Four useful ways to say the same redox story

A redox reaction is a chemical reaction in which oxidation and reduction occur together. The old classroom phrase still works: you can’t have one without the other, because electrons, oxygen or hydrogen are being redistributed between species.

Oxidation is a chemical change in which a species loses electrons, increases its oxidation state, gains oxygen, or loses hydrogen. Reduction is a chemical change in which a species gains electrons, decreases its oxidation state, loses oxygen, or gains hydrogen. Use electron transfer as your first definition. The oxygen and hydrogen versions, though, are especially useful in organic chemistry and in some older inorganic examples.

For example, magnesium is oxidized when it burns:

$2Mg(s) + O_2(g) \to 2MgO(s)$

Copper(II) oxide is reduced by hydrogen because it loses oxygen. Hydrogen is oxidized because it gains oxygen:

$CuO(s) + H_2(g) \to Cu(s) + H_2O(g)$

In the formation of sodium chloride, sodium atoms lose electrons and chlorine molecules gain them:

$2Na(s) \to 2Na^+ + 2e^-$

$Cl_2(g) + 2e^- \to 2Cl^-$

Oxidation states: bookkeeping, not tiny real charges

An oxidation state is a formal number assigned to an atom that represents the charge it would have if bonding electrons were assigned to the more electronegative atom. Treat it as a model, not a direct measurement of charge. That’s why it’s useful: it helps us track redox changes in covalent and ionic substances. The trap is thinking that atoms in molecules literally carry those full charges.

The main rules you need are these:

Transition element ions commonly have variable oxidation states, so Roman numerals are used in names such as iron(II) sulfate and iron(III) chloride. Many main-group non-metals also show variable oxidation states: sulfur is $-2$ in $H_2S$, $+4$ in $SO_2$, and $+6$ in $SO_4^{2-}$; chlorine is $-1$ in $Cl^-$, $0$ in $Cl_2$, $+1$ in $ClO^-$, $+5$ in $ClO_3^-$, and $+7$ in $ClO_4^-$. In names such as chlorate(V) and manganate(VII), the Roman numeral gives the oxidation state of the named element.

Oxidized species, reduced species, agents

An oxidizing agent is a reactant that causes another species to be oxidized by accepting electrons from it; the oxidizing agent is itself reduced. A reducing agent is a reactant that causes another species to be reduced by donating electrons to it; the reducing agent is itself oxidized.

In:

$Fe(s) + 2HBr(aq) \to FeBr_2(aq) + H_2(g)$

iron changes from $0$ in $Fe(s)$ to $+2$ in $FeBr_2(aq)$, so iron is oxidized and is the reducing agent. Hydrogen changes from $+1$ in $HBr(aq)$ to $0$ in $H_2(g)$, so $HBr(aq)$ is reduced and is the oxidizing agent. When naming agents, say the species, not just the atom.

Surface oxidation of metals is often called corrosion, a redox process in which a metal is oxidized by substances in its environment. The effects go beyond appearance: corrosion weakens structures, damages vehicles and pipelines, contaminates products, and creates large economic and safety costs.

Why we split redox equations

A half-equation shows just one side of a redox reaction: either oxidation or reduction, with the electrons lost or gained included. In an oxidation half-equation, the electrons go on the right. In a reduction half-equation, they go on the left.

For sodium reacting with chlorine:

$Na(s) \to Na^+ + e^-$ oxidation

$Cl_2(g) + 2e^- \to 2Cl^-$ reduction

You only get the full equation once the electrons lost and gained are equal. The electrons then cancel; they must not be left in the final overall equation.

Balancing in acidic and neutral solution

Use this routine. It is dull, but it works.

1. Identify what is oxidized and what is reduced.
2. Write the two skeletal half-equations.
3. Balance the atoms whose oxidation states change.
4. Add electrons to account for the oxidation-state change.
5. Multiply half-equations so the electron numbers match.
6. Add and cancel electrons.
7. In acidic solution, balance oxygen with $H_2O(l)$ and hydrogen with $H^+(aq)$.
8. In neutral solution, water and hydroxide ions may be needed; check atoms and charge carefully.
9. Finish by checking both atoms and total charge.

