm{Zn^{2+}/Zn}}=-0.76\ \mathrm{V}$: here zinc actually reduces $\mathrm{H^{+}}$ ions, and the SHE works as the cathode. A negative sign simply means "a weaker oxidising agent than $\mathrm{H^{+}}$", or equivalently, "the metal is a stronger reducing agent than $\mathrm{H_{2}}$".

📐 Relation 2.2 — Standard EMF of a cell

For a cell written in standard notation (anode on the left, cathode on the right),

$$E^{\circ}_{cell}=E^{\circ}_{R}-E^{\circ}_{L}=E^{\circ}_{\text{cathode}}-E^{\circ}_{\text{anode}},$$

where both values are the standard reduction potentials taken straight from the table. Do not change the sign of the anode value before subtracting, and do not multiply $E^{\circ}$ by the coefficients used to balance electrons: an electrode potential is an intensive property, like temperature, which does not depend on how much substance reacts.

If $E^{\circ}_{cell}\gt0$, the cell reaction as written is spontaneous under standard conditions and the cell works as drawn. If $E^{\circ}_{cell}\lt0$, the reaction goes the other way: the electrode written on the right is really the anode. Daniell cell: $E^{\circ}_{cell}=0.34-(-0.76)=1.10\ \mathrm{V}$.

Measured against the SHE in this way, the standard potentials of hundreds of couples have been tabulated. Here is a short table that you will use throughout this chapter, listed from the strongest oxidising agent at the top to the strongest reducing agent at the bottom.

Reduction half-reaction$E^{\circ}$ / V at 298 K
$\mathrm{Cl_{2}(g)+2e^{-}\rightarrow 2Cl^{-}(aq)}$$+1.36$
$\mathrm{Ag^{+}(aq)+e^{-}\rightarrow Ag(s)}$$+0.80$
$\mathrm{Fe^{3+}(aq)+e^{-}\rightarrow Fe^{2+}(aq)}$$+0.77$
$\mathrm{Cu^{2+}(aq)+2e^{-}\rightarrow Cu(s)}$$+0.34$
$\mathrm{2H^{+}(aq)+2e^{-}\rightarrow H_{2}(g)}$$0.00$
$\mathrm{Fe^{2+}(aq)+2e^{-}\rightarrow Fe(s)}$$-0.44$
$\mathrm{Zn^{2+}(aq)+2e^{-}\rightarrow Zn(s)}$$-0.76$
$\mathrm{Al^{3+}(aq)+3e^{-}\rightarrow Al(s)}$$-1.66$
$\mathrm{Mg^{2+}(aq)+2e^{-}\rightarrow Mg(s)}$$-2.36$
$\mathrm{Na^{+}(aq)+e^{-}\rightarrow Na(s)}$$-2.71$

Further values used in this section: $E^{\circ}_{\mathrm{I_{2}/I^{-}}}=+0.54\ \mathrm{V}$, $E^{\circ}_{\mathrm{Ni^{2+}/Ni}}=-0.25\ \mathrm{V}$, $E^{\circ}_{\mathrm{Sn^{2+}/Sn}}=-0.14\ \mathrm{V}$, $E^{\circ}_{\mathrm{Pb^{2+}/Pb}}=-0.13\ \mathrm{V}$.

