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Electrolysis uses an external source of electrical work to drive redox chemical reactions at the electrodes of an electrolytic cell. Its characteristic application is forcing a chemical reaction in a direction that is nonspontaneous under the specified conditions: a discharging galvanic cell supplies electrical work, whereas electrolysis consumes electrical work. This makes electrolysis useful for producing otherwise difficult-to-obtain chemical substances, charging a rechargeable battery, and depositing a metal coating on a conducting object.^[ref]

Always identify the anode by oxidation and the cathode by reduction. In ordinary driven two-terminal operation, the source withdraws Electrons from the positive anode and delivers Electrons to the negative cathode. In a discharging galvanic cell, the anode is negative and the cathode positive. Thus terminal signs change with operating mode, while the reaction definitions remain valid. Through the electrolyte, ions transport electric charge; they need not all be consumed at an electrode. The electrolyte and electrolytic cell articles explain this transport and its conventional current direction.

For a net chemical equation transferring $z$ Electrons per reaction event, the molar reaction Gibbs free energy obeyswhere $F$ is the Faraday constant and $E_{\mathrm{chem}}$ is the reversible cell electromotive force for the reaction in the written direction. A nonspontaneous direction has ${\Delta }_{r}G>0$ and $E_{\mathrm{chem}}<0$ at fixed temperature and pressure. The ideal opposing voltage magnitude is ${\Delta }_{r}G/(zF)$; sustaining a finite electric current generally needs additional drive because of electrical resistance and finite electrode reaction rates. This thermodynamic voltage concerns the actual composition, rather than necessarily standard conditions.^[ref] Not every use of electrolysis increases bulk chemical Gibbs free energy: transferring copper between equivalent surfaces can have zero bulk ${\Delta }_{r}G$ while the source still supplies dissipated electrical work. See the existing electrolytic cell example.

A particularly revealing experiment measures copper transfer instead of merely looking for a coating. Two separated copper electrodes stand in aqueous $\mathrm{CuS}{\mathrm{O}}_4$ electrolyte in a vessel, with insulated leads to a regulated direct-current power supply, a series ammeter and a switch. A balance reads each electrode's mass before and after a timed run. The Institute of Physics describes copper cells with ammeters, adjustable current and removable cathodes; its comparison uses three similar cells at $1\mathrm{A}$ for ten minutes, $0.5\mathrm{A}$ for ten minutes, and $0.5\mathrm{A}$ for twenty minutes. The specified observation to test is that deposited copper mass is proportional to electric current times time. These are documented experimental settings, not measurements generated here.^[ref]

The ideal half-reactions areEach deposited copper chemical atom requires two Electrons, so $z=2$. The anode loses copper and the cathode gains it; ideally their mass changes have equal magnitudes, and the bulk copper cation inventory stays constant. Local concentrations near the electrodes can nevertheless differ. The sulfate anions participate in ionic transport without appearing in these net half-reactions; electronic electric current in the wires becomes ionic electric current in the electrolyte through the two interfaces.^[ref]

Clean, rinse, dry and identify the electrodes before their initial weighing. After interrupting the electric current, rinse away adhering electrolyte, dry again and weigh without rubbing off the deposit. KIT's university practical explicitly calls for distilled-water rinsing, careful drying and weighing both electrodes; it warns that loose copper grains can be lost during handling.^[ref] Useful additional controls are repeated drying and weighing until the reading stabilizes, and a matched unpowered immersion followed by the same rinse/dry cycle. These controls test retained liquid, residual dissolved salts, handling loss and chemical changes unrelated to passed electric charge; they are proposed checks, not additional observations reported by KIT. Keep immersed electrode area, spacing, electrolyte composition and temperature comparable across runs, and keep the electrodes from touching.

