Electrochemistry
Turning chemistry into electricity and back — cells, electrode potentials, the Nernst equation, conductance, and the laws of electrolysis
- The difference between a galvanic cell (chemistry → electricity) and an electrolytic cell (electricity → chemistry).
- Electrode potential, the standard hydrogen electrode, and the electrochemical series.
- The EMF of a cell and the Nernst equation for non-standard conditions.
- The links \(\Delta G^\circ=-nFE^\circ\) and \(E^\circ=\tfrac{0.0591}{n}\log K\).
- Conductivity, molar conductivity, and Kohlrausch's law.
- Faraday's laws of electrolysis, and how batteries, fuel cells and corrosion work.
Two Kinds of Electrochemical Cell
Electrochemistry studies the two-way conversion between chemical and electrical energy. A galvanic (voltaic) cell uses a spontaneous redox reaction to produce electricity; an electrolytic cell uses an external current to drive a non-spontaneous reaction. In both, oxidation occurs at the anode and reduction at the cathode — only the sign of the electrodes flips between the two.
| Feature | Galvanic cell | Electrolytic cell |
|---|---|---|
| Energy change | chemical → electrical | electrical → chemical |
| Reaction | spontaneous \((\Delta G<0)\) | non-spontaneous \((\Delta G>0)\) |
| Anode | negative \((-)\), oxidation | positive \((+)\), oxidation |
| Cathode | positive \((+)\), reduction | negative \((-)\), reduction |
The Galvanic Cell & Cell Notation
The classic galvanic cell is the Daniell cell: a zinc electrode in \(\ce{ZnSO4}\) and a copper electrode in \(\ce{CuSO4}\), joined by a salt bridge that completes the circuit while keeping the solutions from mixing. Zinc dissolves (oxidation); copper deposits (reduction); electrons travel through the external wire from zinc to copper.
Electrode Potential & the SHE
Each electrode has a tendency to gain or lose electrons, measured as its electrode potential. Only differences are observable, so we fix a reference: the standard hydrogen electrode (SHE) is assigned \(E^\circ=0\ \text{V}\). Potentials measured against it, under standard conditions (1 M, 1 bar, 298 K), are standard electrode potentials and are tabulated as reduction potentials.
Both values are reduction potentials. A positive \(E^\circ_{\text{cell}}\) means the cell reaction is spontaneous as written. For the Daniell cell, \(E^\circ_{\text{cell}}=0.34-(-0.76)=1.10\ \text{V}\).
The Electrochemical Series
Arranging electrodes by standard reduction potential gives the electrochemical series. A more positive \(E^\circ\) marks a stronger oxidising agent (more readily reduced); a more negative \(E^\circ\) marks a stronger reducing agent. The series predicts displacement, cell EMF, and reactivity at a glance.
| Reduction half-reaction | \(E^\circ\) (V) |
|---|---|
| \(\ce{F2 + 2e^- -> 2F^-}\) | \(+2.87\) |
| \(\ce{Cu^2+ + 2e^- -> Cu}\) | \(+0.34\) |
| \(\ce{2H+ + 2e^- -> H2}\) | \(0.00\) (SHE) |
| \(\ce{Zn^2+ + 2e^- -> Zn}\) | \(-0.76\) |
| \(\ce{Li+ + e^- -> Li}\) | \(-3.04\) |
EMF & the Nernst Equation
Standard potentials assume 1 M concentrations. The Nernst equation corrects the EMF for any concentration, showing how a cell's voltage falls as it discharges toward equilibrium.
EMF, Gibbs Energy & the Equilibrium Constant
The EMF is electrical work per unit charge, so it ties directly to the free-energy change of the reaction — the bridge between this chapter and thermodynamics. Setting \(E_{\text{cell}}=0\) at equilibrium links \(E^\circ\) to the equilibrium constant.
A positive \(E^\circ\) ⇒ negative \(\Delta G^\circ\) ⇒ \(K>1\): all three say "spontaneous." The Nernst, thermodynamic, and equilibrium pictures are one statement seen three ways.
Conductance of Electrolytes
Electrolytic solutions conduct by ion movement. The resistance of a column of solution gives its conductivity \(\kappa\) (conductance of a unit cube), and dividing by concentration gives the molar conductivity \(\Lambda_m\) — the conductance of all the ions from one mole of electrolyte.
Kohlrausch's Law
At infinite dilution every ion moves independently, so the limiting molar conductivity \(\Lambda^\circ_m\) is just the sum of the ionic contributions. This is Kohlrausch's law of independent migration, and it lets us find \(\Lambda^\circ_m\) for weak electrolytes — which can never be reached directly by dilution.
Electrolysis & Faraday's Laws
In electrolysis, a current forces a non-spontaneous redox reaction — depositing metals, refining copper, producing chlorine and aluminium. Faraday's two laws quantify how much is transformed.
First law: mass deposited is proportional to the charge \(Q=It\). Second law: for the same charge, masses of different substances are proportional to their equivalent weights \(E=M/n\). One mole of electrons is one faraday, \(F=96500\ \text{C}\).
