Redox Reactions and Electrochemistry

The Daniell cell runs on this reaction, and its standard potential is 1.10 V.

$\mathrm{Zn}(\mathrm{s})+\mathrm{C}{\mathrm{u}}^{2+}(\mathrm{aq})\to \mathrm{Z}{\mathrm{n}}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s}),E^{\circ }=1.10\text{V}$

Now write the same reaction doubled.

$2Zn(\mathrm{s})+2C{\mathrm{u}}^{2+}(\mathrm{aq})\to 2Z{\mathrm{n}}^{2+}(\mathrm{aq})+2Cu(\mathrm{s}),E^{\circ }=?$

Nearly everyone writes 2.20 V.

It is still 1.10 V. Doubling a cell reaction does not change its potential at all.

Look at what happened to the two quantities. The Gibbs energy really did double, because twice as much reaction released twice as much energy. But $n$ doubled as well, and

$E^{\circ }=-\frac{\Delta G^{\circ }}{nF}$

divides one by the other. $\Delta G^{\circ }$ is extensive; $E^{\circ }$ is intensive. Potential is energy per unit charge, and per-unit quantities do not care how much you have.

You already knew this from the shop counter. A AAA cell and a giant D cell are both 1.5 V. The D cell is not more powerful in voltage, only in how long it can keep supplying it.

That distinction runs through the whole chapter, and it is the reason electrode potentials can be tabulated at all. A number that changed every time you rewrote the equation would be useless in a table.

The rest is one idea seen from two positions.

Nothing else changed. The chemistry is identical and the energy released is identical. Only the route differs, which is why the cell potential and the Gibbs energy change of the reaction are two ways of saying the same thing.

1. Oxidation, Reduction and Oxidation Number

The electronic definitions are the ones that generalise beyond oxygen.

The naming sounds backwards until you notice that an agent does something to somebody else. The reducing agent reduces the other one, and gets oxidised in the process.

Electrons cannot appear from nowhere, so oxidation and reduction always occur together and in matched amounts. That matching is what makes the half-reaction method work.

Assigning oxidation numbers

Oxidation number. The charge an atom would carry if every bond in the species were fully ionic.

It is a bookkeeping device, not a real charge, which is exactly why it is allowed to take values no real ion ever holds.

Fractional oxidation numbers are perfectly legitimate, because the number is an average over atoms that the formula cannot tell apart. Iron in $\mathrm{F}{\mathrm{e}}_3{\mathrm{O}}_4$ averages $+8/3$, reflecting one $\mathrm{F}{\mathrm{e}}^{2+}$ and two $\mathrm{F}{\mathrm{e}}^{3+}$ per formula unit.

In thiosulphate, ${\mathrm{S}}_2{\mathrm{O}}_3^{2-}$, the average sulphur is $+2$, but the two sulphur atoms are genuinely different: one central and one terminal.

Trap. Rule 4 is a default, not a law, and rule 6 outranks it. Whenever rule 4 produces an impossible answer, the oxygen is not behaving as $-2$.

Illustration 1

Find the oxidation number of chromium in $\mathrm{Cr}{\mathrm{O}}_5$, and of each carbon in acetic acid, $\mathrm{C}{\mathrm{H}}_3\mathrm{COOH}$.

Apply rule 4 blindly to $\mathrm{Cr}{\mathrm{O}}_5$: five oxygens at $-2$ give $-10$, so chromium would be $+10$.

That is impossible. Chromium has 6 electrons outside argon and cannot lose 10. So the assumption is wrong, and the structure says why: chromium peroxide holds four peroxide oxygens at $-1$ and one doubly bonded oxygen at $-2$.

$4(-1)+1(-2)=-6\Rightarrow \text{Cr}=+6$

For acetic acid, the overall average carbon oxidation number is 0, which tells you nothing useful. Split the molecule instead and count the bonds at each carbon.

Bonds to the same element count zero, bonds to hydrogen count $-1$ each, and bonds to oxygen count $+1$ each. The average of $-3$ and $+3$ is 0, which is why the whole-molecule method concealed both numbers. Organic redox questions need the per-atom count.

Types of redox reaction

Combination, decomposition and displacement are the straightforward cases.

Disproportionation. One species is simultaneously oxidised and reduced.

It requires an element in an intermediate oxidation state, with both a higher and a lower state available.

$2{\mathrm{H}}_2{\mathrm{O}}_2\to 2{\mathrm{H}}_2\mathrm{O}+{\mathrm{O}}_2$

Oxygen at $-1$ goes to $-2$ and to $0$ at the same time. Fluorine can never disproportionate, because $-1$ is its lowest state and it has no positive state to reach.