Record the electric current as a function of time, together with the exact switch-on and switch-off times. Taking $I(t)\ge 0$ in the chosen deposition direction, the transported electric charge isFor constant electric current, $Q=I(t_1-t_0)$; with sampled readings, use numerical integration rather than assuming that one ammeter reading remained constant. The two equal-charge Institute of Physics settings each give $600\mathrm{C}$, while its shorter $0.5\mathrm{A}$ setting gives $300\mathrm{C}$. With $M_{\mathrm{Cu}}=63.546\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$, the mass per amount of substance tabulated by KIT, and $F\approx 96485.33212\mathrm{C}\mathrm{mo}{\mathrm{l}}^{-1}$, ideal Faraday's laws of electrolysis predictThese are calculated predictions for complete conversion of the measured electric charge into retained copper, not experimental results. For a stable process, doubling the electric charge doubles the deposited mass; equal electric charge gives equal deposited mass even when the electric current and duration differ.

The balance reading is a test of the whole model. Competing reduction of water or other chemical species can divert electric current from copper deposition; transient electrode capacitance can also consume part of a short pulse without producing copper. A fragile deposit can detach even when its formation used the expected electric charge. Conversely, retained electrolyte or salts can increase an apparent mass gain. Comparing anode loss with cathode gain helps detect such problems but does not identify every cause. A slope below $M_{\mathrm{Cu}}/(2F)$ alone cannot distinguish competing chemical reactions from lost deposit. Current efficiency measures the fraction of transported electric charge assigned to a specified electrode reaction, rather than the fraction of electrical energy stored chemically.^[ref]

The beautiful connection is that an electrical measurement counts chemical events: integrating electric current and reading a balance provide two ways of following the same Electron stoichiometry. This makes electrolysis useful for controlling coating quantity and for testing a charge-to-amount of substance relation. It does not determine coating uniformity, adhesion or the required voltage, which depend on transport and electrode chemical kinetics as well.

Energy coupling links an endergonic reaction to an exergonic reaction through a reaction mechanism, so that the combined transformation has a favourable change in Gibbs free energy. ATP hydrolysis is a common driving chemical reaction. A shared reaction intermediate, produced in one step and consumed in another, can physically connect the transformations; merely placing two independent chemical reactions in the same chemical solution does not establish energy coupling. See Alberts et al., Molecular Biology of the Cell: coupled reactions and activated intermediates.

For a computed example, stipulate an endergonic reaction $\mathrm{A}⟶\mathrm{B}$ with $\Delta G_1^{\circ \prime }=+20\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ under precisely the same biochemical standard state conditions as the quoted ATP hydrolysis value. Assume a reaction mechanism couples one ATP hydrolysis to each A-to-B conversion. The formal sum isThe sum $+20+(-30.5)=-10.5$ gives an exergonic reaction. The $+20$ is a chosen teaching value, not a measured property of an identified substance. Outside these reference conditions, add the actual $\Delta G_1$ and $\Delta G_2$, each evaluated using consistent thermodynamic activities; mixing a standard value for one chemical reaction with an unrelated cellular value for the other gives no justified prediction.

Why do the changes in Gibbs free energy add? Gibbs free energy is a state function. Adding consistent reaction changes cancels any shared reaction intermediate and leaves the difference between the combined reactants and combined reaction products. This explains the thermodynamic accounting; it does not construct the reaction mechanism. In many ATP-dependent routes, phosphorylation transfers a phosphate group to a substrate or an enzyme, and subsequent transformation consumes that modified material. The shared material is what connects the steps. Other reaction mechanisms use ATP differently. See Alberts et al., Molecular Biology of the Cell: ATP-driven biosynthesis.

The useful achievement is that a common regenerated nucleotide can connect many different processes in metabolism. The net negative change in Gibbs free energy permits the coupled conversion, but supplies no reaction rate: enzymes can lower activation energy barriers without changing the endpoint difference. Energy coupling changes which overall chemical reaction is being considered; catalysis accelerates an available route. See OpenStax Biology 2e, 6.2: free-energy changes and reaction barriers.