✏️ Example 3 — Standard EMFs from the table
Find $E^{\circ}_{cell}$ and say whether each cell works as written under standard conditions: (a) $\mathrm{Mg(s)\,|\,Mg^{2+}(aq)\,\|\,Ag^{+}(aq)\,|\,Ag(s)}$; (b) $\mathrm{Zn(s)\,|\,Zn^{2+}(aq)\,\|\,Fe^{3+}(aq),\,Fe^{2+}(aq)\,|\,Pt(s)}$; (c) $\mathrm{Cu(s)\,|\,Cu^{2+}(aq)\,\|\,Zn^{2+}(aq)\,|\,Zn(s)}$.
(a) $E^{\circ}_{cell}=E^{\circ}_{\mathrm{Ag^{+}/Ag}}-E^{\circ}_{\mathrm{Mg^{2+}/Mg}}=0.80-(-2.36)=3.16\ \mathrm{V}$. Positive, so it works as written. The equation $\mathrm{Mg(s)+2Ag^{+}(aq)\rightarrow Mg^{2+}(aq)+2Ag(s)}$ has a coefficient $2$, but $0.80$ is not doubled.
(b) $E^{\circ}_{cell}=0.77-(-0.76)=1.53\ \mathrm{V}$. Positive: zinc reduces iron(III) ions to iron(II).
(c) $E^{\circ}_{cell}=-0.76-0.34=-1.10\ \mathrm{V}$. Negative, so the cell as written is not spontaneous. The real cell runs the other way: zinc is the anode and the Daniell cell gives $+1.10\ \mathrm{V}$.
Check Reversing a cell changes only the sign of $E^{\circ}_{cell}$, never its size.
✏️ Example 4 — Finding an unknown standard potential
(a) The cell $\mathrm{Cd(s)\,|\,Cd^{2+}(aq,\ 1\ M)\,\|\,Ag^{+}(aq,\ 1\ M)\,|\,Ag(s)}$ has an EMF of $1.20\ \mathrm{V}$ at $298\ \mathrm{K}$, with silver positive. Find $E^{\circ}_{\mathrm{Cd^{2+}/Cd}}$.
(b) A metal M dips into $1\ \mathrm{M}$ $\mathrm{M^{2+}(aq)}$ and is connected to an SHE. The voltmeter reads $0.25\ \mathrm{V}$, and the hydrogen electrode is the positive terminal. Find $E^{\circ}_{\mathrm{M^{2+}/M}}$ and identify M from the values listed under the table.
(a) Set up Silver is positive, so it is the cathode (right). $1.20=E^{\circ}_{\mathrm{Ag^{+}/Ag}}-E^{\circ}_{\mathrm{Cd^{2+}/Cd}}=0.80-E^{\circ}_{\mathrm{Cd^{2+}/Cd}}$.
(a) Answer $E^{\circ}_{\mathrm{Cd^{2+}/Cd}}=0.80-1.20=-0.40\ \mathrm{V}$.
(b) The SHE is positive, so it is the cathode and M is the anode: $0.25=0-E^{\circ}_{\mathrm{M^{2+}/M}}$, giving $E^{\circ}_{\mathrm{M^{2+}/M}}=-0.25\ \mathrm{V}$. The metal is nickel.
Check A metal that is the negative terminal against the SHE must have a negative $E^{\circ}$, and it can release hydrogen from a $1\ \mathrm{M}$ acid. Nickel does dissolve slowly in dilute acids.

Reading the Table of Standard Electrode Potentials

A list of standard reduction potentials in order is called the electrochemical series. It is one of the most useful tables in chemistry, because a single subtraction answers a whole family of questions: Which metal will displace which? Which oxidising agent is stronger? Will a metal dissolve in acid? Can this solution be stored in that container?

📐 Rules 2.3 — What the electrochemical series predicts (standard conditions)
  1. Oxidising agents. The higher (more positive) $E^{\circ}$, the stronger the species on the left of the half-reaction is as an oxidising agent. $\mathrm{F_{2}}$ and $\mathrm{Cl_{2}}$ are near the top.
  2. Reducing agents. The lower (more negative) $E^{\circ}$, the stronger the species on the right is as a reducing agent. Li, Na and Mg are near the bottom.
  3. Direction of reaction. The oxidised form of a couple with higher $E^{\circ}$ reacts with the reduced form of a couple with lower $E^{\circ}$. In a cell made of the two, the higher couple is the cathode and $E^{\circ}_{cell}=E^{\circ}_{\text{higher}}-E^{\circ}_{\text{lower}}\gt0$.
  4. Displacement of metals. A metal displaces from solution the ions of any metal that lies above it in the series.
  5. Hydrogen from acids. Metals with negative $E^{\circ}$ (below hydrogen) can liberate $\mathrm{H_{2}}$ from $1\ \mathrm{M}$ $\mathrm{H^{+}}$; metals with positive $E^{\circ}$ (Cu, Ag, Au) cannot.

These rules tell us what is thermodynamically possible. They do not say how fast it happens. Some favourable reactions are very slow, and changing the concentrations (Section 2.2) can move a borderline result.