For example, in acidic solution, iron(II) ions reduce dichromate(VI) ions:

Oxidation:

$Fe^{2+}(aq) \to Fe^{3+}(aq) + e^-$

Reduction after balancing atoms, oxygen, hydrogen and charge:

$Cr_2O_7^{2-}(aq) + 14H^+(aq) + 6e^- \to 2Cr^{3+}(aq) + 7H_2O(l)$

Overall:

$6Fe^{2+}(aq) + Cr_2O_7^{2-}(aq) + 14H^+(aq) \to 6Fe^{3+}(aq) + 2Cr^{3+}(aq) + 7H_2O(l)$

Redox titrations and self-indication

A redox titration is a titration where the analyte and titrant react by electron transfer. Some redox titrations are self-indicating: one reactant has a strong colour, so the endpoint appears as a lasting colour change without a separate indicator.

A typical example uses acidified manganate(VII), $MnO_4^-(aq)$, reacting with iron(II), $Fe^{2+}(aq)$. Purple $MnO_4^-(aq)$ is reduced to very pale $Mn^{2+}(aq)$, and $Fe^{2+}(aq)$ is oxidized to $Fe^{3+}(aq)$. While iron(II) is still present, any added manganate(VII) is decolourized. The first permanent faint pink means there is a tiny excess of manganate(VII). That’s why, in the lab, the white tile and careful dropwise addition near the end matter.

Metals: easier oxidation down the reactive groups

The relative ease of oxidation compares how readily species lose electrons. In Group 1, atoms lose their outer electron more readily as you go down the group, so oxidation becomes easier from lithium to caesium. The structure explains the trend: the outer electron sits further from the nucleus and is more shielded, so the attraction is weaker.

A metal that oxidizes more easily acts as a stronger reducing agent. If metal $M$ displaces metal ions $N^+$ from solution, then $M$ has been oxidized, so it is the more reactive metal in that pair.

For example:

$Zn(s) + Cu^{2+}(aq) \to Zn^{2+}(aq) + Cu(s)$

$Zn(s)$ is oxidized, so zinc is more easily oxidized than copper. In the lab, the zinc surface would change and reddish-brown copper would form; the blue colour from $Cu^{2+}(aq)$ fades as copper(II) ions are removed.

Metal displacement results used to rank ease of oxidation.

Examination questions will give the metal/metal ion data if you need a reactivity order. Read the table carefully: a “reaction” shows that the solid metal is more easily oxidized than the metal whose ions are in solution.

Halogens: easier reduction up the group

The relative ease of reduction compares how readily species gain electrons. Halogens are reduced from $X_2$ to $X^-$, and their oxidizing power increases up Group 17. Fluorine is the most easily reduced halogen; iodine is the least easily reduced of the common halogens.

So chlorine can oxidize bromide ions:

$Cl_2(aq) + 2Br^-(aq) \to 2Cl^-(aq) + Br_2(aq)$

but bromine cannot oxidize chloride ions under the same conditions. In halogen/halide mixtures, the observations often come down to colour changes. Chlorine, bromine and iodine solutions have distinctive colours, and if a new halogen appears in solution, that is evidence of displacement.

Non-metal reactivity in a group often decreases down the group because gaining an electron becomes less favourable when the outer shell is further from the nucleus and more shielded. That structural point explains the halogen trend.

Metal + acid gives salt + hydrogen

Reactive metals react with dilute acids by being oxidized. The acid provides $H^+(aq)$ ions, and these ions gain electrons to form hydrogen gas:

$2H^+(aq) + 2e^- \to H_2(g)$

For a metal that forms $M^{2+}$ ions, the reaction follows this pattern:

$M(s) + 2H^+(aq) \to M^{2+}(aq) + H_2(g)$

With hydrochloric acid:

$Mg(s) + 2HCl(aq) \to MgCl_2(aq) + H_2(g)$

$Zn(s) + 2HCl(aq) \to ZnCl_2(aq) + H_2(g)$

With dilute sulfuric acid:

$Mg(s) + H_2SO_4(aq) \to MgSO_4(aq) + H_2(g)$

$Zn(s) + H_2SO_4(aq) \to ZnSO_4(aq) + H_2(g)$

The metal acts as the reducing agent because it loses electrons. The acid acts as the oxidizing agent because $H^+(aq)$ gains electrons. Metals below hydrogen in the reactivity series, such as copper and silver, don't react with dilute hydrochloric or sulfuric acid in this way.

Hydrogen is identified using the pop test: a lit splint ignites hydrogen gas, giving a squeaky pop as the hydrogen reacts rapidly with oxygen in the air.

