Separate the two halves of a redox reaction and the electrons must travel through a wire: the Daniell cell, electrode potentials measured against the standard hydrogen electrode, and the electrochemical series that predicts which redox reactions can happen.
When a zinc strip reacts with copper(II) sulphate solution (Section 7.1), the solution warms up: the energy of the redox reaction is released as heat, because each electron jumps directly from a zinc atom to a copper ion touching it. In 1836 the English chemist John Daniell found a way to make the same electrons do useful work. He kept the zinc and the copper ions in separate vessels, so that the only path for the electrons was a wire joining the two metals. The wire carried a steady electric current, and the Daniell cell powered the telegraph networks of the nineteenth century.
Every battery since — the dry cell in a torch, the lead–acid battery of a car, the lithium-ion cell in a phone — rests on the same idea: a spontaneous redox reaction with its two half-reactions held apart. This section builds the Daniell cell, defines the electrode potential that measures how strongly each half-reaction pulls on electrons, and arranges these potentials in the electrochemical series, a single table that predicts the outcome of thousands of redox reactions. (Class 12 Chemistry, Ch. 2 — Electrochemistry — develops cells quantitatively.)
The Daniell Cell: a Redox Reaction That Drives a Current
In a Daniell cell a zinc rod dips into zinc sulphate solution and a copper rod into copper(II) sulphate solution, each at $1\ \mathrm{mol\,L^{-1}}$. The two rods are joined through a voltmeter, and the two solutions are connected by a salt bridge — a U-tube filled with a jelly (agar–agar) containing a concentrated solution of an inert salt such as KCl or $\mathrm{KNO_{3}}$. The voltmeter reads about $1.1\ \mathrm{V}$, and the following changes happen while current flows:
$$\text{zinc rod (oxidation):}\quad \mathrm{Zn(s)\rightarrow Zn^{2+}(aq)+2e^{-}}$$
$$\text{copper rod (reduction):}\quad \mathrm{Cu^{2+}(aq)+2e^{-}\rightarrow Cu(s)}$$
The zinc rod slowly dissolves, copper is deposited on the copper rod, and the blue colour of the copper sulphate fades. The net change is exactly the reaction of Section 7.1, $\mathrm{Zn(s)+Cu^{2+}(aq)\rightarrow Zn^{2+}(aq)+Cu(s)}$ — but now the electrons travel from zinc to copper through the external wire.
📘 Definition 7.9 — Galvanic cell, half-cells and electrodes
A galvanic cell (voltaic or electrochemical cell) is a device in which a spontaneous redox reaction produces electrical energy, the oxidation and reduction half-reactions taking place in separate half-cells. Each half-cell contains a redox couple — the oxidised and reduced forms of one species, written oxidised/reduced, e.g. $\mathrm{Zn^{2+}/Zn}$, $\mathrm{Cu^{2+}/Cu}$. The metal or other conductor at which electrons enter or leave the solution is an electrode:
the anode is the electrode at which oxidation occurs; in a galvanic cell it is the negative electrode, because it releases electrons into the wire (zinc in the Daniell cell);
the cathode is the electrode at which reduction occurs; in a galvanic cell it is the positive electrode (copper).
Electrons flow in the external circuit from anode to cathode. A cell is written in shorthand with the anode on the left, a single line for each phase boundary and a double line for the salt bridge: $\mathrm{Zn(s)\,|\,Zn^{2+}(aq)\,\|\,Cu^{2+}(aq)\,|\,Cu(s)}$.
📐 Principle 7.5 — The role of the salt bridge
The salt bridge (i) completes the circuit, letting ions carry the current between the two solutions, and (ii) keeps each solution electrically neutral. As $\mathrm{Zn^{2+}}$ ions form in the anode compartment, negative ions ($\mathrm{Cl^{-}}$ or $\mathrm{NO_{3}^{-}}$) move out of the bridge into it; as $\mathrm{Cu^{2+}}$ ions are removed from the cathode compartment, positive ions ($\mathrm{K^{+}}$) move into it. Without the bridge, a positive charge would build up around the zinc and a negative charge around the copper within a fraction of a second, and the current would stop. The ions of the bridge must not react with either solution (KCl cannot be used with $\mathrm{Ag^{+}}$, which would precipitate as AgCl; $\mathrm{KNO_{3}}$ is used instead).
✏️ Example 1 — Describing galvanic cells
(a) For the Daniell cell, name the anode and the cathode, give their signs, and state the direction of flow of electrons in the wire and of anions in the salt bridge. (b) A cell is made from a copper rod in $\mathrm{Cu(NO_{3})_{2}(aq)}$ and a silver rod in $\mathrm{AgNO_{3}(aq)}$. Copper is the negative electrode. Write the half-reactions, the overall cell reaction and the cell notation, and say why the salt bridge should contain $\mathrm{KNO_{3}}$ rather than KCl. (c) What happens to the masses of the two rods in (b) as the cell is used?
💡 Show solution▶
(a) Anode: zinc (oxidation; negative). Cathode: copper (reduction; positive). Electrons flow from Zn to Cu through the wire; anions in the bridge move towards the zinc (anode) compartment, cations towards the copper compartment.
