Redox Reactions and Electrochemistry

Take two copper electrodes and dip them into two copper sulphate solutions of different concentrations, joined by a salt bridge. Same metal, same ion, same reaction at both ends — so $E_{cell}^{\circ }=0$. Does the cell produce any voltage at all?

It does. Not much, but a real, measurable, usable voltage.

$E^{\circ }$ is zero because the standard reaction is identical on both sides. But a cell is driven by the difference in free energy, and free energy depends on concentration. The dilute side wants to become more concentrated and the concentrated side wants to dilute, and the cell does electrical work while that happens:

$E=\frac{0.0591}{n}\log \frac{C_{cathode}}{C_{anode}}$

The cell runs until the two concentrations are equal, at which point $E=0$ and it is dead. Nothing about the chemistry changed — only the concentrations.

That is the theme of the chapter. Advanced examines the Nernst equation and its consequences far more than it examines standard potentials, and the questions turn on concentration, pH, and which of several possible reactions actually happens.

1. Balancing redox reactions

Before any electrochemistry, the reaction must be balanced, and the ion-electron method does it mechanically.

In basic solution the protons must be removed at the end: add an equal number of hydroxide ions to both sides, combine each proton with a hydroxide to give water, and cancel any water appearing on both sides.

Disproportionation is the special case where one species is both oxidised and reduced, which requires an element in an intermediate oxidation state. Chlorine in cold alkali gives chloride and hypochlorite; copper(I) disproportionates to copper metal and copper(II) in water.

Illustration 1

Balance the reaction of permanganate with oxalate in acidic solution.

Reduction: ${\text{MnO}}_4^{-}+8{\text{H}}^{+}+5e^{-}\to {\text{Mn}}^{2+}+4{\text{H}}_2\text{O}$

Oxidation: ${\text{C}}_2{\text{O}}_4^{2-}\to 2{\text{CO}}_2+2e^{-}$

Multiplying by $2$ and $5$ to match ten electrons and adding:

$2{\text{MnO}}_4^{-}+5{\text{C}}_2{\text{O}}_4^{2-}+16{\text{H}}^{+}\to 2{\text{Mn}}^{2+}+10{\text{CO}}_2+8{\text{H}}_2\text{O}$

Check the charge as well as the atoms: $-2-10+16=+4$ on the left and $+4$ on the right, so the equation balances in both senses.

Illustration 2

Balance the reaction of permanganate with iodide in basic solution, giving manganese dioxide and iodate.

Reduction: ${\text{MnO}}_4^{-}+2{\text{H}}_2\text{O}+3e^{-}\to {\text{MnO}}_2+4{\text{OH}}^{-}$

Oxidation: ${\text{I}}^{-}+6{\text{OH}}^{-}\to {\text{IO}}_3^{-}+3{\text{H}}_2\text{O}+6e^{-}$

Doubling the first to match six electrons and adding, then cancelling water and hydroxide that appear on both sides:

$2{\text{MnO}}_4^{-}+{\text{I}}^{-}+{\text{H}}_2\text{O}\to 2{\text{MnO}}_2+{\text{IO}}_3^{-}+2{\text{OH}}^{-}$

Note the product change with medium. Permanganate goes to manganese dioxide here rather than to ${\text{Mn}}^{2+}$, which is exactly why its n-factor is $3$ in basic solution and $5$ in acid.

2. Cell potential, free energy and the equilibrium constant

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

using reduction potentials for both, and never reversing a sign. The link to thermodynamics is direct:

$\Delta G^{\circ }=-nF{E}^{\circ },E^{\circ }=\frac{0.0591}{n}\log K\text{at }298\text{K}$

with $F=96500$ C mol${}^{-1}$. A positive $E^{\circ }$ therefore means a negative $\Delta G^{\circ }$ and a $K$ greater than one, which is the fastest route from a table of potentials to an equilibrium constant.

Potentials are intensive. Multiplying a half-reaction through by two doubles $n$ and doubles $\Delta G^{\circ }$, but leaves $E^{\circ }$ exactly unchanged.

Illustration 3

For the Daniell cell, $E^{\circ }\left({\text{Cu}}^{2+}/\text{Cu}\right)=+0.34$ V and $E^{\circ }\left({\text{Zn}}^{2+}/\text{Zn}\right)=-0.76$ V. Find $E_{cell}^{\circ }$, $\Delta G^{\circ }$ and $K$.