Anode and cathode are named by reaction, not by sign

An electrochemical cell is a device that uses redox reactions to interconvert chemical energy and electrical energy. The electrode names come from the reaction happening at each electrode.

An anode is an electrode at which oxidation occurs. A cathode is an electrode at which reduction occurs. Keep it simple: anode = oxidation; cathode = reduction. “Red Cat” is a useful memory trick: reduction at the cathode.

The sign of each electrode changes with the type of cell:

Don’t learn “anode is positive” or “anode is negative” as a universal rule. That’s the trap. First name the electrode from the process, then use the type of cell to assign the sign.

Building a primary cell

A primary cell is an electrochemical cell that uses a spontaneous redox reaction to produce electrical energy and is not designed to be recharged. A voltaic cell is an electrochemical cell in which a spontaneous redox reaction drives electron flow through an external circuit.

In a simple metal/metal ion cell, there are two half-cells. Each one has a metal electrode placed in a solution containing ions of that metal. An external wire joins the half-cells, and so does a salt bridge, an ion-conducting connection that completes the circuit while limiting mixing of the two solutions.

In a zinc–copper cell:

Anode, oxidation:

$Zn(s) \to Zn^{2+}(aq) + 2e^-$

Cathode, reduction:

$Cu^{2+}(aq) + 2e^- \to Cu(s)$

Overall:

$Zn(s) + Cu^{2+}(aq) \to Zn^{2+}(aq) + Cu(s)$

Electrons move through the external circuit from the zinc anode to the copper cathode. Conventional current is described in the opposite direction, but in chemistry we usually follow the electrons.

The salt bridge is doing real work. As zinc is oxidized, positive charge builds up in the anode half-cell, so anions from the salt bridge move toward the anode. As copper(II) ions are reduced, the cathode half-cell would become relatively negative, so cations from the salt bridge move toward the cathode. Without this ion movement, charge separation stops the reaction.

A cell diagram puts the anode on the left and the cathode on the right. A single vertical line represents a phase boundary; a double vertical line represents the salt bridge:

$Zn(s) \mid Zn^{2+}(aq) \mid\mid Cu^{2+}(aq) \mid Cu(s)$

Electrical energy can come from combustion indirectly—burn fuel, heat water, turn turbines—or directly from electrochemical reactions. Both involve redox and energy transfer, but electrochemical cells keep oxidation and reduction separate, letting electrons do useful work in an external circuit instead of releasing energy mainly as heat.

Reversible electrode reactions

A secondary cell is an electrochemical cell where the discharge redox reactions can be reversed by supplying electrical energy. On discharge, it behaves like a voltaic cell. On charging, it behaves like an electrolytic cell being driven in the opposite direction.

When discharge half-equations are given, you get the charging half-equations by reversing them. Electrons move to the other side, reactants and products change places, and the overall equation reverses too. No new chemistry is happening here; it’s Le Châtelier’s principle with an electrical push, as the imposed current drives the system away from the discharged state.

For a simplified lithium-ion example during discharge:

Anode:

$Li(s) \to Li^+ + e^-$

Cathode:

$Li^+ + e^- + CoO_2(s) \to LiCoO_2(s)$

During charging:

$Li^+ + e^- \to Li(s)$

$LiCoO_2(s) \to Li^+ + e^- + CoO_2(s)$

Primary cells, secondary cells and fuel cells

A fuel cell is an electrochemical cell that produces electrical energy from a continuous supply of fuel and oxidant. You don’t recharge it like a secondary cell; instead, fresh reactants are supplied.

In a hydrogen fuel cell, hydrogen is oxidized at the anode and oxygen is reduced at the cathode:

$H_2(g) \to 2H^+(aq) + 2e^-$

$O_2(g) + 4H^+(aq) + 4e^- \to 2H_2O(l)$

Overall:

$2H_2(g) + O_2(g) \to 2H_2O(l)$

A proton exchange membrane lets $H^+$ pass, but stops electrons and gases from crossing directly. Electrons therefore have to travel through the external circuit, giving the useful electrical output.

Electrolysis: forcing a redox reaction

An electrolytic cell is an electrochemical cell where electrical energy drives a non-spontaneous redox reaction. Electrolysis is the process of using an electric current to cause a chemical change.

Most setups use one container with an electrolyte, a substance containing mobile ions that conduct charge, plus two electrodes joined to a direct current power source. Ionic compounds work as electrolytes when molten or dissolved in water because their ions can move. In a solid ionic lattice, they don’t conduct, since the ions are fixed in position.