(b) Anode: $\mathrm{Cu(s)\rightarrow Cu^{2+}(aq)+2e^{-}}$. Cathode: $\mathrm{2Ag^{+}(aq)+2e^{-}\rightarrow 2Ag(s)}$. Overall: $\mathrm{Cu(s)+2Ag^{+}(aq)\rightarrow}$ $\mathrm{Cu^{2+}(aq)+2Ag(s)}$. Notation: $\mathrm{Cu(s)\,|\,Cu^{2+}(aq)\,\|\,Ag^{+}(aq)\,|\,Ag(s)}$. Chloride would precipitate AgCl in the silver half-cell.
(c) The copper anode loses mass (63.5 g per 2 mol of electrons) and the silver cathode gains mass (2 × 107.9 g per 2 mol of electrons) — the cathode gains more mass than the anode loses.
Interactive 1 — Build a Galvanic Cell
Choose the electrode for the left and right half-cells (all solutions $1\ \mathrm{mol\,L^{-1}}$, $298\ \mathrm{K}$; "H₂ (SHE)" is the standard hydrogen electrode). Press ▶ Play to watch electrons (yellow) flow through the wire from the anode to the cathode while ions move through the salt bridge — cations towards the cathode, anions towards the anode. Untick the salt bridge and see what happens to the current. The voltmeter shows $E^{\circ}_{\text{cell}}=E^{\circ}_{\text{cathode}}-E^{\circ}_{\text{anode}}$.
anode (oxidation)—
cathode (reduction)—
E°cell—
cell reaction—
Electrode Potential and the Standard Hydrogen Electrode
Why does the Daniell cell produce a voltage at all? When a metal dips into a solution of its own ions, two opposite tendencies compete at the surface: metal atoms leave as ions, leaving electrons behind in the metal, and ions from the solution take electrons from the metal and deposit as atoms. A zinc rod loses ions faster than it gains them and becomes slightly negative; a copper rod gains copper ions and becomes slightly positive. Each half-cell develops its own potential difference between the metal and the solution.
📘 Definition 7.10 — Electrode potential and standard electrode potential
The electrode potential of a half-cell is the potential difference that develops between the electrode and the solution of its redox couple. When all species are in their standard states — solutes at $1\ \mathrm{mol\,L^{-1}}$, gases at $1\ \mathrm{bar}$, at a stated temperature (usually $298\ \mathrm{K}$) — it is the standard electrode potential, $E^{\circ}$. By IUPAC convention every $E^{\circ}$ is written for the half-reaction as a reduction, $\mathrm{M^{n+}(aq)+ne^{-}\rightarrow M(s)}$ (a standard reduction potential). A more positive $E^{\circ}$ means a greater tendency for the oxidised form to accept electrons.
The potential of a single electrode cannot be measured on its own: a voltmeter needs two connections, so it always measures the difference between two half-cells. Chemists therefore agree on one reference half-cell and give it the value zero.
📘 Definition 7.11 — The standard hydrogen electrode
The standard hydrogen electrode is a platinum electrode coated with finely divided platinum black, dipping into a solution with $[\mathrm{H^{+}}]=1\ \mathrm{mol\,L^{-1}}$, with pure hydrogen gas at $1\ \mathrm{bar}$ bubbled over it at $298\ \mathrm{K}$. Its half-reaction and potential are, by definition,
$$\mathrm{2H^{+}(aq)+2e^{-}\rightarrow H_{2}(g)};\qquad E^{\circ}=0.00\ \mathrm{V}\ \text{at all temperatures}$$
Platinum is inert: it only carries electrons to and from the $\mathrm{H^{+}/H_{2}}$ couple. Cell notation: $\mathrm{Pt(s)\,|\,H_{2}(g,1\ bar)\,|\,H^{+}(aq,1\ M)}$.
To find the standard potential of any other couple, it is connected to the standard hydrogen electrode and the cell voltage is measured. Zinc against the hydrogen electrode gives $0.76\ \mathrm{V}$, with zinc as the negative electrode (anode): zinc reduces $\mathrm{H^{+}}$, so $E^{\circ}(\mathrm{Zn^{2+}/Zn})=-0.76\ \mathrm{V}$. Copper gives $0.34\ \mathrm{V}$ with copper as the positive electrode (cathode): $\mathrm{Cu^{2+}}$ is reduced and $\mathrm{H_{2}}$ oxidised, so $E^{\circ}(\mathrm{Cu^{2+}/Cu})=+0.34\ \mathrm{V}$.
📐 Rule 7.6 — Standard EMF of a cell
The standard electromotive force (EMF) of a galvanic cell, the cell voltage when no current is drawn and all species are in their standard states, is
$$E^{\circ}_{\text{cell}}=E^{\circ}_{\text{cathode}}-E^{\circ}_{\text{anode}}=E^{\circ}_{\text{right}}-E^{\circ}_{\text{left}}$$
using the standard reduction potentials of both electrodes as tabulated (do not change the sign of the anode value or multiply by stoichiometric coefficients — $E^{\circ}$ is an intensive property). A positive $E^{\circ}_{\text{cell}}$ means the cell reaction, as written with the left electrode as anode, is spontaneous; a negative value means the reaction runs in the opposite direction. Daniell cell: $E^{\circ}_{\text{cell}}=0.34-(-0.76)=1.10\ \mathrm{V}$.