A historical landmark is Fritz Lipmann's ^[ref]. Its sections on the phosphate cycle, generation and utilization of phosphate-bond energy, and group transfer organized these processes around a reusable chemical carrier. The historical terminology should be read with the modern distinction between net hydrolysis and the cost of breaking a chemical bond.

An enzyme is a biological catalyst: it accelerates a chemical reaction and is regenerated after a successful catalytic cycle. Most enzymes are proteins, but catalytic RNA molecules, called ribozymes, also exist. Many reactions needed by a biological cell would proceed too slowly under its ordinary conditions without enzymes. See Cooper, The Cell: the central role of enzymes as biological catalysts.

The biological beauty is selective speed. A biological cell contains many possible reactants; its enzymes help determine which transformations happen on a useful timescale. In metabolism, one step's reaction product can become another step's substrate, so selective catalysts support an organized flow of material.

The ribosome is a striking biological example: the machinery that builds proteins has RNA at its catalytic centre. The complete ribosome contains both RNA and proteins, but its bond-forming active site is built from RNA. Structural evidence established this distinction in Nissen et al., The structural basis of ribosome activity in peptide bond synthesis, Science 289, 920–930 (2000).

Selection occurs at an active site, the region where substrate binding and catalytic chemistry take place. Shape and chemical interactions favour some substrates over others; binding can also reshape the enzyme, an induced fit. Recognition alone is insufficient: holding a substrate extremely tightly need not make it react quickly. Catalysis requires a favourable route through the transition state, the high-barrier configuration between reactants and reaction products. Positioning reactive groups and stabilizing this configuration relative to the bound substrate can lower that barrier. Specificity is often strong, but need not mean exactly one acceptable substrate. See Cooper, The Cell: mechanisms of enzymatic catalysis.

Read the diagram numerically. Its chosen Gibbs free energies are $0$ for the reactants and $-10\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ for the reaction products. The uncatalysed peak is $60\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, whereas the catalysed peak is $40\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. The forward free-energy barriers are therefore $60$ and $40\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$; viewed from the reaction products, the reverse barriers are $70$ and $50\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Both decrease by $20\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$, while the endpoint difference stays $-10\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. These are constructed values, not measurements of a particular enzyme.

This separates chemical kinetics from chemical equilibrium. An enzyme changes how rapidly the system approaches chemical equilibrium; it does not change the equilibrium constant for the same overall chemical reaction under the same conditions. Faster catalysis therefore supplies no extra driving energy. In a biological cell, continuing supply of reactants and removal of reaction products can nevertheless sustain a net flow: the surroundings keep that system away from chemical equilibrium. See IUBMB, Symbolism and terminology in enzyme kinetics: basic definitions.

Some enzymes require a cofactor, an additional nonprotein component such as a metal ion or a small organic molecule. Organic helpers are often called coenzymes; they can assist transfers that the protein alone cannot readily perform. The cofactor may remain bound or exchange with the surroundings, depending on the system. A coenzyme can be chemically changed in one reaction and regenerated by another: regeneration of the enzyme does not mean every helper is unchanged in that individual reaction. Some enzymes need no additional cofactor. See Alberts et al., Molecular Biology of the Cell: small molecules associated with proteins and Alberts et al., activated carriers and their regeneration.

How much faster does adding substrate make a reaction? Enzyme kinetics gives a quantitative answer when a model's assumptions fit. For an illustrative reaction obeying Michaelis-Menten kinetics, use the Michaelis-Menten equation with substrate concentration $s$, limiting reaction rate $V_{\max }$ and Michaelis constant $K_M$. At $s/K_M=0.1$, substitution gives $v/V_{\max }=0.1/1.1=1/11\approx 0.0909$. At $s/K_M=1$, it gives $1/2$; at $s/K_M=10$, it gives $10/11\approx 0.9091$. Thus a hundredfold increase in substrate between the first and last examples yields only a tenfold increase in reaction rate.