✏️ Example 5 — Can this solution be stored in that vessel?
Use the table to decide, giving $E^{\circ}_{cell}$ for the possible reaction: (a) Can copper(II) sulphate solution be stored in an iron vessel? (b) Can silver nitrate solution be stored in a copper vessel? (c) Can zinc sulphate solution be stored in a copper vessel? (d) Which of Mg, Al, Zn, Fe, Cu and Ag liberate hydrogen from $1\ \mathrm{M}$ hydrochloric acid?
(a) Possible reaction $\mathrm{Fe(s)+Cu^{2+}(aq)\rightarrow Fe^{2+}(aq)+Cu(s)}$: $E^{\circ}_{cell}=0.34-(-0.44)=0.78\ \mathrm{V}\gt0$. The vessel would dissolve and copper would deposit, so no.
(b) $\mathrm{Cu(s)+2Ag^{+}(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)}$: $E^{\circ}_{cell}=0.80-0.34=0.46\ \mathrm{V}\gt0$. Copper would be attacked, so no.
(c) $\mathrm{Cu(s)+Zn^{2+}(aq)\rightarrow Cu^{2+}(aq)+Zn(s)}$: $E^{\circ}_{cell}=-0.76-0.34=-1.10\ \mathrm{V}\lt0$. No reaction, so yes, it is safe.
(d) Mg, Al, Zn and Fe have negative $E^{\circ}$ and liberate $\mathrm{H_{2}}$; e.g. $\mathrm{Zn(s)+2H^{+}(aq)\rightarrow Zn^{2+}(aq)+H_{2}(g)}$ has $E^{\circ}_{cell}=0-(-0.76)=0.76\ \mathrm{V}$. Cu ($+0.34$) and Ag ($+0.80$) do not.
✏️ Example 6 — Ranking oxidising and reducing agents
(a) Arrange $\mathrm{Cl_{2}}$, $\mathrm{Ag^{+}}$, $\mathrm{Fe^{3+}}$, $\mathrm{Cu^{2+}}$ and $\mathrm{Zn^{2+}}$ in order of increasing oxidising power. (b) Arrange Mg, Al, Fe, Cu and Ag in order of increasing reducing power. (c) Can $\mathrm{Fe^{3+}}$ ions oxidise iodide ions to iodine? Can they oxidise chloride ions to chlorine? Write the balanced equation for any reaction that occurs.
(a) Oxidising power rises with $E^{\circ}$: $\mathrm{Zn^{2+}}\ (-0.76)\lt\mathrm{Cu^{2+}}\ (+0.34)$ $\lt\mathrm{Fe^{3+}}\ (+0.77)\lt\mathrm{Ag^{+}}\ (+0.80)$ $\lt\mathrm{Cl_{2}}\ (+1.36)$.
(b) Reducing power rises as $E^{\circ}$ falls: $\mathrm{Ag}\lt\mathrm{Cu}\lt\mathrm{Fe}\lt\mathrm{Al}\lt\mathrm{Mg}$.
(c) Iodide $E^{\circ}_{cell}=E^{\circ}_{\mathrm{Fe^{3+}/Fe^{2+}}}-E^{\circ}_{\mathrm{I_{2}/I^{-}}}=0.77-0.54=0.23\ \mathrm{V}\gt0$, so yes: $$\mathrm{2Fe^{3+}(aq)+2I^{-}(aq)\rightarrow 2Fe^{2+}(aq)+I_{2}(s)}$$
(c) Chloride $E^{\circ}_{cell}=0.77-1.36=-0.59\ \mathrm{V}\lt0$, so no. In fact the reverse happens: chlorine oxidises $\mathrm{Fe^{2+}}$ to $\mathrm{Fe^{3+}}$.
⚠️ Common mistakes with electrode potentials

1. Flipping a sign and then subtracting as well. Use reduction potentials for both electrodes in $E^{\circ}_{R}-E^{\circ}_{L}$. If you prefer "reduction potential + oxidation potential", change the anode's sign once and add; never do both.

2. Multiplying $E^{\circ}$ by coefficients. Balancing $\mathrm{2Ag^{+}(aq)+2e^{-}\rightarrow 2Ag(s)}$ leaves $E^{\circ}=+0.80\ \mathrm{V}$. You will see in Section 2.2 that $\Delta_{r}G^{\circ}$ doubles while $E^{\circ}$ does not.

3. Mixing up the signs of the electrodes. In a galvanic cell the anode is negative. In an electrolytic cell (Section 2.4) the anode is connected to the positive terminal of the supply. In both, oxidation happens at the anode.

4. Salt bridge carries electrons. It does not. Electrons flow only through the metal wire; ions carry the charge inside the solutions and the bridge.

5. "Feasible" means "fast". A positive $E^{\circ}_{cell}$ says a reaction can happen, not that it happens quickly.

Interactive 2 — Metal in a Solution: Will It Displace?

Dip a metal strip into a $1\ \mathrm{M}$ solution of another metal's ions (or an acid). The ladder on the right is the electrochemical series. If the metal's couple lies below the ion's couple, $E^{\circ}_{cell}\gt0$ and the reaction goes: press Play to watch the strip get coated (or bubble with hydrogen) while the solution changes. If the metal lies above, nothing happens however long you wait.

possible reaction—
$E^{\circ}_{cell}=E^{\circ}_{\text{ion}}-E^{\circ}_{\text{metal}}$—
progress—