2. Balancing Redox Equations

The half-reaction method is more reliable than the oxidation number method, and it is the one to learn properly.

In basic medium, do all six steps as though the medium were acidic, then add enough $\mathrm{O}{\mathrm{H}}^{-}$ to both sides to neutralise every ${\mathrm{H}}^{+}$, and simplify the water that results. Balancing directly in basic medium is possible and much slower.

Always finish by checking both the atom count and the total charge. Charge is the check that catches the errors the atom count misses, because a wrong electron count leaves the atoms perfectly balanced.

Illustration 2

Balance the reaction of permanganate with iodide in basic medium, giving manganese dioxide and iodine.

Reduction half, acidic first. Manganese goes from $+7$ to $+4$, gaining 3 electrons.

$\mathrm{Mn}{\mathrm{O}}_4^{-}+4{\mathrm{H}}^{+}+3{\mathrm{e}}^{-}\to \mathrm{Mn}{\mathrm{O}}_2+2{\mathrm{H}}_2\mathrm{O}$

Oxidation half. Iodide goes from $-1$ to $0$.

$2{\mathrm{I}}^{-}\to {\mathrm{I}}_2+2{\mathrm{e}}^{-}$

Scale by 2 and 3 to match at 6 electrons, then add.

$2Mn{\mathrm{O}}_4^{-}+6{\mathrm{I}}^{-}+8{\mathrm{H}}^{+}\to 2Mn{\mathrm{O}}_2+3{\mathrm{I}}_2+4{\mathrm{H}}_2\mathrm{O}$

Now convert to basic. There are 8 ${\mathrm{H}}^{+}$, so add 8 $\mathrm{O}{\mathrm{H}}^{-}$ to both sides. On the left they combine with the ${\mathrm{H}}^{+}$ to give 8 ${\mathrm{H}}_2\mathrm{O}$; on the right they stay as they are.

$2Mn{\mathrm{O}}_4^{-}+6{\mathrm{I}}^{-}+8{\mathrm{H}}_2\mathrm{O}\to 2Mn{\mathrm{O}}_2+3{\mathrm{I}}_2+4{\mathrm{H}}_2\mathrm{O}+8O{\mathrm{H}}^{-}$

Cancel 4 ${\mathrm{H}}_2\mathrm{O}$ from each side.

$2Mn{\mathrm{O}}_4^{-}+6{\mathrm{I}}^{-}+4{\mathrm{H}}_2\mathrm{O}\to 2Mn{\mathrm{O}}_2+3{\mathrm{I}}_2+8O{\mathrm{H}}^{-}$

Check the charge: left is $-2-6=-8$; right is $-8$. Check the oxygens: left is $8+4=12$; right is $4+8=12$. Balanced.

3. Galvanic Cells

Separate the two half-reactions into different containers and the electron transfer becomes a current.

Oxidation happens at the anode and reduction at the cathode. In a galvanic cell the anode is negative and the cathode positive, and electrons flow from anode to cathode through the external wire.

The salt bridge completes the circuit, and its more important job is keeping both solutions electrically neutral. Without it the anode compartment accumulates positive charge within moments and the reaction stops dead.

Cell notation

By convention the anode goes on the left and the cathode on the right, a single vertical line marks a phase boundary and a double line marks the salt bridge.

$\mathrm{Zn}(\mathrm{s})|\mathrm{Z}{\mathrm{n}}^{2+}(\mathrm{aq})||\mathrm{C}{\mathrm{u}}^{2+}(\mathrm{aq})|\mathrm{Cu}(\mathrm{s})$

Read it left to right and it tells you the whole cell: zinc is oxidised, copper ion is reduced, electrons run left to right outside.

Electrode potential

A single electrode potential cannot be measured, because any measurement needs a second electrode to complete a circuit. Only differences are accessible.

Standard hydrogen electrode. Hydrogen gas at 1 bar bubbling over platinised platinum in 1 M ${\mathrm{H}}^{+}$ at 298 K, assigned a potential of exactly zero by convention.

Every other standard potential is measured against it, and by IUPAC convention all are quoted as reduction potentials.

$E_{cell}^{\circ }=E_{cathode}^{\circ }-E_{anode}^{\circ }$

with both taken as reduction potentials. A positive cell potential means the reaction as written is spontaneous.