For a chosen $V_{\max }=110\mu \mathrm{M}\mathrm{mi}{\mathrm{n}}^{-1}$, those fractions mean reaction rates of $10$, $55$ and $100\mu \mathrm{M}\mathrm{mi}{\mathrm{n}}^{-1}$. Doubling substrate from $10K_M$ to $20K_M$ raises the normalized rate only from $10/11$ to $20/21\approx 0.9524$, a relative increase of about $4.76\%$. In this simplest model, the catalytic machinery is already nearly fully occupied. Doubling the amount of active enzyme instead doubles $V_{\max }$ and the predicted reaction rates at fixed free substrate concentration, provided the other conditions are unchanged. This distinction helps decide whether additional substrate or additional catalytic capacity could improve throughput.

The interpretation has limits. The simplest Michaelis-Menten kinetics assumes a single substrate, fixed active enzyme amount, a suitable quasi-steady-state approximation, and initial measurements with negligible reverse reaction from accumulated reaction product. Substantial depletion, interacting binding sites, or inhibition by excess substrate can invalidate this model. Within it, $K_M$ is the substrate concentration giving half the limiting reaction rate; it is not generally the binding dissociation constant $K_d$. See IUBMB, enzyme kinetics: steady-state approximation and substrate dissociation and IUBMB, Michaelis-Menten and non-Michaelis-Menten kinetics. Measuring activity also requires specified temperature, pH and necessary cofactor availability: a weak signal can reflect unsuitable conditions rather than little enzyme. See Brooks et al., Assay Guidance Manual: basics of enzymatic assays for HTS.

An exergonic reaction has negative molar reaction change in Gibbs free energy under the specified conditions. At fixed temperature and pressure, its forward direction is thermodynamically favourable; its reaction rate remains a separate question. See OpenStax Biology 2e, 6.2: exergonic reactions.

In induced fit, a substrate binds to an enzyme and the resulting enzyme-substrate complex subsequently changes its protein structure. This adjustment can position interacting groups at the active site for catalysis. The mechanistic claim concerns order: binding precedes the relevant change. In conformational selection, the unbound enzyme already visits an alternative protein structure, which the substrate preferentially binds. Both processes can occur in one system. Let $E$ and $E^{\ast }$ denote two structures of the same enzyme, and $S$ a substrate. The limiting routes areThe star labels a structure, not a different protein primary structure, and does not by itself mean catalytic activity. Each reversible arrow abbreviates a step, not the whole catalytic cycle. See Gordon G. Hammes, Yu-Chu Chang and Terrence G. Oas, Conformational selection or induced fit: a flux description of reaction mechanism, PNAS 106 (2009), 13737–13741.

A rigid lock-and-key cartoon usefully suggests selective substrate recognition, but it suppresses movement of the protein and chemical changes at the active site. Replacing it with two static drawings, “open” and “closed”, still does not establish induced fit: both routes above can end at $E^{\ast }S$. Also, if complete closure prevents a substrate entering, the pre-existing structure selected in conformational selection might be partly open and undergo further induced fit after binding. The usefulness of the distinction is that it turns an attractive shape story into questions about enzyme kinetics. Relative traffic through the routes depends on populations, rate constants and free substrate concentration, rather than affinity alone. See Hammes, Chang and Oas, 2009: coupled pathways, their concentrations and the limitations of endpoint structures.

For this analytical illustration, hold the free substrate concentration $s$ constant and assume independent waiting times with exponential distributions of constant rates. The law of mass action gives the binding rate $a=k_{\mathrm{on}}s$; the competing change has rate $b$. The probability density function for binding to happen first at time $t$ is $a{e}^{-(a+b)t}$, because neither event has happened earlier. Its integral isWith the stated invented rate constants, $s=0.1,1,10\mu \mathrm{M}$ gives $1/11,1/2,10/11$, respectively. Raising free substrate concentration makes binding more likely to win this initial competition. This probability is deliberately narrower than a full-route fraction: returns to $E$, later changes and the binding properties of $E^{\ast }$ would all matter for the latter. The illustration explains why timing and concentration must accompany structural evidence, without assigning a binding mechanism to a real enzyme.