Exercises

I
Galvanic cells and cell notation. ($\mathrm{Zn}=65.4$, $\mathrm{Cu}=63.5\ \mathrm{g\,mol^{-1}}$.)
  • 1 Give two differences between a galvanic cell and an electrolytic cell.
  • 2 For the cell $\mathrm{Mg(s)\,|\,Mg^{2+}(aq)\,\|\,Cu^{2+}(aq)\,|\,Cu(s)}$, name the anode, state the sign of each electrode and give the direction of electron flow in the wire.
  • 3 Write the cell notation for a galvanic cell based on $\mathrm{Zn(s)+2Ag^{+}(aq)\rightarrow Zn^{2+}(aq)+2Ag(s)}$.
  • 4 Write the cell notation for a cell in which the reaction is $\mathrm{H_{2}(g)+Cl_{2}(g)\rightarrow 2H^{+}(aq)+2Cl^{-}(aq)}$, using platinum electrodes.
  • 5 Write the two half-reactions and the balanced cell reaction for $\mathrm{Al(s)\,|\,Al^{3+}(aq)\,\|\,Ni^{2+}(aq)\,|\,Ni(s)}$.
  • 6 State two functions of a salt bridge. Why is a $\mathrm{KNO_{3}}$ bridge, rather than a $\mathrm{KCl}$ bridge, used in a cell containing $\mathrm{Ag^{+}(aq)}$?
  • 7 In a Daniell cell, $0.0400\ \mathrm{mol}$ of zinc dissolves. How many moles of electrons pass through the wire, and what mass of copper is deposited?
  • 8 What happens to the voltmeter reading of a Daniell cell if the salt bridge is lifted out? Explain.
1 A galvanic cell turns the chemical energy of a spontaneous reaction into electrical energy ($\Delta_{r}G\lt0$); an electrolytic cell uses electrical energy to drive a non-spontaneous reaction. In a galvanic cell the anode is negative and the cathode positive; in an electrolytic cell the anode is joined to the positive terminal of the supply. (Usually a galvanic cell has two separate compartments with a salt bridge; an electrolytic cell often has a single vessel.)
2 Mg (lower $E^{\circ}$) is oxidised: it is the anode and is negative. Cu is the cathode and is positive. Electrons flow through the wire from Mg to Cu.
3 $\mathrm{Zn(s)\,|\,Zn^{2+}(aq)\,\|\,Ag^{+}(aq)\,|\,Ag(s)}$.
4 $\mathrm{H_{2}}$ is oxidised (left), $\mathrm{Cl_{2}}$ reduced (right): $\mathrm{Pt(s)\,|\,H_{2}(g)\,|\,H^{+}(aq)\,\|\,Cl^{-}(aq)\,|\,Cl_{2}(g)\,|\,Pt(s)}$.
5 Anode: $\mathrm{Al(s)\rightarrow Al^{3+}(aq)+3e^{-}}$ (×2). Cathode: $\mathrm{Ni^{2+}(aq)+2e^{-}\rightarrow Ni(s)}$ (×3). Cell: $\mathrm{2Al(s)+3Ni^{2+}(aq)\rightarrow 2Al^{3+}(aq)+3Ni(s)}$, with $6e^{-}$ transferred; charge $+6$ on both sides.
6 It completes the circuit by letting ions move between the solutions, and it keeps each solution electrically neutral (it also stops the solutions mixing). $\mathrm{Cl^{-}}$ from a $\mathrm{KCl}$ bridge would precipitate $\mathrm{AgCl(s)}$ with the $\mathrm{Ag^{+}}$ ions, so a bridge with a non-reacting anion ($\mathrm{NO_{3}^{-}}$) is used.
7 Each Zn gives $2e^{-}$: $0.0800\ \mathrm{mol}$ of electrons. They reduce $0.0400\ \mathrm{mol}$ of $\mathrm{Cu^{2+}}$: $0.0400\times63.5=2.54\ \mathrm{g}$ of copper.
8 It drops to zero. With no ionic path between the solutions the circuit is open; also, charge would build up at once in each compartment and stop the electrode reactions.
II
Standard EMFs. Use the table of standard electrode potentials and the values listed under it.
  • 1 $\mathrm{Zn(s)\,|\,Zn^{2+}(aq)\,\|\,Ag^{+}(aq)\,|\,Ag(s)}$: find $E^{\circ}_{cell}$.
  • 2 $\mathrm{Al(s)\,|\,Al^{3+}(aq)\,\|\,Cu^{2+}(aq)\,|\,Cu(s)}$: find $E^{\circ}_{cell}$.
  • 3 $\mathrm{Fe(s)\,|\,Fe^{2+}(aq)\,\|\,Cl^{-}(aq)\,|\,Cl_{2}(g)\,|\,Pt(s)}$: find $E^{\circ}_{cell}$.
  • 4 $\mathrm{Pt(s)\,|\,H_{2}(g)\,|\,H^{+}(aq)\,\|}$ $\mathrm{Fe^{3+}(aq),\,Fe^{2+}(aq)\,|\,Pt(s)}$: find $E^{\circ}_{cell}$.
  • 5 $\mathrm{Mg(s)\,|\,Mg^{2+}(aq)\,\|\,Zn^{2+}(aq)\,|\,Zn(s)}$: find $E^{\circ}_{cell}$.