In an electrolytic cell, the cathode connects to the negative terminal of the power supply. Cations move to the cathode and are reduced. Anions move to the anode and are oxidized. Electrons carry current through the wires; ions carry it through the electrolyte.

For molten sodium chloride, only $Na^+$ and $Cl^-$ are present:

Cathode:

$$Na^+(l) + e^- \to Na(l)$$

Anode:

$$2Cl^-(l) \to Cl_2(g) + 2e^-$$

Overall:

$$2NaCl(l) \to 2Na(l) + Cl_2(g)$$

The same reasoning applies to any molten salt: the metal cation is reduced to the metal, while the non-metal anion is oxidized to the element. In molten lead(II) bromide, for example, lead forms at the cathode and bromine forms at the anode.

Oxidizing alcohol functional groups

A functional group is a particular atom, or group of atoms, in an organic molecule that gives the molecule its typical chemical reactions. For alcohol oxidation, look at the carbon carrying the $-OH$ group: how many carbon atoms are attached to it, and does it still have a hydrogen attached?

In organic equations, $[O]$ is used as shorthand for an oxidizing agent. The names and formulas of particular oxidizing agents are not assessed here, and neither is the mechanism.

A primary alcohol oxidizes first to an aldehyde, then to a carboxylic acid:

$RCH_2OH + [O] \to RCHO + H_2O$

$RCHO + [O] \to RCOOH$

Overall:

$RCH_2OH + 2[O] \to RCOOH + H_2O$

A secondary alcohol oxidizes to a ketone:

$RCH(OH)R' + [O] \to RCOR' + H_2O$

A tertiary alcohol is not oxidized under similar conditions because the carbon bonded to the $-OH$ has no hydrogen atom to remove. That’s a structural reason, not a magic exception.

Reflux and distillation

Reflux is a heating technique where vapour condenses and returns to the reaction flask, so the mixture can be heated for a long time without losing volatile reactants or products. Use it when the oxidation needs to continue, for example when converting a primary alcohol fully to a carboxylic acid, or when converting a secondary alcohol to a ketone. For complete oxidation of a primary alcohol, the oxidizing agent is in excess.

Distillation is a separation technique where a volatile product is vaporized, condensed and collected. It is used to obtain an aldehyde from a primary alcohol, because the aldehyde can be removed from the hot oxidizing mixture before further oxidation changes it into a carboxylic acid. In this case, the alcohol is usually in excess.

Changing the functional group changes intermolecular forces, so physical properties such as boiling point change as well. Alcohols and carboxylic acids can hydrogen bond strongly. Aldehydes and ketones have polar carbonyl groups but no $O-H$ bond for donating hydrogen bonds. Combustion of an alcohol is complete reaction with oxygen to form carbon dioxide and water, whereas controlled oxidation changes the functional group while keeping the carbon skeleton.

Reducing carbonyl and carboxyl groups

In organic equations, $[H]$ is shorthand for a reducing agent. Many reducing agents supply the hydride ion, an anion, $H^-, which donates a pair of electrons to an electron-poor carbon atom. You don’t need to know the names of specific reducing agents or the mechanisms, but you should know that hydride delivery is the key chemical role.

A ketone is reduced to a secondary alcohol:

$RCOR' + 2[H] \to RCH(OH)R'$

A carboxylic acid is reduced to a primary alcohol via an aldehyde:

$RCOOH + 2[H] \to RCHO + H_2O$

$RCHO + 2[H] \to RCH_2OH$

Overall:

$RCOOH + 4[H] \to RCH_2OH + H_2O$

Oxidation states show carbon becoming more oxidised in this sequence:

$CH_4 \to CH_3OH \to HCHO \to HCOOH \to CO_2$

For the carbon atom, the oxidation states are $-4$, $-2$, $0$, $+2$ and $+4$ respectively. Notice the pattern: more bonds from carbon to oxygen, or fewer bonds from carbon to hydrogen, usually gives carbon a higher oxidation state.

Hydrogenation lowers unsaturation

An unsaturated compound is an organic compound with at least one carbon–carbon double or triple bond that can add atoms across the multiple bond. Hydrogenation is an addition reaction where hydrogen is added across a multiple bond, reducing the compound.