Couple (reduction)
$\mathrm{F_{2}/F^{-}}$
$\mathrm{MnO_{4}^{-}/Mn^{2+}}$
$\mathrm{Cl_{2}/Cl^{-}}$
$\mathrm{Cr_{2}O_{7}^{2-}/Cr^{3+}}$
$\mathrm{Br_{2}/Br^{-}}$
$\mathrm{Ag^{+}/Ag}$
$\mathrm{Fe^{3+}/Fe^{2+}}$
$\mathrm{I_{2}/I^{-}}$
$\mathrm{Cu^{2+}/Cu}$
$\mathrm{H^{+}/H_{2}}$
$\mathrm{Pb^{2+}/Pb}$
$\mathrm{Ni^{2+}/Ni}$
$\mathrm{Fe^{2+}/Fe}$
$\mathrm{Zn^{2+}/Zn}$
$\mathrm{Al^{3+}/Al}$
$\mathrm{Mg^{2+}/Mg}$
$\mathrm{Na^{+}/Na}$
$\mathrm{Li^{+}/Li}$
$E^{\circ}$ / V
+2.87
+1.51
+1.36
+1.33
+1.09
+0.80
+0.77
+0.54
+0.34
0.00
−0.13
−0.25
−0.44
−0.76
−1.66
−2.37
−2.71
−3.05
Standard reduction potentials at $298\ \mathrm{K}$ (rounded literature values; $\mathrm{MnO_{4}^{-}}$ and $\mathrm{Cr_{2}O_{7}^{2-}}$ in acid, with $[\mathrm{H^{+}}]=1\ \mathrm{mol\,L^{-1}}$). Use them throughout this section.
✏️ Example 2 — Calculating cell EMFs
Using the table, write the cell reaction and find $E^{\circ}_{\text{cell}}$ for: (a) $\mathrm{Mg(s)\,|\,Mg^{2+}(aq)\,\|\,Ag^{+}(aq)\,|\,Ag(s)}$; (b) $\mathrm{Fe(s)\,|\,Fe^{2+}(aq)\,\|\,Cu^{2+}(aq)\,|\,Cu(s)}$; (c) $\mathrm{Cu(s)\,|\,Cu^{2+}(aq)\,\|\,Zn^{2+}(aq)\,|\,Zn(s)}$. Comment on the sign in (c).
💡 Show solution▶
(a) $\mathrm{Mg(s)+2Ag^{+}(aq)\rightarrow}$ $\mathrm{Mg^{2+}(aq)+2Ag(s)}$; $E^{\circ}_{\text{cell}}=0.80-(-2.37)=3.17\ \mathrm{V}$. (The 2 in front of $\mathrm{Ag^{+}}$ does not double $E^{\circ}$.)
(c) As written, copper would be the anode: $\mathrm{Cu(s)+Zn^{2+}(aq)\rightarrow Cu^{2+}(aq)+Zn(s)}$; $E^{\circ}_{\text{cell}}=-0.76-0.34=-1.10\ \mathrm{V}$.
Sign The negative value says this reaction is not spontaneous — it is the reverse of the Daniell cell reaction, which is. Copper does not reduce $\mathrm{Zn^{2+}}$, exactly as the activity series of Section 7.1 predicted. If the cell is actually built, zinc is the anode and the voltmeter reads $1.10\ \mathrm{V}$.
✏️ Example 3 — Finding an unknown standard potential
(a) The cell $\mathrm{Ni(s)\,|\,Ni^{2+}(aq)\,\|\,Cu^{2+}(aq)\,|\,Cu(s)}$ has $E^{\circ}_{\text{cell}}=0.59\ \mathrm{V}$. Find $E^{\circ}(\mathrm{Ni^{2+}/Ni})$. (b) A cell made from a metal M in $\mathrm{M^{2+}(aq)}$ and the standard hydrogen electrode gives $0.13\ \mathrm{V}$, the hydrogen electrode being positive. Find $E^{\circ}(\mathrm{M^{2+}/M})$ and identify M from the table. (c) Will M liberate hydrogen from $1\ \mathrm{mol\,L^{-1}}$ HCl under standard conditions?
💡 Show solution▶
(a) $0.59=E^{\circ}_{\text{cathode}}-E^{\circ}_{\text{anode}}=0.34-E^{\circ}(\mathrm{Ni^{2+}/Ni})$, so $E^{\circ}(\mathrm{Ni^{2+}/Ni})=-0.25\ \mathrm{V}$.
(b) The hydrogen electrode is the cathode, so M is the anode: $0.13=0.00-E^{\circ}(\mathrm{M^{2+}/M})$, $E^{\circ}=-0.13\ \mathrm{V}$: lead.
(c) Yes in principle: $E^{\circ}_{\text{cell}}$ for $\mathrm{Pb+2H^{+}\rightarrow Pb^{2+}+H_{2}}$ is $0.00-(-0.13)=+0.13\ \mathrm{V}$. (In practice the reaction with HCl is very slow, partly because insoluble $\mathrm{PbCl_{2}}$ coats the metal — a reminder that $E^{\circ}$ says what can happen, not how fast.)
The Electrochemical Series and What It Predicts
When the standard reduction potentials of many couples are listed in order — most positive at the top, most negative at the bottom — the list is called the electrochemical series. The table above is a short version of it. Its top left holds the strongest oxidising agents ($\mathrm{F_{2}}$, $\mathrm{MnO_{4}^{-}}$, $\mathrm{Cl_{2}}$), which are eager to gain electrons; its bottom right holds the strongest reducing agents (Li, Na, Mg), which lose electrons most easily. The activity series of metals in Section 7.1 is simply the metal part of this list read upwards.