  • 6 Find $E^{\circ}_{cell}$ for $\mathrm{Cu(s)\,|\,Cu^{2+}(aq)\,\|\,Fe^{2+}(aq)\,|\,Fe(s)}$. Is the cell spontaneous as written? If not, write the correct notation.
  • 7 The cell $\mathrm{X(s)\,|\,X^{2+}(aq)\,\|\,Cu^{2+}(aq)\,|\,Cu(s)}$ has $E^{\circ}_{cell}=0.47\ \mathrm{V}$. Find $E^{\circ}_{\mathrm{X^{2+}/X}}$ and identify X from the values listed under the table.
  • 8 Find $E^{\circ}_{cell}$ for $\mathrm{Sn(s)\,|\,Sn^{2+}(aq)\,\|\,H^{+}(aq)\,|\,H_{2}(g)\,|\,Pt(s)}$. Which electrode is positive?
  • 9 Which two half-cells from the table would give the largest standard EMF? Find it, and suggest why such a cell is not practical with aqueous solutions.
Check: $E^{\circ}_{cell}$ from two reduction potentials
$E^{\circ}_{cell}=E^{\circ}_{R}-E^{\circ}_{L}$—
as written—
1 $0.80-(-0.76)=1.56\ \mathrm{V}$.
2 $0.34-(-1.66)=2.00\ \mathrm{V}$.
3 $1.36-(-0.44)=1.80\ \mathrm{V}$.
4 $0.77-0.00=0.77\ \mathrm{V}$.
5 $-0.76-(-2.36)=1.60\ \mathrm{V}$.
6 $-0.44-0.34=-0.78\ \mathrm{V}$: not spontaneous as written. The working cell is $\mathrm{Fe(s)\,|\,Fe^{2+}(aq)\,\|\,Cu^{2+}(aq)\,|\,Cu(s)}$ with $E^{\circ}_{cell}=+0.78\ \mathrm{V}$.
7 $0.47=0.34-E^{\circ}_{\mathrm{X^{2+}/X}}$, so $E^{\circ}_{\mathrm{X^{2+}/X}}=-0.13\ \mathrm{V}$: X is lead.
8 $0.00-(-0.14)=0.14\ \mathrm{V}$. The hydrogen electrode (cathode) is positive.
9 The highest and lowest couples: $\mathrm{Cl_{2}/Cl^{-}}$ and $\mathrm{Na^{+}/Na}$, $E^{\circ}_{cell}=1.36-(-2.71)=4.07\ \mathrm{V}$. Sodium reacts violently with water, so it cannot be used as an electrode in an aqueous cell.
III
Predictions from the electrochemical series (standard conditions).
  • 1 Name the strongest oxidising agent and the strongest reducing agent in the table.
  • 2 Does zinc displace iron from iron(II) sulphate solution? Give $E^{\circ}_{cell}$ and the ionic equation if it does.
  • 3 Can silver metal reduce $\mathrm{Cu^{2+}(aq)}$ ions? Justify with $E^{\circ}_{cell}$.
  • 4 Can chlorine oxidise $\mathrm{Fe^{2+}(aq)}$ to $\mathrm{Fe^{3+}(aq)}$? Write the balanced equation if so.
  • 5 Will copper dissolve in $1\ \mathrm{M}$ hydrochloric acid with the release of hydrogen? Explain.
  • 6 Arrange Zn, Fe, Mg, Al and Ag in order of decreasing reducing power.
  • 7 Can a nickel spatula be used to stir (a) silver nitrate solution, (b) zinc sulphate solution? Give $E^{\circ}_{cell}$ for each possible reaction.
  • 8 Electronics. Copper tracks on printed circuit boards are etched away with iron(III) chloride solution. Show that the reaction is feasible and write its equation.
1 Strongest oxidising agent: $\mathrm{Cl_{2}}$ (highest $E^{\circ}$, $+1.36\ \mathrm{V}$). Strongest reducing agent: Na (lowest $E^{\circ}$, $-2.71\ \mathrm{V}$).
2 $E^{\circ}_{cell}=-0.44-(-0.76)=0.32\ \mathrm{V}\gt0$, so yes: $\mathrm{Zn(s)+Fe^{2+}(aq)\rightarrow Zn^{2+}(aq)+Fe(s)}$.
3 $\mathrm{2Ag(s)+Cu^{2+}(aq)\rightarrow 2Ag^{+}(aq)+Cu(s)}$ would have $E^{\circ}_{cell}=0.34-0.80=-0.46\ \mathrm{V}\lt0$. No.
4 $E^{\circ}_{cell}=1.36-0.77=0.59\ \mathrm{V}\gt0$, so yes: $\mathrm{Cl_{2}(g)+2Fe^{2+}(aq)\rightarrow 2Cl^{-}(aq)+2Fe^{3+}(aq)}$.
5 No. $E^{\circ}_{cell}=0.00-0.34=-0.34\ \mathrm{V}$: $\mathrm{H^{+}}$ is too weak an oxidising agent for copper. (Copper does dissolve in nitric acid, where the nitrate ion is the oxidising agent.)
6 $\mathrm{Mg}\gt\mathrm{Al}\gt\mathrm{Zn}\gt\mathrm{Fe}\gt\mathrm{Ag}$ (in order of increasingly positive $E^{\circ}$).