Trap. Do not reverse the sign of the anode's tabulated value and then also subtract. The formula already does the reversing. Subtract the tabulated numbers exactly as they appear in the table.

Types of electrode

The syllabus names four kinds, and questions often turn on recognising which is in play.

The calomel electrode matters in practice because the hydrogen electrode is fragile and inconvenient. Calomel has a fixed, reproducible potential of 0.2444 V and serves as the secondary reference in almost every real laboratory, including inside every pH meter.

The electrochemical series

Arranging standard reduction potentials in order produces a table of enormous predictive power.

A more positive reduction potential means a stronger oxidising agent. Fluorine at $+2.87$ V is the strongest common oxidising agent; lithium at $-3.05$ V is the strongest reducing agent.

Two predictions follow, and between them they answer most feasibility questions in the paper.

Illustration 3

Will $\mathrm{F}{\mathrm{e}}^{3+}$ oxidise iodide to iodine? Will it oxidise bromide to bromine? Take $E^{\circ }$ values of $+0.77$ V for $\mathrm{F}{\mathrm{e}}^{3+}/F{\mathrm{e}}^{2+}$, $+0.54$ V for ${\mathrm{I}}_2/{\mathrm{I}}^{-}$ and $+1.09$ V for $\mathrm{B}{\mathrm{r}}_2/B{\mathrm{r}}^{-}$.

For $\mathrm{F}{\mathrm{e}}^{3+}$ to act as the oxidising agent it must be the cathode.

$\text{with iodide: }{E}_{cell}^{\circ }=0.77-0.54=+0.23\text{V}\text{spontaneous}$

$\text{with bromide: }{E}_{cell}^{\circ }=0.77-1.09=-0.32\text{V}\text{not spontaneous}$

Iron(III) chloride solution turns brown with potassium iodide and does nothing with potassium bromide. The threshold sits between 0.54 and 1.09 V, and $\mathrm{F}{\mathrm{e}}^{3+}$ falls squarely between them.

This is the whole technique. Put the candidate oxidising agent at the cathode, subtract, and read the sign. No intuition about which looks stronger is needed or wanted.

4. Cell Potential, Gibbs Energy and the Nernst Equation

The electrical work a cell can deliver is charge times potential, and the maximum non-expansion work available is the Gibbs energy change.

$\Delta G^{\circ }=-nF{E}_{cell}^{\circ }$

with $n$ the moles of electrons transferred and $F$ the Faraday constant, 96500 C mol${}^{-1}$.

The minus sign makes a positive cell potential correspond to a negative Gibbs energy change, which is to say a spontaneous reaction. This single relation ties electrochemistry to the whole of thermodynamics, and it is also the tool for the hook this chapter opened on.

Illustration 4

Given $E^{\circ }=-0.036$ V for $\mathrm{F}{\mathrm{e}}^{3+}/Fe$ and $E^{\circ }=-0.44$ V for $\mathrm{F}{\mathrm{e}}^{2+}/Fe$, find $E^{\circ }$ for $\mathrm{F}{\mathrm{e}}^{3+}/F{\mathrm{e}}^{2+}$.

The obvious move is to subtract the two potentials. That is wrong, and it is wrong for exactly the reason the chapter opened with: potentials are intensive and do not add.

Gibbs energies do add. Convert, combine, convert back.

$\mathrm{F}{\mathrm{e}}^{3+}+3{\mathrm{e}}^{-}\to \mathrm{Fe}:\Delta G_1^{\circ }=-3F(-0.036)=+0.108F$

$\mathrm{F}{\mathrm{e}}^{2+}+2{\mathrm{e}}^{-}\to \mathrm{Fe}:\Delta G_2^{\circ }=-2F(-0.44)=+0.88F$

The target is the first minus the second.

$\mathrm{F}{\mathrm{e}}^{3+}+{\mathrm{e}}^{-}\to \mathrm{F}{\mathrm{e}}^{2+}:\Delta G_3^{\circ }=0.108F-0.88F=-0.772F$

$E_3^{\circ }=-\frac{\Delta G_3^{\circ }}{n_{3}F}=-\frac{-0.772F}{1\times F}=+0.772\text{V}$

The measured value is $+0.77$ V. Subtracting the potentials directly would have given $-0.036-(-0.44)=+0.404$ V, which is not close to anything.

The general rule: potentials may be subtracted only when the two half-reactions combine into a full cell, where the electrons cancel. When they combine into another half-reaction, electrons remain and you must go through $\Delta G^{\circ }$.