📐 Rule 7.7 — Predictions from the electrochemical series
For species in their standard states:
The more positive $E^{\circ}$, the stronger the oxidised form as an oxidising agent; the more negative $E^{\circ}$, the stronger the reduced form as a reducing agent.
The oxidised form of a couple can oxidise the reduced form of any couple with a lower (less positive) $E^{\circ}$; the reaction is spontaneous because $E^{\circ}_{\text{cell}}=$ $E^{\circ}(\text{oxidant couple})-E^{\circ}(\text{reductant couple})$ $\gt0$.
A metal with a negative $E^{\circ}$ can displace hydrogen from a dilute acid ($[\mathrm{H^{+}}]=1\ \mathrm{mol\,L^{-1}}$); metals with positive $E^{\circ}$ (Cu, Ag, Au) cannot.
A metal displaces from solution the ions of any metal with a more positive $E^{\circ}$.
In a galvanic cell made from two couples, the couple with the lower $E^{\circ}$ is the anode (negative electrode).
These predictions concern the direction of reaction (thermodynamics), not its rate: a reaction allowed by the series may still be too slow to observe.
✏️ Example 4 — Who can oxidise whom?
Use the table to predict whether a reaction occurs under standard conditions. If it does, write the balanced ionic equation and find $E^{\circ}_{\text{cell}}$.
(a) $\mathrm{Fe^{3+}(aq)}$ with $\mathrm{I^{-}(aq)}$; (b) $\mathrm{Fe^{3+}(aq)}$ with $\mathrm{Br^{-}(aq)}$; (c) $\mathrm{Cl_{2}(aq)}$ with $\mathrm{Br^{-}(aq)}$; (d) acidified $\mathrm{MnO_{4}^{-}}$ with $\mathrm{Cl^{-}}$ — and what does (d) mean for permanganate titrations (Section 7.2)?
(b) $+0.77\lt+1.09$: no. Indeed the reverse happens: bromine oxidises $\mathrm{Fe^{2+}}$ to $\mathrm{Fe^{3+}}$.
(c) $+1.36\gt+1.09$: yes. $\mathrm{Cl_{2}(aq)+2Br^{-}(aq)\rightarrow}$ $\mathrm{2Cl^{-}(aq)+Br_{2}(aq)}$; $E^{\circ}_{\text{cell}}=0.27\ \mathrm{V}$ — the halogen displacement of Section 7.2.
(d) $+1.51\gt+1.36$: yes, $E^{\circ}_{\text{cell}}=0.15\ \mathrm{V}$:
$$\mathrm{2MnO_{4}^{-}(aq)+16H^{+}(aq)+10Cl^{-}(aq)\rightarrow}$$ $$\mathrm{2Mn^{2+}(aq)+5Cl_{2}(g)+8H_{2}O(l)}$$
This is why permanganate titrations are acidified with $\mathrm{H_{2}SO_{4}}$, not HCl.
✏️ Example 5 — Metals, acids and protective coatings
(a) Which of Mg, Zn, Fe, Cu and Ag liberate hydrogen from dilute hydrochloric acid? (b) Can copper(II) sulphate solution be stored in a zinc-coated (galvanised) container? (c) Iron sheets are galvanised (coated with zinc) to stop rusting. When the coating is scratched, the iron still does not rust; tin-plated iron, once scratched, rusts faster than bare iron. Explain using $E^{\circ}(\mathrm{Sn^{2+}/Sn})=-0.14\ \mathrm{V}$ and the table.
💡 Show solution▶
(a) Mg ($-2.37$), Zn ($-0.76$) and Fe ($-0.44$) have negative $E^{\circ}$ and liberate $\mathrm{H_{2}}$; Cu ($+0.34$) and Ag ($+0.80$) do not.
(b) No: $\mathrm{Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu}$ has $E^{\circ}_{\text{cell}}=+1.10\ \mathrm{V}$; the zinc would dissolve and copper would deposit.
(c) Zinc ($-0.76$) is a stronger reducing agent than iron ($-0.44$): where both are exposed, zinc is oxidised in preference and supplies electrons to the iron, so the iron is protected (sacrificial protection). Tin ($-0.14$) is a weaker reducing agent than iron, so at a scratch the iron becomes the anode and corrodes, faster than it would alone.
✏️ Example 6 — Ranking oxidising and reducing agents
(a) Arrange $\mathrm{Cl_{2}}$, $\mathrm{Fe^{3+}}$, $\mathrm{Ag^{+}}$, $\mathrm{Cr_{2}O_{7}^{2-}}$ (in acid) and $\mathrm{I_{2}}$ in decreasing order of oxidising power. (b) Arrange $\mathrm{Br^{-}}$, Zn, Fe, $\mathrm{I^{-}}$ and Li in decreasing order of reducing power. (c) Which species in (a) can oxidise Ag to $\mathrm{Ag^{+}}$?
💡 Show solution▶
(a) Order of $E^{\circ}$: $\mathrm{Cl_{2}}$ ($1.36$) $\gt$ $\mathrm{Cr_{2}O_{7}^{2-}}$ ($1.33$) $\gt$ $\mathrm{Ag^{+}}$ ($0.80$) $\gt$ $\mathrm{Fe^{3+}}$ ($0.77$) $\gt$ $\mathrm{I_{2}}$ ($0.54$).