7 (a) $\mathrm{Ni(s)+2Ag^{+}(aq)\rightarrow Ni^{2+}(aq)+2Ag(s)}$: $E^{\circ}_{cell}=0.80-(-0.25)=1.05\ \mathrm{V}$; the spatula would be attacked, so no. (b) $\mathrm{Ni(s)+Zn^{2+}(aq)\rightarrow Ni^{2+}(aq)+Zn(s)}$: $E^{\circ}_{cell}=-0.76-(-0.25)=-0.51\ \mathrm{V}$; no reaction, so yes.
8 $E^{\circ}_{cell}=E^{\circ}_{\mathrm{Fe^{3+}/Fe^{2+}}}-E^{\circ}_{\mathrm{Cu^{2+}/Cu}}=0.77-0.34=0.43\ \mathrm{V}\gt0$: $\mathrm{2Fe^{3+}(aq)+Cu(s)\rightarrow 2Fe^{2+}(aq)+Cu^{2+}(aq)}$.
IV
Thinking with electrode potentials.
  • 1 Why can the potential of a single electrode not be measured? How does the standard hydrogen electrode get round this problem?
  • 2 Iron sheets are often coated with zinc (galvanised). Using $E^{\circ}$ values, explain why the iron is protected even where the zinc coating is scratched.
  • 3 From Mg, Zn, Cu and Ag (each with a $1\ \mathrm{M}$ solution of its ions), which pair gives the cell with the highest standard EMF, and which the lowest? Give both values.
  • 4 Kitchen science. A zinc-plated nail and a copper coin pushed into a lemon light a small LED weakly. There are almost no $\mathrm{Cu^{2+}}$ ions in lemon juice. Suggest the cathode reaction, and estimate the standard EMF of this cell.
  • 5 $E^{\circ}_{cell}$ is $1.10\ \mathrm{V}$ for $\mathrm{Zn|Zn^{2+}\|Cu^{2+}|Cu}$ and $0.46\ \mathrm{V}$ for $\mathrm{Cu|Cu^{2+}\|Ag^{+}|Ag}$. Without using the table, find $E^{\circ}_{cell}$ for $\mathrm{Zn|Zn^{2+}\|Ag^{+}|Ag}$, and explain why the two EMFs may simply be added here.
1 A voltmeter measures a potential difference, so it needs two electrodes; any second contact with the solution is itself an electrode. The SHE is chosen as a reference and assigned $E^{\circ}=0$ exactly. Every other electrode is measured against it, so its tabulated value is really "the EMF of a cell with the SHE on the left".
2 $E^{\circ}_{\mathrm{Zn^{2+}/Zn}}=-0.76\ \mathrm{V}$ is lower than $E^{\circ}_{\mathrm{Fe^{2+}/Fe}}=-0.44\ \mathrm{V}$, so zinc is the stronger reducing agent. Where the two metals touch in moisture, zinc becomes the anode and is oxidised; iron is the cathode and is not. (Section 2.4 returns to this sacrificial protection.)
3 The six pairs give: Mg–Zn $1.60$, Mg–Cu $2.70$, Mg–Ag $3.16$, Zn–Cu $1.10$, Zn–Ag $1.56$, Cu–Ag $0.46\ \mathrm{V}$. Highest: Mg and Ag, $0.80-(-2.36)=3.16\ \mathrm{V}$ (the two couples farthest apart). Lowest: Cu and Ag, $0.80-0.34=0.46\ \mathrm{V}$ (the two closest together).
4 Zinc is the anode. With no $\mathrm{Cu^{2+}}$ present, the species reduced at the copper surface is $\mathrm{H^{+}}$ from the citric acid: $\mathrm{2H^{+}(aq)+2e^{-}\rightarrow H_{2}(g)}$. The copper is only an electrode. Estimate: $E^{\circ}_{cell}=0.00-(-0.76)=0.76\ \mathrm{V}$. The real voltage differs because the juice is not at standard conditions and its acid is weak.
5 $E^{\circ}(\mathrm{Zn|Ag})=E^{\circ}_{\mathrm{Ag}}-E^{\circ}_{\mathrm{Zn}}$ $=(E^{\circ}_{\mathrm{Ag}}-E^{\circ}_{\mathrm{Cu}})+(E^{\circ}_{\mathrm{Cu}}-E^{\circ}_{\mathrm{Zn}})$ $=0.46+1.10=1.56\ \mathrm{V}$. Adding works because each EMF is a difference of two electrode potentials, and the copper term cancels. (This is different from adding two half-cell potentials to get a third one, which needs the Gibbs energy method of Section 2.2.)
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks (a) Describe the standard hydrogen electrode and write its notation. (b) Why is a reference electrode needed? (c) The cell $\mathrm{Zn(s)\,|\,Zn^{2+}(aq,\ 1\ M)\,\|}$ $\mathrm{H^{+}(aq,\ 1\ M)\,|\,H_{2}(g,\ 1\ bar)\,|\,Pt(s)}$ has an EMF of $0.76\ \mathrm{V}$. Find $E^{\circ}_{\mathrm{Zn^{2+}/Zn}}$ and state which electrode is negative.