A motivating experiment tested the crystal interpretation in solution. McDonald, Steitz and Engelman prepared solutions of monomeric yeast hexokinase B and compared the unbound preparation with sugar-bound preparations. Their apparatus measured small-angle scattering of X-rays; the spatial extent inferred from that readout was the radius of gyration, rather than a catalytic reaction rate. Binding glucose reduced the radius of gyration by $0.95\pm 0.24\mathring{\mathrm{A}}$; binding the phosphorylated glucose product reduced it by $1.25\pm 0.28\mathring{\mathrm{A}}$, as reported by the authors. The glucose change agreed with the reduction calculated from the crystal coordinates. See Robert C. McDonald, Thomas A. Steitz and Donald M. Engelman, Yeast hexokinase in solution exhibits a large conformational change upon binding glucose or glucose 6-phosphate, Biochemistry 18 (1979), 338–342: primary abstract.

The comparison without added substrate supplies a baseline within the same form of hexokinase; the second bound preparation tests another binding condition. Together with the crystal comparison, solution compaction supports closure beyond crystal packing. It still gives an average spatial readout, not a movie of individual binding events, and product-associated compaction does not demonstrate ongoing catalysis. Neither experiment alone proves that binding precedes movement. To distinguish induced fit from conformational selection, structural comparisons must be combined with measurements sensitive to timing and with a model of the relevant populations and enzyme kinetics.

Light initiates methane chlorination by photodissociation of $\mathrm{C}{\mathrm{l}}_2$ through homolysis. The elementary bookkeeping isHere $h\nu$ denotes the photon energy. The first propagation step replaces a chlorine radical with a methyl radical; the second regenerates a chlorine radical. Adding the two propagation equations cancels both free radicals and gives $\mathrm{C}{\mathrm{H}}_4+\mathrm{C}{\mathrm{l}}_2⟶\mathrm{C}{\mathrm{H}}_3\mathrm{Cl}+\mathrm{HCl}$: chloromethane and hydrogen chloride are made while the chain carrier is regenerated.^[ref]

For example, recombination $2{\mathrm{C}{\mathrm{H}}_3}^{\bullet }⟶{\mathrm{C}}_2{\mathrm{H}}_6$ of methyl radicals makes ethane and a covalent bond from two unpaired Electrons, removing two carriers. Recombination of two ${\mathrm{Cl}}^{\bullet }$ also terminates carriers; in the gas phase, stabilization of the product requires disposal of released energy, for example to a third collision partner. The figure omits that partner. Termination competes with propagation, and further chlorination can consume chloromethane, so this cycle alone does not predict a pure product or a measured reaction rate.^[ref]

A rate constant is the coefficient in a specified rate law, held fixed under specified conditions. It can describe an empirical overall chemical reaction, as well as an elementary reaction. Fixed temperature matters: changing temperature can change $k$, even for the same chemical reaction. See IUPAC Gold Book: order of reaction and rate constants and IUPAC Gold Book: rate coefficient.

For $v=k[\mathrm{A}{]}^n$, dimensional consistency requiresA zero-order reaction therefore uses $\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$; a first-order reaction uses ${\mathrm{s}}^{-1}$; a second-order reaction uses $\mathrm{L}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{s}}^{-1}$. Comparing bare numerical values of $k$ across different orders of reaction is meaningless without their units and concentrations. Even for the same order of reaction, state whether $k$ multiplies the normalized reaction rate $v$ or a particular reactant disappearance rate: a stoichiometric coefficient can change that convention.