(b) Reducing power increases as $E^{\circ}$ becomes more negative: Li ($-3.05$) $\gt$ Zn ($-0.76$) $\gt$ Fe ($-0.44$) $\gt$ $\mathrm{I^{-}}$ ($+0.54$) $\gt$ $\mathrm{Br^{-}}$ ($+1.09$).
(c) Only couples above $\mathrm{Ag^{+}/Ag}$ ($+0.80$): $\mathrm{Cl_{2}}$ and acidified $\mathrm{Cr_{2}O_{7}^{2-}}$. ($\mathrm{Fe^{3+}}$, at $0.77$, is just below: under standard conditions it cannot.)
🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — EMF and Gibbs energy
Class 12 work, not in the Class 11 syllabus. CBSE teaches it in Class 12 electrochemistry, and JEE Main, NEET and JEE Advanced all list the link between cell potential and Gibbs energy.
The electrical work a cell can do links its EMF to the Gibbs energy of Section 5.4: $\Delta_{r}G^{\circ}=-nFE^{\circ}_{\text{cell}}$, where $n$ is the number of moles of electrons transferred in the equation as written and $F=96\,485\ \mathrm{C\,mol^{-1}}$ is the Faraday constant (the charge on one mole of electrons). A positive $E^{\circ}_{\text{cell}}$ therefore means a negative $\Delta_{r}G^{\circ}$ — a spontaneous reaction. For the Daniell cell ($n=2$): $\Delta_{r}G^{\circ}=-2\times96\,485\times1.10$ $=-2.12\times10^{5}\ \mathrm{J\,mol^{-1}}$ $=-212\ \mathrm{kJ\,mol^{-1}}$. With $\Delta_{r}G^{\circ}=-RT\ln K$, this gives $K\approx1.6\times10^{37}$ at $298\ \mathrm{K}$: the displacement of copper by zinc goes essentially to completion. (Class 12 Chemistry develops these relations, and the Nernst equation for non-standard concentrations.)
Interactive 2 — Electrochemical Series: Will It React?
Pick an oxidising agent (an oxidised form, left of a couple) and a reducing agent (a reduced form, right of a couple). On the ladder of standard reduction potentials the reaction is spontaneous when the oxidising agent's couple lies above the reducing agent's couple — the electron arrow from the reductant's row to the oxidant's row then points upwards (green); a downward arrow (red, dashed) means no reaction. The lab writes the balanced equation from the two half-reactions and gives $E^{\circ}_{\text{cell}}$. ▶ Play tours a set of pairs.
E° (oxidant couple)—
E° (reductant couple)—
E°cell—
reaction—
Exercises
I
Galvanic cells and the salt bridge.
1 Define the anode and the cathode of a galvanic cell and give the sign of each.
2 State two functions of a salt bridge.
3 Describe the Daniell cell and write it in cell notation.
4 For $\mathrm{Zn(s)\,|\,Zn^{2+}(aq)\,\|\,Ag^{+}(aq)\,|\,Ag(s)}$, write the two half-reactions and the overall cell reaction, and give the direction of electron flow in the wire.
5 Why can the potential of a single electrode not be measured on its own?
6 As a Daniell cell is used, what happens to the concentrations of $\mathrm{Zn^{2+}}$ and $\mathrm{Cu^{2+}}$?
7 In a copper–silver cell, the silver electrode gains $1.079\ \mathrm{g}$. How many moles of electrons passed through the wire, and what mass of copper dissolved? ($M$: Ag $107.9$, Cu $63.55\ \mathrm{g\,mol^{-1}}$)
💡 Show solutions▶
1 Anode: the electrode where oxidation occurs — negative in a galvanic cell. Cathode: where reduction occurs — positive.
2 It completes the electrical circuit (ions carry the charge between the solutions) and keeps both solutions electrically neutral; it also stops the two solutions from mixing.
3 Zn rod in $1\ \mathrm{M}$ $\mathrm{ZnSO_{4}}$ and Cu rod in $1\ \mathrm{M}$ $\mathrm{CuSO_{4}}$, joined by a wire and a salt bridge; $1.10\ \mathrm{V}$. $\mathrm{Zn(s)\,|\,Zn^{2+}(aq)\,\|\,Cu^{2+}(aq)\,|\,Cu(s)}$.
4 $\mathrm{Zn\rightarrow Zn^{2+}+2e^{-}}$; $\mathrm{2Ag^{+}+2e^{-}\rightarrow 2Ag}$; $\mathrm{Zn(s)+2Ag^{+}(aq)\rightarrow}$ $\mathrm{Zn^{2+}(aq)+2Ag(s)}$. Electrons flow from Zn to Ag.
5 A voltmeter measures a difference between two points; connecting it to an electrode and its solution creates a second electrode–solution interface. Only differences between two half-cells can be measured, hence a reference electrode.
6 $[\mathrm{Zn^{2+}}]$ increases and $[\mathrm{Cu^{2+}}]$ decreases (the blue colour fades); the EMF falls slowly.
7 $n(\mathrm{Ag})=1.079/107.9=0.01000\ \mathrm{mol}$, so $0.01000\ \mathrm{mol}$ of electrons; copper dissolved $=0.00500\ \mathrm{mol}\times63.55=0.318\ \mathrm{g}$.
II
Electrode potentials and cell EMF (use the table).