The Nernst equation

Standard potentials assume unit concentrations, which no real cell ever has. The Nernst equation corrects for the actual composition.

$E=E^{\circ }-\frac{2.303RT}{nF}\log Q$

At 298 K the constants collapse to a number worth memorising.

$E=E^{\circ }-\frac{0.0591}{n}\log Q$

Two consequences follow at once. As a cell discharges, $Q$ rises towards $K$ and the potential falls towards zero, so a dead battery is a cell that has reached equilibrium. And setting $E=0$ with $Q=K$ gives the bridge to the previous chapter.

$\log K=\frac{n{E}_{cell}^{\circ }}{0.0591}$

A standard potential of about 1 V with two electrons corresponds to $K$ near $10^{34}$. Small potentials mean enormous equilibrium constants, because the relation is logarithmic.

Illustration 5

Find the potential of a hydrogen electrode at 298 K in a solution of pH 4.0, with hydrogen at 1 bar.

The half-reaction is $2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2$, with $E^{\circ }=0$ and $n=2$.

$Q=\frac{p_{{\mathrm{H}}_2}}{[{\mathrm{H}}^{+}{]}^2}=\frac{1}{[{\mathrm{H}}^{+}{]}^2}$

$E=0-\frac{0.0591}{2}\log \frac{1}{[{\mathrm{H}}^{+}{]}^2}=-\frac{0.0591}{2}\left(-2\log [{\mathrm{H}}^{+}]\right)=-0.0591\mathrm{pH}$

At pH 4.0 this gives $E=-0.236$ V.

The 2 in the denominator and the 2 in the exponent cancelled exactly, leaving a potential that is linear in pH with a slope of $-59.1$ mV per unit. That is not a curiosity. It is the working principle of every pH meter ever built: measure a voltage against a calomel reference, divide by 59.1 mV, and read off the pH.

Concentration cells

A cell can be built from two identical electrodes in the same solution at two different concentrations. Its standard potential is zero, so the entire potential comes from the logarithmic term.

Such a cell runs until the two concentrations equalise, which is diffusion doing electrical work.

Illustration 6

Find the potential of $\mathrm{Cu}|\mathrm{C}{\mathrm{u}}^{2+}(0.0010\mathrm{M})||\mathrm{C}{\mathrm{u}}^{2+}(0.10\mathrm{M})|\mathrm{Cu}$ at 298 K.

Both electrodes are copper, so $E^{\circ }=0$ exactly. The cell reaction moves copper ions from the concentrated side to the dilute side, so $Q$ is dilute over concentrated.

$E=0-\frac{0.0591}{2}\log \frac{0.0010}{0.10}=-\frac{0.0591}{2}(-2)=+0.0591\text{V}$

A cell built from two pieces of the same metal in two strengths of the same solution, driving current with no net chemistry at all. Copper dissolves on one side and plates on the other, and when the concentrations meet the cell is dead.

Note the pattern: a hundredfold ratio with $n=2$ gives exactly 59.1 mV, the same number as the pH slope. Both are $0.0591\times (\text{decades})/n$.

5. Electrolytic Cells and Faraday's Laws

An electrolytic cell reverses the logic. Electrical energy is supplied from outside to drive a non-spontaneous reaction.

Oxidation still happens at the anode and reduction still at the cathode. The definition never changes. Only the sign does, and it changes because in electrolysis the external supply is now the thing setting the polarity.

Faraday's first law. The mass deposited is proportional to the charge passed.

Faraday's second law. Equal charges deposit masses proportional to their equivalent masses.

$Q=It,n_{e^{-}}=\frac{Q}{F},m=\frac{QM}{nF}$

One faraday, 96500 C, deposits one mole of a singly charged ion, half a mole of a doubly charged one, and so on.

Illustration 7

A current of 5.0 A is passed through acidified water for 20 minutes. Find the volumes of hydrogen and oxygen liberated at STP.

$Q=It=5.0\times 1200=6000\text{C}\Rightarrow n_{e^{-}}=\frac{6000}{96500}=0.0622\text{mol}$

At the cathode, $2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2$, so 2 electrons per molecule.

$n_{{\mathrm{H}}_2}=\frac{0.0622}{2}=0.0311\text{mol}\Rightarrow V=0.0311\times 22400=697\text{mL}$

At the anode, $2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{O}}_2+4{\mathrm{H}}^{+}+4{\mathrm{e}}^{-}$, so 4 electrons per molecule.