(a) A platinum electrode coated with platinum black, dipped in a solution with $[\mathrm{H^{+}}]=1\ \mathrm{mol\,L^{-1}}$, with pure $\mathrm{H_{2}}$ gas at $1\ \mathrm{bar}$ bubbled over it; $\mathrm{2H^{+}(aq)+2e^{-}\rightarrow H_{2}(g)}$, $E^{\circ}=0.00\ \mathrm{V}$. Notation: $\mathrm{Pt(s)\,|\,H_{2}(g,\ 1\ bar)\,|\,H^{+}(aq,\ 1\ M)}$.
(b) Only potential differences can be measured, so one electrode must be given an agreed value (zero) and all others measured against it.
(c) $0.76=E^{\circ}_{R}-E^{\circ}_{L}=0-E^{\circ}_{\mathrm{Zn^{2+}/Zn}}$, so $E^{\circ}_{\mathrm{Zn^{2+}/Zn}}=-0.76\ \mathrm{V}$. Zinc is the anode, so the zinc electrode is negative.
Marking tip One mark per part. In (a) the three conditions ($1\ \mathrm{M}$ $\mathrm{H^{+}}$, $1\ \mathrm{bar}$ $\mathrm{H_{2}}$, platinised Pt) must all be stated.

E2. JEE Main Given $E^{\circ}$: $\mathrm{Fe^{3+}/Fe^{2+}}=+0.77\ \mathrm{V}$, $\mathrm{I_{2}/I^{-}}=+0.54\ \mathrm{V}$, $\mathrm{Ag^{+}/Ag}=+0.80\ \mathrm{V}$, $\mathrm{Ni^{2+}/Ni}=-0.25\ \mathrm{V}$. The strongest reducing agent among the following is (A) $\mathrm{Fe^{2+}}$  (B) $\mathrm{I^{-}}$  (C) Ag  (D) Ni

Idea The listed species are all reduced forms. The reduced form of the couple with the lowest $E^{\circ}$ gives up electrons most readily.
Order Ni $(-0.25)$, $\mathrm{I^{-}}$ $(+0.54)$, $\mathrm{Fe^{2+}}$ $(+0.77)$, Ag $(+0.80)$: reducing power decreases along this list.
Answer (D) Ni.

E3. NEET Copper(II) sulphate solution can be stored safely in a vessel made of (A) zinc  (B) iron  (C) silver  (D) aluminium

Idea The vessel is safe only if its metal lies above $\mathrm{Cu^{2+}/Cu}$ ($+0.34\ \mathrm{V}$) in the series, so that it cannot reduce $\mathrm{Cu^{2+}}$.
Check Zn $(-0.76)$, Fe $(-0.44)$ and Al $(-1.66)$ would all displace copper. Ag $(+0.80)$: $E^{\circ}_{cell}=0.34-0.80=-0.46\ \mathrm{V}$, no reaction.
Answer (C) silver.

E4. JEE Main Numerical value. At $298\ \mathrm{K}$, $E^{\circ}_{cell}=0.36\ \mathrm{V}$ for $\mathrm{Zn|Zn^{2+}\|Cd^{2+}|Cd}$ and $0.74\ \mathrm{V}$ for $\mathrm{Cd|Cd^{2+}\|Cu^{2+}|Cu}$. The standard EMF of $\mathrm{Zn|Zn^{2+}\|Cu^{2+}|Cu}$, in units of $10^{-2}\ \mathrm{V}$, is ______.

Idea $E^{\circ}_{\mathrm{Cd}}-E^{\circ}_{\mathrm{Zn}}=0.36$ and $E^{\circ}_{\mathrm{Cu}}-E^{\circ}_{\mathrm{Cd}}=0.74$. Adding, the cadmium terms cancel.
Sum $E^{\circ}_{\mathrm{Cu}}-E^{\circ}_{\mathrm{Zn}}=0.36+0.74=1.10\ \mathrm{V}$.
Answer $\mathbf{110}$. (Consistent with $E^{\circ}_{\mathrm{Cd^{2+}/Cd}}=-0.40\ \mathrm{V}$ from Example 4.)

E5. JEE Advanced One or more correct. For the cell $\mathrm{Al(s)\,|\,Al^{3+}(aq)\,\|\,Fe^{3+}(aq),\,Fe^{2+}(aq)\,|\,Pt(s)}$ under standard conditions ($E^{\circ}$: $\mathrm{Al^{3+}/Al}=-1.66\ \mathrm{V}$, $\mathrm{Fe^{3+}/Fe^{2+}}=+0.77\ \mathrm{V}$), which statements are correct? (A) $E^{\circ}_{cell}=2.43\ \mathrm{V}$  (B) three electrons are transferred in the balanced cell reaction  (C) electrons flow through the wire from the platinum to the aluminium  (D) the concentration of $\mathrm{Al^{3+}}$ rises as the cell discharges

(A) $E^{\circ}_{cell}=0.77-(-1.66)=2.43\ \mathrm{V}$. Correct.
(B) $\mathrm{Al(s)+3Fe^{3+}(aq)\rightarrow Al^{3+}(aq)+3Fe^{2+}(aq)}$: $n=3$. Correct.
(C) Aluminium is the anode, so electrons flow from Al to Pt. Incorrect.
(D) Al is oxidised to $\mathrm{Al^{3+}}$, so its concentration rises. Correct.
Answer (A), (B) and (D).