$E_{cell}^{\circ }=0.34-\left(-0.76\right)=1.10$ V

$\Delta G^{\circ }=-2\times 96500\times 1.10=-2.12\times 10^5$ J mol${}^{-1}$

$\log K=\frac{2\times 1.10}{0.0591}=37.2\Rightarrow K=1.6\times 10^{37}$

Essentially complete. A cell potential of just over a volt already corresponds to an equilibrium constant of $10^{37}$, which is why redox reactions with modest potentials go so far to completion.

3. The Nernst equation

$E=E^{\circ }-\frac{0.0591}{n}\log Q\text{at }298\text{K}$

Three habits make it reliable. Write $Q$ with products over reactants exactly as for any equilibrium; include only species whose concentration can vary, so pure solids and pure liquids are omitted; and remember that gases enter as partial pressures.

Because ${\text{H}}^{+}$ appears in many half-reactions, pH changes potentials. The permanganate couple in acid consumes eight protons, so its potential falls steeply as the solution becomes less acidic — which is why permanganate is a much weaker oxidant in neutral solution.

Illustration 4

Find the potential of the hydrogen electrode at pH $4$ and $1$ atm of hydrogen.

$2{\text{H}}^{+}+2e^{-}\to {\text{H}}_2$, with $E^{\circ }=0$ by definition.

$E=0-\frac{0.0591}{2}\log \frac{1}{{\left[{\text{H}}^{+}\right]}^2}=0.0591\log \left[{\text{H}}^{+}\right]=-0.0591\times 4$

$E=-0.236$ V

The hydrogen electrode's potential is $-0.0591\text{pH}$ exactly, which is what makes it usable as a pH sensor and why the glass electrode that replaced it obeys the same relation.

Illustration 5

For ${\text{MnO}}_4^{-}+8{\text{H}}^{+}+5e^{-}\to {\text{Mn}}^{2+}+4{\text{H}}_2\text{O}$, $E^{\circ }=+1.51$ V. Find the potential at pH $3$ with all other species at unit concentration.

$E=1.51-\frac{0.0591}{5}\log \frac{1}{{\left[{\text{H}}^{+}\right]}^8}=1.51+\frac{0.0591\times 8}{5}\log \left[{\text{H}}^{+}\right]$

$=1.51-\frac{0.0591\times 8\times 3}{5}=1.51-0.284=1.23$ V

A drop of nearly $0.3$ V for three pH units. Because eight protons appear, the potential falls by about $0.095$ V per pH unit, which is why permanganate titrations are always done in strongly acidic solution.

4. Concentration cells

Two kinds exist. In an electrolyte concentration cell the electrodes are identical and only the solution concentrations differ, as in the hook. In an electrode concentration cell the solutions match and the electrodes differ, for instance two hydrogen electrodes at different pressures.

$E=\frac{0.0591}{n}\log \frac{C_{cathode}}{C_{anode}}$

The maximum potential such a cell can produce is small — a tenfold ratio with $n=1$ gives only $59$ mV — but the relationship is exact and is the basis of every ion-selective electrode.

Illustration 6

Find the potential of a cell made from silver electrodes in $0.01$ M and $1.0$ M silver nitrate.

The dilute half-cell is the anode, since silver dissolves where the ion concentration is low.

$E=\frac{0.0591}{1}\log \frac{1.0}{0.01}=0.0591\times 2=0.118$ V

One hundred and eighteen millivolts from no chemistry at all. The driving force is entirely the entropy of mixing, which is why the cell stops the instant the two concentrations become equal.

5. Conductance: Kohlrausch and degree of dissociation

Conductivity $\kappa$ measures a solution's ability to carry current; molar conductivity divides that by concentration:

${\Lambda }_m=\frac{1000\kappa }{C}$

with $\kappa$ in S cm${}^{-1}$ and $C$ in mol L${}^{-1}$. On dilution, both electrolytes behave differently, and the contrast is the point of the topic.

A strong electrolyte's molar conductivity falls linearly with the square root of concentration, so ${\Lambda }_m^{\circ }$ is obtained by extrapolating to zero. A weak electrolyte's rises steeply near zero and cannot be extrapolated at all, so its limiting value must come from Kohlrausch's law:

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

Ionic conductivities being additive, the value for acetic acid is assembled from those of hydrochloric acid, sodium acetate and sodium chloride. The degree of dissociation then follows:

$\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^{\circ }},K_a=\frac{C{\alpha }^2}{1-\alpha }$

Illustration 7

The molar conductivity of $0.001$ M acetic acid is $49.1$ S cm${}^2$ mol${}^{-1}$, and its limiting value is $390.7$. Find the degree of dissociation and $K_a$.