$n_{{\mathrm{O}}_2}=\frac{0.0622}{4}=0.0156\text{mol}\Rightarrow V=349\text{mL}$

The ratio comes out at 2:1, which it had to, since the water being decomposed is ${\mathrm{H}}_2\mathrm{O}$. That agreement is the free check on this kind of question, and it catches a dropped factor instantly.

Predicting the products of electrolysis

Which species is discharged is decided by electrode potentials, not by what is most abundant.

Molten sodium chloride has only two species available, so it gives sodium at the cathode and chlorine at the anode.

Aqueous sodium chloride is a different problem, because water competes at both electrodes. Water is far easier to reduce than the sodium ion, so hydrogen appears at the cathode rather than sodium.

Illustration 8

At the anode of an aqueous sodium chloride cell, compare $E^{\circ }=+1.23$ V for oxygen from water with $E^{\circ }=+1.36$ V for chlorine from chloride. Which is released, and why does the answer depend on concentration?

Lower potential means easier to oxidise, so thermodynamics says water should go first and oxygen should be released. In very dilute solution that is exactly what happens.

Run the same cell on concentrated brine and you get chlorine. Two effects push it there.

The 0.13 V gap is small enough for both to overturn it. This is not a footnote: the entire chlor-alkali industry, and therefore most of the world's chlorine and sodium hydroxide, depends on that reversal.

Copper sulphate makes the same point differently. Between platinum electrodes it gives copper at the cathode and oxygen at the anode. Replace the anode with copper and the anode dissolves instead, because oxidising the electrode is easier than oxidising water. That substitution is exactly how copper is refined to 99.99 per cent purity.

Trap. The products depend on the electrode material and the concentration, not only on the salt. Exam questions change one of those two and keep everything else the same.

6. Conductance in Electrolytic Solutions

Conductivity $\kappa$. The conductance of a unit cube of solution, in S cm${}^{-1}$. It falls on dilution, simply because there are fewer ions per unit volume.

Molar conductivity ${\Lambda }_m$. Conductivity corrected for how much solute is actually there.

${\Lambda }_m=\frac{\kappa \times 1000}{c}$

with $c$ in mol L${}^{-1}$, giving S cm${}^2$ mol${}^{-1}$. It rises on dilution, because each ion is freer to move.

Both statements are true at once and they are not in tension. $\kappa$ asks how well this beaker conducts; ${\Lambda }_m$ asks how well each mole conducts. Dilution reduces the first and improves the second.

Two very different curves

Strong electrolytes are already fully ionised, so dilution only reduces interionic interference. Molar conductivity rises slowly and linearly in $\sqrt{c}$, extrapolating cleanly.

${\Lambda }_m={\Lambda }_m^{\circ }-A\sqrt{c}$

Weak electrolytes ionise more as they are diluted, so their molar conductivity climbs steeply near zero concentration and never settles. Their limiting value cannot be reached by extrapolation at all.

Kohlrausch's law

At infinite dilution each ion contributes independently of whatever it arrived with.

${\Lambda }_m^{\circ }={\nu }_{+}{\lambda }_{+}^{\circ }+{\nu }_{-}{\lambda }_{-}^{\circ }$

That solves the weak electrolyte problem. Acetic acid's limiting value is assembled from sodium acetate, hydrochloric acid and sodium chloride, all strong, all extrapolable.

Two further applications follow. The degree of dissociation is $\alpha ={\Lambda }_m/{\Lambda }_m^{\circ }$, and feeding that into Ostwald's dilution law gives the ionisation constant from conductivity alone.

Illustration 9

A saturated solution of silver chloride has conductivity $3.40\times 10^{-6}$ S cm${}^{-1}$, and the water used has $1.60\times 10^{-6}$ S cm${}^{-1}$. Given ${\lambda }^{\circ }=61.9$ for $\mathrm{A}{\mathrm{g}}^{+}$ and 76.3 for $\mathrm{C}{\mathrm{l}}^{-}$ in S cm${}^2$ mol${}^{-1}$, find $K_{sp}$.

Subtract the water's own conductivity, which is not negligible here.