1 Describe the standard hydrogen electrode and state the conditions under which its potential is taken as zero.
2 Find $E^{\circ}_{\text{cell}}$ for $\mathrm{Zn(s)\,|\,Zn^{2+}(aq)\,\|\,Ag^{+}(aq)\,|\,Ag(s)}$.
3 Write the balanced cell reaction and find $E^{\circ}_{\text{cell}}$ for $\mathrm{Al(s)\,|\,Al^{3+}(aq)\,\|\,Cu^{2+}(aq)\,|\,Cu(s)}$.
4 Find $E^{\circ}_{\text{cell}}$ for $\mathrm{Pb(s)\,|\,Pb^{2+}(aq)\,\|\,Ag^{+}(aq)\,|\,Ag(s)}$.
5 A magnesium half-cell is connected to the standard hydrogen electrode. Which electrode is the anode, and what does the voltmeter read?
6 A cell is built from $\mathrm{Fe^{2+}/Fe}$ and $\mathrm{Ag^{+}/Ag}$. Identify the anode, write the cell notation and find $E^{\circ}_{\text{cell}}$.
7 The cell $\mathrm{Cd(s)\,|\,Cd^{2+}(aq)\,\|\,Cu^{2+}(aq)\,|\,Cu(s)}$ has $E^{\circ}_{\text{cell}}=0.74\ \mathrm{V}$. Find $E^{\circ}(\mathrm{Cd^{2+}/Cd})$.
8 In part 3 the copper half-reaction is multiplied by 3. Why is $E^{\circ}(\mathrm{Cu^{2+}/Cu})$ not multiplied by 3?
💡 Show solutions▶
1 A platinised platinum electrode in a solution of $[\mathrm{H^{+}}]=1\ \mathrm{mol\,L^{-1}}$, with $\mathrm{H_{2}}$ gas at $1\ \mathrm{bar}$ bubbling over it ($298\ \mathrm{K}$); for $\mathrm{2H^{+}+2e^{-}\rightarrow H_{2}}$, $E^{\circ}=0$ by convention.
5 Magnesium is the anode ($-2.37$ is below $0$); the voltmeter reads $2.37\ \mathrm{V}$.
6 Iron is the anode: $\mathrm{Fe(s)\,|\,Fe^{2+}(aq)\,\|\,Ag^{+}(aq)\,|\,Ag(s)}$; $E^{\circ}_{\text{cell}}=0.80-(-0.44)=1.24\ \mathrm{V}$.
7 $0.74=0.34-E^{\circ}(\mathrm{Cd^{2+}/Cd})$, so $E^{\circ}=-0.40\ \mathrm{V}$.
8 $E^{\circ}$ is an intensive property — a potential (energy per unit charge), which does not depend on how much substance reacts. Tripling the equation triples the charge transferred and the energy, but not the energy per coulomb.
III
Predictions from the electrochemical series (standard conditions).
1 Name the strongest oxidising agent and the strongest reducing agent in the table.
2 Can $\mathrm{Fe^{3+}}$ oxidise copper metal? If so, write the equation and give $E^{\circ}_{\text{cell}}$.
3 Can iodine oxidise $\mathrm{Fe^{2+}}$ to $\mathrm{Fe^{3+}}$?
4 Can $\mathrm{Ag^{+}}$ oxidise $\mathrm{Fe^{2+}}$? Comment on the size of $E^{\circ}_{\text{cell}}$.
5 Will a nickel rod displace lead from lead(II) nitrate solution? Give $E^{\circ}_{\text{cell}}$.
6 Which of Ni, Ag, Al and Pb can displace hydrogen from $1\ \mathrm{mol\,L^{-1}}$ acid?
7 Arrange Mg, Fe, Cu, Al and Ag in increasing order of reducing power.
8 Which of $\mathrm{Cl_{2}}$, $\mathrm{Br_{2}}$ and $\mathrm{Fe^{3+}}$ can oxidise $\mathrm{I^{-}}$ but not $\mathrm{Br^{-}}$?
3 No: $0.54\lt0.77$ (the reverse reaction, $\mathrm{Fe^{3+}}$ oxidising $\mathrm{I^{-}}$, is the spontaneous one).
4 Yes, but only just: $E^{\circ}_{\text{cell}}=0.80-0.77=0.03\ \mathrm{V}$. A small change in concentrations could reverse the direction.
5 Yes: Ni ($-0.25$) is below Pb ($-0.13$). $\mathrm{Ni(s)+Pb^{2+}(aq)\rightarrow Ni^{2+}(aq)+Pb(s)}$; $E^{\circ}_{\text{cell}}=0.12\ \mathrm{V}$.
6 Ni, Al and Pb (negative $E^{\circ}$); not Ag.
7 Ag $\lt$ Cu $\lt$ Fe $\lt$ Al $\lt$ Mg.
8 $\mathrm{Br_{2}}$ ($+1.09$: above $\mathrm{I_{2}/I^{-}}$, and it cannot oxidise its own ion) and $\mathrm{Fe^{3+}}$ ($+0.77$: between $+0.54$ and $+1.09$). $\mathrm{Cl_{2}}$ oxidises both.
IV
Applications.
1 Explain why zinc protects iron from rusting even where a zinc coating is scratched.
2 Blocks of zinc or magnesium are bolted to the steel hulls of ships. Why would blocks of copper be a disastrous choice?
3 Printed circuit boards are etched with iron(III) chloride solution, which dissolves unwanted copper. Write the ionic equation and use $E^{\circ}$ values to show that it is spontaneous.