$\alpha =\frac{49.1}{390.7}=0.1257$

$K_a=\frac{0.001\times {\left(0.1257\right)}^2}{1-0.1257}=\frac{1.58\times 10^{-5}}{0.874}=1.81\times 10^{-5}$

Which is the accepted value for acetic acid. Conductance gives an independent route to an acid constant, needing no pH measurement at all.

Illustration 8

Given ${\Lambda }_m^{\circ }$ values of $426.2$ for $\text{HCl}$, $91.0$ for ${\text{CH}}_3\text{COONa}$ and $126.5$ for $\text{NaCl}$, find ${\Lambda }_m^{\circ }$ for acetic acid.

${\Lambda }_m^{\circ }\left({\text{CH}}_3\text{COOH}\right)={\Lambda }_m^{\circ }\left(\text{HCl}\right)+{\Lambda }_m^{\circ }\left({\text{CH}}_3\text{COONa}\right)-{\Lambda }_m^{\circ }\left(\text{NaCl}\right)$

$=426.2+91.0-126.5=390.7$ S cm${}^2$ mol${}^{-1}$

Sodium and chloride cancel exactly, leaving hydrogen and acetate — which is the whole content of Kohlrausch's law of independent ionic migration.

6. Electrolysis: Faraday's laws and selective discharge

$w=\frac{M}{nF}\times It$

so the mass deposited is proportional to charge passed, and for the same charge through different cells the masses are in the ratio of their equivalent masses.

The harder question is which ion discharges when several could. Three factors decide it:

Electrode potential — the species easiest to reduce goes first at the cathode, the easiest to oxidise at the anode. Concentration, through the Nernst equation. And overpotential, the extra voltage a gas needs above its thermodynamic value, which is why hydrogen appears from brine rather than sodium, and why chlorine is released in preference to oxygen despite oxygen's more favourable potential.

Illustration 9

A current of $2$ A is passed for $30$ minutes through molten aluminium chloride. Find the mass of aluminium deposited.

Charge $=2\times 1800=3600$ C

$n=3$ for ${\text{Al}}^{3+}$

$w=\frac{27}{3\times 96500}\times 3600=\frac{27\times 3600}{289500}=0.336$ g

Aluminium is expensive to extract for exactly this reason. Three faradays per mole is three times the charge iron would need, and the process runs continuously.

Illustration 10

Explain why electrolysis of aqueous sodium chloride gives hydrogen and chlorine rather than sodium and oxygen.

At the cathode, sodium would require $-2.71$ V while water reduction to hydrogen requires only $-0.83$ V at pH $7$. Hydrogen wins on potential.

At the anode, oxygen from water is thermodynamically easier ($+0.82$ V at pH $7$) than chlorine ($+1.36$ V), so oxygen ought to appear.

Chlorine appears instead because oxygen evolution carries a large overpotential on most electrode materials, and because concentrated brine shifts the chlorine potential favourably.

This is the one place where kinetics overrules thermodynamics in the syllabus, and the entire chlor-alkali industry depends on it.

7. Batteries, fuel cells and corrosion

A primary cell is not rechargeable; a secondary cell is. The lead storage battery is the standard secondary example, delivering about $2.0$ V per cell:

$\text{Pb}+{\text{PbO}}_2+2{\text{H}}_2{\text{SO}}_4\to 2{\text{PbSO}}_4+2{\text{H}}_2\text{O}$

Both electrodes become lead sulphate on discharge, and the reaction is reversed on charging. Because sulphuric acid is consumed, the electrolyte density falls as the battery discharges, which is how its state of charge is measured.

A fuel cell is not a store but a converter, supplied continuously with reactants. The hydrogen-oxygen cell gives $E^{\circ }=1.23$ V and, being unconstrained by the Carnot limit, reaches efficiencies far above any heat engine.

Corrosion is an electrochemical cell operating on a single piece of metal: iron oxidises at one point and oxygen is reduced at another, with the water film as electrolyte. Protection therefore works by breaking that cell — by coating, by cathodic protection with a more reactive metal, or by galvanising.

Illustration 11

Explain why zinc protects iron from rusting even when the coating is scratched, while tin does not.