Alkenes react with hydrogen to form alkanes:

$RCH=CHR' + H_2 \to RCH_2CH_2R'$

Alkynes can be reduced to alkenes using one mole of hydrogen:

$RC\equiv CR' + H_2 \to RCH=CHR'$

With excess hydrogen, alkynes are reduced further to alkanes:

$RC\equiv CR' + 2H_2 \to RCH_2CH_2R'$

This reaction is both reduction and addition. It counts as reduction because the carbon atoms gain hydrogen and their oxidation states decrease. It counts as addition because atoms are added across a carbon–carbon multiple bond. Later, in alkene chemistry, some reactions are described as electrophilic addition because the mechanism starts with attack by an electrophile; here, hydrogenation is classified by the redox change.

The standard hydrogen electrode is the reference point

A standard electrode potential is the potential difference between a half-cell and the standard hydrogen electrode under standard conditions, written as a reduction potential. Its symbol is

You’ll find standard reduction potentials in the data booklet.

The hydrogen half-cell is given this value by convention:

$$H^+(\text{aq}) + e^- \rightleftharpoons \frac{1}{2}H_2(\text{g}) \quad E^\circ = 0.00\ \mathrm{V}$$

A standard hydrogen electrode is a reference half-cell where hydrogen gas at standard pressure is in equilibrium with hydrogen ions at standard concentration on an inert platinum electrode. Platinum is used because there is no solid hydrogen electrode; it gives a conducting surface for electron transfer.

Standard conditions for electrode potentials are 298 K, 100 kPa for gases, and $1.0\ \mathrm{mol\,dm^{-3}}$ for aqueous ions.

Interpreting values

A more positive $E^\circ$ shows that the species on the left of the reduction half-equation is reduced more easily, so it is a stronger oxidizing agent. A more negative $E^\circ$ shows that the reduced form on the right is oxidized more easily, so it is a stronger reducing agent.

Take a half-cell with a very negative value, such as a Group 1 metal ion/metal half-cell. It matches a metal that is readily oxidized. A halogen half-cell with a very positive value matches a halogen that is readily reduced. These numbers are the reactivity trends you met earlier in electrochemical form.

Calculating cell potential

A standard cell potential is the potential difference between two standard half-cells in an electrochemical cell. Its symbol is $E^\circ_{\text{cell}}$, where $E^\circ_{\text{cell}}$ is the standard cell potential (V).

Use:

A positive $E^\circ_{\text{cell}}$ shows that the reaction is spontaneous in the direction written for the cell. A negative $E^\circ_{\text{cell}}$ shows that the reverse direction is spontaneous.

For a copper–silver cell:

$$\mathrm{Ag^+(aq)} + e^- \rightleftharpoons \mathrm{Ag(s)} \qquad E^\circ = +0.80\ \mathrm{V}$$ $$\mathrm{Cu^{2+}(aq)} + 2e^- \rightleftharpoons \mathrm{Cu(s)} \qquad E^\circ = +0.34\ \mathrm{V}$$

Silver has the more positive reduction potential, so silver ions are reduced at the cathode. Copper is oxidized at the anode:

$$\mathrm{Cu(s)} \to \mathrm{Cu^{2+}(aq)} + 2e^-$$ $$2\mathrm{Ag^+(aq)} + 2e^- \to 2\mathrm{Ag(s)}$$

Overall:

$$\mathrm{Cu(s)} + 2\mathrm{Ag^+(aq)} \to \mathrm{Cu^{2+}(aq)} + 2\mathrm{Ag(s)}$$ $$E^\circ_{\text{cell}} = (+0.80\ \mathrm{V}) - (+0.34\ \mathrm{V}) = +0.46\ \mathrm{V}$$

Don’t multiply $E^\circ$ values when you multiply half-equations. Potentials are not amounts of substance; they are energy per unit charge.

Linking electrochemistry and thermodynamics

The standard change in Gibbs energy is the Gibbs energy change for a reaction under standard conditions. Its symbol is $\Delta G^\circ$, where $\Delta G^\circ$ is standard Gibbs energy change ($\mathrm{J\ mol^{-1}}$). Cell potential links to it through:

You’ll find both the equation and $F$ in the data booklet. The negative sign matters: a positive $E^\circ_{\text{cell}}$ produces a negative $\Delta G^\circ$, so the reaction is thermodynamically spontaneous under standard conditions. That matches the Gibbs energy criterion, where negative $\Delta G^\circ$ indicates a spontaneous process.