4 $E^{\circ}(\mathrm{Al^{3+}/Al})=-1.66\ \mathrm{V}$, yet aluminium drinks cans do not dissolve in acidic soft drinks. Suggest why.
5Applied. A student builds a Daniell cell in the laboratory and measures $1.07\ \mathrm{V}$ instead of $1.10\ \mathrm{V}$. Give two possible reasons.
💡 Show solutions▶
1 $E^{\circ}(\mathrm{Zn^{2+}/Zn})=-0.76$ is below $E^{\circ}(\mathrm{Fe^{2+}/Fe})=-0.44$, so zinc is the stronger reducing agent; in contact with iron it becomes the anode and is oxidised instead of the iron (sacrificial protection).
2 Copper ($+0.34$) lies above iron: in seawater the iron would become the anode and corrode faster. Zinc and magnesium lie below iron and corrode in its place.
4 The can is covered by a thin, tightly bound layer of aluminium oxide (and an inner lacquer), which stops the reaction; $E^{\circ}$ predicts what is possible, not the rate.
5 Any two: the solutions are not exactly $1\ \mathrm{mol\,L^{-1}}$; the temperature is not $298\ \mathrm{K}$; the voltmeter draws some current, so part of the EMF is lost across the internal resistance of the cell and bridge; the metal surfaces are oxidised or impure.
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E1.NEET Given $E^{\circ}$: $\mathrm{Zn^{2+}/Zn}$ $-0.76\ \mathrm{V}$, $\mathrm{Cu^{2+}/Cu}$ $+0.34\ \mathrm{V}$, $\mathrm{Ag^{+}/Ag}$ $+0.80\ \mathrm{V}$, $\mathrm{Mg^{2+}/Mg}$ $-2.37\ \mathrm{V}$. The strongest reducing agent is (A) $\mathrm{Ag}$ (B) $\mathrm{Mg}$ (C) $\mathrm{Mg^{2+}}$ (D) $\mathrm{Ag^{+}}$
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Idea Reducing agents are the reduced forms (metals here); the most negative $E^{\circ}$ marks the strongest. $\mathrm{Mg^{2+}}$ and $\mathrm{Ag^{+}}$ are oxidised forms; $\mathrm{Ag^{+}}$ is the strongest oxidising agent of the list.
Answer(B).
E2.JEE MainNumerical answer. For the cell $\mathrm{Al(s)\,|\,Al^{3+}(aq)\,\|\,Fe^{2+}(aq)\,|\,Fe(s)}$ under standard conditions, with $E^{\circ}(\mathrm{Al^{3+}/Al})=-1.66\ \mathrm{V}$ and $E^{\circ}(\mathrm{Fe^{2+}/Fe})=-0.44\ \mathrm{V}$, the value of $100\times E^{\circ}_{\text{cell}}$ (in V) is ____.
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Rule 7.6 $E^{\circ}_{\text{cell}}=E^{\circ}_{\text{cathode}}-E^{\circ}_{\text{anode}}=-0.44-(-1.66)=1.22\ \mathrm{V}$. The coefficients of $\mathrm{2Al+3Fe^{2+}\rightarrow 2Al^{3+}+3Fe}$ do not enter.
Answer122.
E3.Board · 3 marks (a) Describe the standard hydrogen electrode with the help of its half-reaction. (b) How is it used to find the standard electrode potential of zinc? (c) Why is a reference electrode needed at all?
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1 mark Platinised Pt electrode in $1\ \mathrm{mol\,L^{-1}}$ $\mathrm{H^{+}}$ with $\mathrm{H_{2}}$ at $1\ \mathrm{bar}$, $298\ \mathrm{K}$; $\mathrm{2H^{+}(aq)+2e^{-}\rightarrow H_{2}(g)}$, $E^{\circ}=0.00\ \mathrm{V}$ by convention.
1 mark Connect it to $\mathrm{Zn\,|\,Zn^{2+}(1\ M)}$ through a salt bridge: the cell reads $0.76\ \mathrm{V}$ with zinc negative (anode), so $0.76=0-E^{\circ}(\mathrm{Zn^{2+}/Zn})$ and $E^{\circ}=-0.76\ \mathrm{V}$.
1 mark Only potential differences between two electrodes can be measured, so one electrode must be assigned an agreed value.
E4.JEE Main Using $E^{\circ}$: $\mathrm{Br_{2}/Br^{-}}$ $+1.09$, $\mathrm{Fe^{3+}/Fe^{2+}}$ $+0.77$, $\mathrm{I_{2}/I^{-}}$ $+0.54$, $\mathrm{Ag^{+}/Ag}$ $+0.80$, $\mathrm{Cu^{2+}/Cu}$ $+0.34\ \mathrm{V}$, the reaction that is not spontaneous under standard conditions is (A) $\mathrm{Cu+2Ag^{+}\rightarrow Cu^{2+}+2Ag}$ (B) $\mathrm{2Fe^{3+}+2I^{-}\rightarrow 2Fe^{2+}+I_{2}}$ (C) $\mathrm{Br_{2}+2Fe^{2+}\rightarrow 2Br^{-}+2Fe^{3+}}$ (D) $\mathrm{2Fe^{3+}+2Br^{-}\rightarrow 2Fe^{2+}+Br_{2}}$
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Check each (A) $0.80-0.34=+0.46$; (B) $0.77-0.54=+0.23$; (C) $1.09-0.77=+0.32$; (D) $0.77-1.09=-0.32$.