For one chemical reaction in a closed system of fixed volume $V$, with uniform concentrations and no competing reactions or appreciable accumulation of unlisted species, choose the chemical equationBrackets denote amount of substance per volume, for example $[\mathrm{A}]=n_{\mathrm{A}}/V$. The stoichiometry gives one instantaneous reaction ratewhere $\xi$ is the existing extent of reaction. The minus signs make forward consumption positive; dividing by the coefficients makes measurements of different chemical species agree. Units can be $\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$. See IUPAC Gold Book: rate of reaction and Flowers et al., Chemistry 2e, §12.1: chemical reaction rates.

For the illustrative chemical equation $2\mathrm{A}+\mathrm{B}⟶3\mathrm{C}$, a measured $d[\mathrm{C}]/dt=0.060\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$ implies $v=0.020\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$, $d[\mathrm{A}]/dt=-0.040\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$, and $d[\mathrm{B}]/dt=-0.020\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$. These numbers express stoichiometry, without supplying a rate law.

Each derivative is the slope of a concentration curve at one instant. Over a finite interval, $-\Delta [\mathrm{A}]/(a\Delta t)$ instead gives an average reaction rate. For A converting reversibly to B, the net reaction rate is $v=d[\mathrm{B}]/dt=J_f-J_r$; both directional reaction rates can remain positive when $v=0$ at chemical equilibrium.

Here is an explicitly stipulated elementary kinetic model, not experimental data. A well-mixed ideal closed system has constant temperature and volume. Alongside an uncatalysed elementary reaction, allow an illustrative catalytic encounter that returns C immediately:Both channels obey the law of mass action. In the second, the forward reaction rate is ${\alpha }_f[\mathrm{A}][\mathrm{C}]$ and the reverse reaction rate is ${\alpha }_r[\mathrm{B}][\mathrm{C}]$. C cancels in each encounter, so its free concentration $c_C$ stays constant without an external supply. This toy elementary reaction encodes regeneration without binding; it is not a claim that real catalytic cycles universally consist of one encounter. A binding cycle can accumulate a reaction intermediate, in which case constant total C does not imply constant free $c_C$.

Adding the parallel reaction rates gives effective first-order reaction rate constantsBecause $k_f^u/k_r^u={\alpha }_f/{\alpha }_r=3$, each channel balances at the same $[\mathrm{B}]/[\mathrm{A}]=3$, and so does their sum. At $c_C=0$ the pair is $(0.030,0.010){\mathrm{s}}^{-1}$; at $c_C=0.001\mathrm{mol}{\mathrm{L}}^{-1}$ it is $(0.300,0.100){\mathrm{s}}^{-1}$. Both give $K=3$ with ideal thermodynamic activities and a common concentration standard state. At $T=300\mathrm{K}$ their compatible standard reaction change in Gibbs free energy is ${\Delta }_r{G}^{\circ }=-RT\ln 3\approx -2.7403\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Choosing an unrelated reverse rate constant would break this consistency.

Let $S=[\mathrm{A}]+[\mathrm{B}]$ be the conserved concentration, and define the dimensionless quantity $x=[\mathrm{B}]/S$. ThenSolving this first-order linear differential equation givesThe chemical relaxation time is $25\mathrm{s}$ without C and $2.5\mathrm{s}$ with the chosen C. From either $x_0=0$ or $x_0=1$, the exponential decay of the deviation leads to the same chemical equilibrium $x=3/4$. Reducing that deviation to one percent takes $\ln (100)\tau$: about $115.13\mathrm{s}$ versus $11.51\mathrm{s}$.

At this chemical equilibrium, conversion continues in both directions. For $S=1\mathrm{mol}{\mathrm{L}}^{-1}$, the uncatalysed channel has equal forward and reverse reaction rates $0.0075\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$. Adding C gives total directional reaction rates $0.075\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$ each, yet zero net reaction rate. Faster molecular exchange and unchanged composition coexist. This exact calculation explains how catalysis can matter enormously within an observation window while leaving the final equilibrium constant intact.