Batteries, Fuel Cells & Corrosion
Practical cells apply these ideas. Primary batteries (the dry Leclanché cell) are used once; secondary batteries (the lead storage battery, nickel–cadmium) are rechargeable; fuel cells (the \(\ce{H2}\!-\!\ce{O2}\) cell) convert fuel directly to electricity with high efficiency. Corrosion is an unwanted galvanic process — iron rusts where it acts as an anode.
| Device | Type | Note |
|---|---|---|
| Dry (Leclanché) cell | primary | \(\ce{Zn}\) anode, \(\ce{MnO2}\)/C cathode, ≈ 1.5 V |
| Lead storage battery | secondary | \(\ce{Pb}\)/\(\ce{PbO2}\) in \(\ce{H2SO4}\); rechargeable |
| \(\ce{H2}\!-\!\ce{O2}\) fuel cell | fuel cell | continuous fuel feed; water is the only product |
Putting It to Work
Problem. Find \(E^\circ_{\text{cell}}\) for the Daniell cell. \(E^\circ_{\ce{Cu^2+/Cu}}=+0.34,\ E^\circ_{\ce{Zn^2+/Zn}}=-0.76\ \text{V}\).
Solution. Copper is reduced (cathode), zinc oxidised (anode):
Problem. For the Daniell cell with \([\ce{Zn^2+}]=0.1\) and \([\ce{Cu^2+}]=0.01\ \text{M}\), find \(E_{\text{cell}}\).
Solution. Here \(n=2\) and \(Q=[\ce{Zn^2+}]/[\ce{Cu^2+}]\):
Problem. Find \(\Delta G^\circ\) for the Daniell cell \((E^\circ=1.10\ \text{V},\ n=2)\).
Solution. Apply \(\Delta G^\circ=-nFE^\circ\):
Problem. Find \(K\) for the Daniell cell reaction \((E^\circ=1.10\ \text{V},\ n=2)\).
Solution. Use \(E^\circ=\frac{0.0591}{n}\log K\):
Problem. A current of \(2.0\ \text{A}\) flows for \(30\ \text{min}\) through \(\ce{CuSO4}\). Find the mass of copper deposited \((M=63.5,\ n=2)\).
Solution. Charge \(Q=It=2.0\times1800=3600\ \text{C}\):
Problem. Acetic acid has \(\Lambda^\circ_m=390.5\) and, at \(0.1\ \text{M}\), \(\Lambda_m=5.2\ \text{S cm}^2\text{mol}^{-1}\). Find \(\alpha\) and \(K_a\).
Solution. \(\alpha=\Lambda_m/\Lambda^\circ_m\), then Ostwald's law:
Chapter Summary
Galvanic = spontaneous (chemical→electrical); electrolytic = driven. Anode oxidises, cathode reduces.
\(E^\circ_{\text{cell}}=E^\circ_{\text{cathode}}-E^\circ_{\text{anode}}\); positive means spontaneous.
\(E=E^\circ-\tfrac{0.0591}{n}\log Q\); voltage falls as the cell discharges.
\(\Delta G^\circ=-nFE^\circ\) and \(E^\circ=\tfrac{0.0591}{n}\log K\) — one reaction, three views.
\(\Lambda_m=\tfrac{\kappa\times1000}{c}\); Kohlrausch gives \(\Lambda^\circ_m\) and \(\alpha=\Lambda_m/\Lambda^\circ_m\).
Faraday: \(w=\tfrac{MIt}{nF}\); \(F=96500\ \text{C}\) per mole of electrons.
Problems
Identify the cathode and anode half-reactions first, then choose between EMF, Nernst, and Faraday relations. Take \(F=96500\ \text{C mol}^{-1}\) and \(\tfrac{0.0591}{n}\) at \(298\ \text{K}\). Difficulty rises down the list.
- Write the cell notation and \(E^\circ_{\text{cell}}\) for a cell made from \(\ce{Mg^2+/Mg}\,(-2.37)\) and \(\ce{Ag+/Ag}\,(+0.80)\).
- State which is the anode and which the cathode in a galvanic cell and in an electrolytic cell, with the sign of each.
- Predict whether \(\ce{Fe}\) can displace \(\ce{Cu^2+}\) from solution, using \(E^\circ_{\ce{Fe^2+/Fe}}=-0.44\) and \(E^\circ_{\ce{Cu^2+/Cu}}=+0.34\).
- A cell has \(E^\circ=0.46\ \text{V}\) and \(n=2\). Find \(\Delta G^\circ\).
- For the cell in Problem 4, find the equilibrium constant \(K\).
- Find \(E_{\text{cell}}\) for \(\ce{Zn|Zn^2+(0.01\,M)||Cu^2+(0.1\,M)|Cu}\) given \(E^\circ=1.10\ \text{V}\).
- The conductivity of \(0.02\ \text{M}\ \ce{KCl}\) is \(2.77\times10^{-3}\ \text{S cm}^{-1}\). Find its molar conductivity.
- Given \(\lambda^\circ_{\ce{Na+}}=50.1\) and \(\lambda^\circ_{\ce{Cl-}}=76.3\), find \(\Lambda^\circ_m\) of \(\ce{NaCl}\).
- How long must \(5\ \text{A}\) flow to deposit \(9.6\ \text{g}\) of copper \((M=63.5,\ n=2)\)?
- The same charge is passed through \(\ce{AgNO3}\) and \(\ce{CuSO4}\) in series. If \(1.08\ \text{g}\) of \(\ce{Ag}\) is deposited \((M=108)\), find the mass of \(\ce{Cu}\) \((M=63.5)\).
- Explain, using \(E^\circ\) values, why a magnesium block protects a buried iron pipeline from corrosion.
- A weak acid at \(0.05\ \text{M}\) has \(\Lambda_m=7.8\) and \(\Lambda^\circ_m=390\ \text{S cm}^2\text{mol}^{-1}\). Find \(\alpha\) and \(K_a\).