Answer(D) — it is the reverse of (C).
E5.JEE AdvancedOne or more options correct. For a working Daniell cell: (A) electrons flow from zinc to copper in the external circuit; (B) anions of the salt bridge move into the copper half-cell; (C) the mass of the copper electrode increases; (D) $E^{\circ}_{\text{cell}}=1.10\ \mathrm{V}$ and $\Delta_{r}G^{\circ}$ for the cell reaction is negative.
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(A) True: Zn is the anode.
(B) False: anions move towards the anode (zinc) compartment, where $\mathrm{Zn^{2+}}$ is being formed.
(C) True: copper deposits on the cathode.
(D) True: $0.34-(-0.76)=1.10\ \mathrm{V}$, and $\Delta_{r}G^{\circ}=-nFE^{\circ}_{\text{cell}}=-212\ \mathrm{kJ\,mol^{-1}}\lt0$.
Answer(A), (C), (D).
E6.Board · Case study Steel piers and ship hulls in seawater corrode. Engineers attach blocks of a more reactive metal that corrodes instead of the steel. Use $E^{\circ}$: $\mathrm{Zn^{2+}/Zn}$ $-0.76$, $\mathrm{Fe^{2+}/Fe}$ $-0.44$, $\mathrm{Cu^{2+}/Cu}$ $+0.34\ \mathrm{V}$. (i) Which of zinc and copper should be used for the blocks? Explain. (2 marks) (ii) Write the half-reaction at the block. (1 mark) (iii) Find $E^{\circ}_{\text{cell}}$ for a cell made from the block metal and the $\mathrm{Fe^{2+}/Fe}$ couple. (1 mark)
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(i) Zinc. Its $E^{\circ}$ is lower than iron's, so it is the stronger reducing agent and becomes the anode, giving up electrons to the steel, which becomes the cathode and is protected. Copper, with a higher $E^{\circ}$, would make the steel the anode and speed up its corrosion.
(ii) $\mathrm{Zn(s)\rightarrow Zn^{2+}(aq)+2e^{-}}$ (oxidation at the anode).
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes
Research question. Offshore wind turbines, oil platforms, ships and coastal piers are made of steel standing in seawater for 25 years or more. Which metal makes the best sacrificial anode — magnesium, zinc or aluminium — and how much metal does each consume per mole of electrons it supplies?
Why it matters. Corrosion is estimated to cost several per cent of the world's GDP each year. Cathodic protection with sacrificial anodes guards the foundations of the North Sea wind farms, the Mumbai High oil field, the hulls of container ships and buried gas pipelines; a poor choice of anode metal means either wasted metal or rusting steel.
What to do.
From the table, find the "driving voltage" $E^{\circ}(\mathrm{Fe^{2+}/Fe})-E^{\circ}(\mathrm{M^{n+}/M})$ for $\mathrm{M}$ = Mg, Zn and Al.
Write each anode half-reaction and find the mass of metal consumed per mole of electrons supplied ($M$: Mg $24.31$, Zn $65.38$, Al $26.98\ \mathrm{g\,mol^{-1}}$).
Find out why magnesium anodes are used in soil and fresh water but rarely in seawater, and why pure aluminium does not work well (hint: what happens to its surface?). What do industrial "Al–Zn–In" anode alloys achieve?
Rank the three metals for (a) a buried pipeline in dry soil, (b) a ship hull, (c) an offshore wind-turbine foundation.
Tools: Rule 7.7 and the $E^{\circ}$ table, a cathodic-protection standard or manufacturer's data sheet, a spreadsheet.
Hand in: A table of driving voltage and mass per mole of electrons for the three metals, and a 200-word recommendation for one structure.
Going further. Some structures are protected instead by "impressed current": a power supply drives electrons into the steel through an inert anode. Compare the two approaches for a 100-metre pier.
Instructor note: driving voltages $1.93\ \mathrm{V}$ (Mg), $0.32\ \mathrm{V}$ (Zn), $1.22\ \mathrm{V}$ (Al). Mass per mole of electrons: Mg $24.31/2=12.2\ \mathrm{g}$, Zn $65.38/2=32.7\ \mathrm{g}$, Al $26.98/3=8.99\ \mathrm{g}$ — aluminium supplies the most charge per kilogram. Magnesium's large driving voltage makes it corrode rapidly and evolve hydrogen in conductive seawater (wasteful, can damage coatings), but suits high-resistance soil and fresh water. Pure Al passivates (oxide film); alloying with Zn and In keeps it active — the usual choice offshore. Zinc is reliable for ships and is self-limiting.
🧪 Self-check quiz
0/5 correct
Q1. In a galvanic cell, the anode is the electrode at which
Q2. With $E^{\circ}(\mathrm{Ag^{+}/Ag})=+0.80\ \mathrm{V}$ and $E^{\circ}(\mathrm{Cu^{2+}/Cu})=+0.34\ \mathrm{V}$, $E^{\circ}_{\text{cell}}$ for $\mathrm{Cu(s)+2Ag^{+}(aq)\rightarrow}$ $\mathrm{Cu^{2+}(aq)+2Ag(s)}$ is
Q3. The potential of the standard hydrogen electrode is zero because
Q4. Which metal will not liberate hydrogen from $1\ \mathrm{mol\,L^{-1}}$ hydrochloric acid?
Q5. If the salt bridge of a working Daniell cell is removed,