${\kappa }_{\mathrm{AgCl}}=3.40\times 10^{-6}-1.60\times 10^{-6}=1.80\times 10^{-6}\mathrm{S}\mathrm{c}{\mathrm{m}}^{-1}$

${\Lambda }_m^{\circ }=61.9+76.3=138.2\mathrm{S}\mathrm{c}{\mathrm{m}}^2\mathrm{mo}{\mathrm{l}}^{-1}$

A saturated solution of silver chloride is so dilute that ${\Lambda }_m$ is effectively ${\Lambda }_m^{\circ }$ already, so

$c=\frac{1000\kappa }{{\Lambda }_m^{\circ }}=\frac{1000\times 1.80\times 10^{-6}}{138.2}=1.30\times 10^{-5}\mathrm{mol}{\mathrm{L}}^{-1}$

$K_{sp}=s^2={\left(1.30\times 10^{-5}\right)}^2=1.7\times 10^{-10}$

The accepted value is $1.8\times 10^{-10}$. A conductivity bridge has just measured a solubility product too small to weigh, and it agrees with the value obtained by completely different means in the Equilibrium chapter.

7. Batteries, Fuel Cells and Corrosion

Charging a lead accumulator drives the discharge reaction backwards, which is exactly the electrolytic cell logic applied to a galvanic cell.

A fuel cell differs from both in that reactants are supplied continuously rather than stored. It converts chemical energy directly to electrical with no heat step, so it escapes the Carnot limit that caps every combustion engine.

Illustration 10

A garage tests a car battery with a hydrometer, which measures the density of the acid. Why does that reveal the state of charge?

Write the discharge reaction of the lead accumulator.

$\mathrm{Pb}(\mathrm{s})+\mathrm{Pb}{\mathrm{O}}_2(\mathrm{s})+2{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4(\mathrm{aq})\to 2PbS{\mathrm{O}}_4(\mathrm{s})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

Sulphuric acid is consumed and water is produced. Both changes push the same way: the electrolyte gets less concentrated and therefore less dense, from about 1.28 g cm${}^{-3}$ when charged to about 1.18 g cm${}^{-3}$ when flat.

So the acid is not a bystander. It is a reactant, and its concentration is a direct readout of how much of the cell reaction has already happened. A hydrometer is measuring extent of reaction disguised as a density.

Charging reverses the equation, regenerates the acid, and the density climbs back.

Corrosion is an electrochemical process, not a simple chemical one. Rusting sets up tiny galvanic cells across the iron surface, with iron oxidised at anodic patches and oxygen reduced at cathodic ones. Both oxygen and water are required, which is why iron does not rust in dry air or in boiled, deoxygenated water.

Prevention works by blocking one requirement or by supplying electrons more cheaply. Galvanising gives cathodic protection: zinc sits below iron in the series, so it is oxidised preferentially and corrodes instead. The protection continues even where the coating is scratched, which paint can never claim.

Summary

Oxidation is loss of electrons and reduction is gain, and the two always occur together in matched amounts. Oxidation number is a bookkeeping charge, may be fractional because it is an average, and must be computed per atom in organic molecules where the average hides everything.

Disproportionation needs an element in an intermediate oxidation state with both a higher and a lower state available, which is why fluorine can never do it.

Balance by half-reactions, adding water for oxygen and ${\mathrm{H}}^{+}$ for hydrogen, converting to basic medium afterwards with $\mathrm{O}{\mathrm{H}}^{-}$ on both sides, and always checking the total charge, since a wrong electron count leaves the atoms balanced.

A galvanic cell is a redox reaction with its halves separated so electrons must travel through a wire. Oxidation is at the anode and reduction at the cathode in every cell ever built; only the sign flips between galvanic and electrolytic.

Standard potentials are all reduction potentials against the hydrogen electrode, which is zero by convention, and $E_{cell}^{\circ }=E_{cathode}^{\circ }-E_{anode}^{\circ }$ using tabulated values as they stand.

$\Delta G^{\circ }=-nF{E}^{\circ }$ is the bridge to thermodynamics. $\Delta G^{\circ }$ is extensive and $E^{\circ }$ is intensive, so doubling a reaction doubles the first and leaves the second alone, and combining two half-reactions into a third must go through $\Delta G^{\circ }$ rather than through the potentials.

The Nernst equation corrects for real concentrations and falls to zero when the cell reaches equilibrium and dies. At 298 K the factor is $0.0591/n$ per decade, which is also the $-59.1$ mV per pH unit that every pH meter runs on.

Faraday's laws relate deposited mass to charge through 96500 C per mole of electrons, and which species is discharged depends on electrode potentials, overpotential and concentration rather than on abundance.

Conductivity falls on dilution while molar conductivity rises, because one asks about the beaker and the other about the mole. Only strong electrolytes extrapolate in $\sqrt{c}$; Kohlrausch's law supplies the limiting value for weak ones and yields both $\alpha$ and $K_a$, and even a solubility product too small to weigh.