E6. Board NEET Assertion–reason. Assertion (A): The standard potential of the half-reaction $\mathrm{2Ag^{+}(aq)+2e^{-}\rightarrow 2Ag(s)}$ is $+0.80\ \mathrm{V}$, the same as for $\mathrm{Ag^{+}(aq)+e^{-}\rightarrow Ag(s)}$. Reason (R): Electrode potential is an intensive property. (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.

Check Both statements are true. A potential is energy per unit charge, so it does not depend on how many electrons (how much substance) the equation is written for.
Answer (A).
🔬 Research–Teaching NexusResearch-tutored · optional, about 40 minutes

Research question. When two different metals touch in salt water, which one corrodes, and how well does the table of standard electrode potentials predict the voltage between them?

Why it matters. Marine engineers worry about galvanic corrosion: a bronze propeller bolted to a steel hull, or aluminium fittings on a steel mast, form a galvanic cell with seawater as the electrolyte, and the more reactive metal is eaten away. Ship owners in Mumbai, Rotterdam and Singapore all fight it, usually by bolting blocks of zinc or magnesium to the hull to be corroded instead.

What to do.

  1. Collect strips or wires of zinc (a galvanised nail), iron (a clean steel nail), copper (wire or a coin), aluminium (foil) and, if available, magnesium ribbon. Clean each with sandpaper.
  2. Use a multimeter on the 2 V DC range. Push pairs of metals about $2\ \mathrm{cm}$ apart into a lemon, or into a paper towel soaked in $3.5\%$ salt water (about the salinity of seawater). Note the reading and which metal is the negative terminal.
  3. Predict each voltage from the table as $E^{\circ}_{\text{higher}}-E^{\circ}_{\text{lower}}$ and tabulate prediction against measurement.
  4. Now keep one metal fixed (zinc) and change only the other. Does the voltage depend strongly on the second metal? Use your answer to Exercise IV.4 to explain what is really being reduced at the positive metal.

Tools: Multimeter, lemon or salt water, metal samples; Interactive 1 for the predictions.

Hand in: A table of predicted and measured voltages, the order of the metals from "most protected" to "most corroded", and two sentences of advice for a boat builder.

Going further. Look up a "galvanic series in seawater", which ranks alloys such as stainless steel and bronze by potentials measured in seawater. Compare its order with the electrochemical series and suggest why the two differ (oxide films, alloys, non-standard concentrations).

Instructor note: the order (Mg most negative, then Al, Zn, Fe, Cu) matches the series and the more negative metal always corrodes. The values do not match $E^{\circ}$ differences, because no solution contains the metals' own ions at $1\ \mathrm{M}$. The positive metal mainly acts as a surface for reducing $\mathrm{H^{+}}$ or dissolved $\mathrm{O_{2}}$, so voltages depend mostly on the negative metal. Aluminium often reads low because of its protective oxide layer. Expect zinc–copper readings of very roughly $0.8$–$1.0\ \mathrm{V}$ and magnesium–copper readings well above $1\ \mathrm{V}$.

🧪 Self-check quiz

0/5 correct

Q1. In a galvanic cell, the anode is

Q2. Using $E^{\circ}_{\mathrm{Mg^{2+}/Mg}}=-2.36\ \mathrm{V}$ and $E^{\circ}_{\mathrm{Cu^{2+}/Cu}}=+0.34\ \mathrm{V}$, the standard EMF of $\mathrm{Mg|Mg^{2+}\|Cu^{2+}|Cu}$ is

Q3. The main job of the salt bridge in a Daniell cell is to

Q4. Given $E^{\circ}$: $\mathrm{I_{2}/I^{-}}=+0.54$, $\mathrm{Cl_{2}/Cl^{-}}=+1.36\ \mathrm{V}$. Which species can oxidise $\mathrm{I^{-}}$ to $\mathrm{I_{2}}$ but cannot oxidise $\mathrm{Cl^{-}}$ to $\mathrm{Cl_{2}}$ (standard conditions)?

Q5. The cell $\mathrm{X|X^{2+}\|Y^{2+}|Y}$ has $E^{\circ}_{cell}=+0.30\ \mathrm{V}$, and $E^{\circ}_{\mathrm{Y^{2+}/Y}}=-0.14\ \mathrm{V}$. Then $E^{\circ}_{\mathrm{X^{2+}/X}}$ is

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