The units check out too:

$\mathrm{C\ mol^{-1}} \times \mathrm{V} = \mathrm{C\ mol^{-1}} \times \mathrm{J\ C^{-1}} = \mathrm{J\ mol^{-1}}$

For the reaction:

$Zn(s) + 2H^+(aq) \to Zn^{2+}(aq) + H_2(g)$

if $E^\circ_{\text{cell}}$ is $+0.76\ \mathrm{V}$ and two electrons are transferred, then:

$\Delta G^\circ = -2 \times F \times 0.76$

Using $F = 9.65 \times 10^4\ \mathrm{C\ mol^{-1}}$, $\Delta G^\circ$ is about $-1.47 \times 10^5\ \mathrm{J\ mol^{-1}}$, or $-147\ \mathrm{kJ\ mol^{-1}}$.

Why aqueous electrolysis is not the same as molten electrolysis

In a molten salt, the salt’s ions are the only mobile ions present. In aqueous solution, water is there too, and it can be oxidized or reduced. So the electrode reactions can compete.

At the cathode, the metal ion and water both compete for reduction. Water can be reduced:

$H_2O(l) + e^- \to \frac{1}{2}H_2(g) + OH^-(aq) \qquad E^\circ = -0.83\ \text{V}$

If the metal ion has a much more negative reduction potential than this, water is reduced instead, producing hydrogen.

At the anode, the anion competes with water for oxidation. The data booklet gives oxygen reduction as:

$\frac{1}{2}O_2(g) + 2H^+(aq) + 2e^- \to H_2O(l) \qquad E^\circ = +1.23\ \text{V}$

Reversed for oxidation:

$H_2O(l) \to \frac{1}{2}O_2(g) + 2H^+(aq) + 2e^-$

Required cases: water, sodium chloride and copper(II) sulfate

During electrolysis of water, hydrogen forms at the cathode and oxygen forms at the anode. The overall reaction is:

$2H_2O(l) \to 2H_2(g) + O_2(g)$

For concentrated aqueous sodium chloride with inert electrodes, water is reduced at the cathode instead of sodium ions:

$2H_2O(l) + 2e^- \to H_2(g) + 2OH^-(aq)$

At the anode, concentration makes a difference. A concentrated chloride solution mainly produces chlorine:

$2Cl^-(aq) \to Cl_2(g) + 2e^-$

Overall:

$2NaCl(aq) + 2H_2O(l) \to H_2(g) + Cl_2(g) + 2NaOH(aq)$

In dilute sodium chloride, oxygen competes more strongly at the anode, so oxygen may be produced instead, or it may form as a mixture with chlorine depending on concentration.

For aqueous copper(II) sulfate with inert electrodes, copper(II) ions are reduced at the cathode because $Cu^{2+}/Cu$ has a more positive reduction potential than water:

$Cu^{2+}(aq) + 2e^- \to Cu(s)$

At the anode, sulfate ions are not oxidized under these conditions. Water is oxidized:

$H_2O(l) \to \frac{1}{2}O_2(g) + 2H^+(aq) + 2e^-$

Overall, including spectator sulfate:

$2CuSO_4(aq) + 2H_2O(l) \to 2Cu(s) + O_2(g) + 2H_2SO_4(aq)$

The electrode material matters here. Inert electrodes allow water to oxidize; copper electrodes can dissolve at the anode instead.

How electroplating works

Electroplating is an electrolytic process that coats a conducting object with a thin layer of metal by reducing metal ions onto its surface. The object being plated is made the cathode, since reduction and metal deposition happen there.

For copper electroplating of a steel object, use a copper anode and a solution containing $Cu^{2+}(aq)$, such as copper(II) sulfate solution.

At the anode, copper dissolves:

$Cu(s) \to Cu^{2+}(aq) + 2e^-$

At the cathode, copper is deposited on the object:

$Cu^{2+}(aq) + 2e^- \to Cu(s)$

When the anode is the same metal as the plating metal, the metal ion concentration in solution can stay nearly constant: the anode produces ions as the cathode uses them up. The anode loses mass; the plated cathode gains mass.

For silver plating a spoon, the spoon is the cathode and a silver electrode is the anode:

Cathode:

$Ag^{+}(aq) + e^- \to Ag(s)$

Anode:

$Ag(s) \to Ag^{+}(aq) + e^-$

In the lab, you might see a shiny metallic coating forming on the cathode and the anode gradually thinning. The observation is what you can see; calling the solid silver or copper is an inference based on the electrolyte, electrodes and half-equations.