Zinc has $E^{\circ }=-0.76$ V against iron's $-0.44$ V, so zinc is more easily oxidised and corrodes preferentially, protecting the iron cathodically. The protection continues even where the iron is exposed.

Tin has $E^{\circ }=-0.14$ V, making it less easily oxidised than iron. Intact tin plate works as a physical barrier, but once scratched the iron becomes the anode of a cell in which tin is the cathode, and it rusts faster than bare iron would.

A scratch reverses the outcome entirely. This is why galvanised buckets outlast tinned cans that have been dented.

Illustration 12

A lead-acid battery is described as discharged when its electrolyte density falls from $1.28$ to $1.15$ g mL${}^{-1}$. Explain the connection.

The discharge reaction consumes sulphuric acid and produces water at both electrodes.

Since sulphuric acid is much denser than water, the electrolyte becomes progressively more dilute and its density falls.

A hydrometer therefore reads the state of charge directly, without any electrical measurement.

The stoichiometry is unusually convenient. Few cells have an electrolyte whose composition changes measurably with the extent of discharge.

Illustration 13

A cell has $E^{\circ }=+0.46$ V with $n=2$. Find $K$ and $\Delta G^{\circ }$, and state the direction when $Q=10^{20}$.

$\log K=\frac{2\times 0.46}{0.0591}=15.6\Rightarrow K=4\times 10^{15}$

$\Delta G^{\circ }=-2\times 96500\times 0.46=-88.8$ kJ mol${}^{-1}$

At $Q=10^{20}$, $E=0.46-\frac{0.0591}{2}\times 20=0.46-0.591=-0.131$ V

Negative, so the cell runs backwards. Since $Q$ exceeds $K$ by four orders of magnitude, there is too much product and the reaction must reverse — the electrochemical statement of Le Chatelier's principle.

Illustration 14

The same quantity of electricity is passed through solutions of silver nitrate and copper sulphate in series. If $2.16$ g of silver is deposited, find the mass of copper.

Equivalents are equal in a series arrangement.

Silver: $n=1$, so equivalents $=\frac{2.16}{108}=0.02$

Copper: $n=2$, equivalent mass $=\frac{63.5}{2}=31.75$

Mass $=0.02\times 31.75=0.635$ g

Equal charge means equal equivalents, never equal moles. Working in moles here would give twice the correct answer for copper.

Summary

• A concentration cell works with $E^{\circ }=0$: $E=\frac{0.0591}{n}\log \frac{C_{cathode}}{C_{anode}}$, driven purely by mixing.
• Ion-electron method: balance O with water, H with protons, charge with electrons; in base, neutralise the protons with hydroxide afterwards.
• Disproportionation needs an intermediate oxidation state — chlorine in cold alkali, copper(I) in water.
• $E_{cell}^{\circ }=E_{cathode}^{\circ }-E_{anode}^{\circ }$, both as reduction potentials, with no sign reversal.
• $\Delta G^{\circ }=-nF{E}^{\circ }$ and $E^{\circ }=\frac{0.0591}{n}\log K$; potentials are intensive and do not scale with the equation.
• Nernst: $E=E^{\circ }-\frac{0.0591}{n}\log Q$, omitting pure solids and liquids, with gases as partial pressures.
• pH changes potentials: the hydrogen electrode gives exactly $-0.0591\text{pH}$, and permanganate loses about $0.095$ V per pH unit.
• Strong electrolytes: ${\Lambda }_m$ falls linearly in $\sqrt{C}$ and extrapolates. Weak electrolytes: it rises steeply and cannot be extrapolated.
• Kohlrausch: ${\Lambda }_m^{\circ }={\nu }_{+}{\lambda }_{+}^{\circ }+{\nu }_{-}{\lambda }_{-}^{\circ }$, so acetic acid comes from $\text{HCl}+{\text{CH}}_3\text{COONa}-\text{NaCl}$.
• $\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^{\circ }}$ then gives $K_a$ without any pH measurement.
• Faraday: $w=\frac{M}{nF}It$; equal charge deposits equal equivalents, never equal moles.
• Discharge is decided by potential, concentration and overpotential — which is why brine gives hydrogen and chlorine.
• Lead-acid: $2.0$ V per cell, both electrodes becoming ${\text{PbSO}}_4$, and the electrolyte density falls as it discharges.
• Fuel cells are converters not stores, escape the Carnot limit, and give $1.23$ V for hydrogen and oxygen.
• Zinc protects iron cathodically even when scratched; tin protects only as an intact barrier and accelerates rusting once broken.