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Electrochemical Cells, Electrode Potential and the Nernst Equation

By the end you should be able to: Describe an electrochemical cell and single electrode potential, apply the Nernst equation to a single electrode and to a complete cell, and use the electrochemical series.

The electrochemical cell

A galvanic (voltaic) cell converts chemical energy into electrical energy through a spontaneous redox reaction, with the two half-reactions physically separated so the electrons must travel through an external wire.

That separation is the entire idea. Drop zinc into copper sulphate and the same reaction happens, releasing the same energy — but as heat, because the electrons pass directly between atoms in contact. Separate the halves and the electrons have to go the long way round, and you can put a motor in their path.

The Daniell cell:

  • Anode (negative in a galvanic cell): zinc in ZnSO₄. Oxidation. Zn→Zn2++2e−\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^{-}
  • Cathode (positive): copper in CuSO₄. Reduction. Cu2++2e−→Cu\text{Cu}^{2+} + 2e^{-} \rightarrow \text{Cu}
  • Salt bridge completes the circuit and keeps both solutions electrically neutral.

Oxidation is always at the anode and reduction always at the cathode. The signs reverse between galvanic and electrolytic cells, but that rule does not — which is why it is the one to hold onto.

Cell notation, by convention anode on the left:

Zn ∣ Zn2+(a1) ∣∣ Cu2+(a2) ∣ Cu\text{Zn}\,|\,\text{Zn}^{2+}(a_1)\,||\,\text{Cu}^{2+}(a_2)\,|\,\text{Cu}

A single bar is a phase boundary, a double bar the salt bridge.

Why the salt bridge matters. Without it, zinc ions accumulate on one side and sulphate on the other; the charge imbalance builds within moments and stops the reaction. The bridge lets ions migrate to neutralise it, and the cell keeps running.

Electrode potential

Any metal dipped in a solution of its ions develops a potential difference across the interface, as the metal's tendency to release ions balances against the solution's tendency to deposit them.

Only differences are measurable. A single electrode potential cannot be measured in isolation, because any measurement requires a second contact — which is itself an electrode. The solution is to pick one electrode, define it as zero, and measure everything against it. That is the standard hydrogen electrode (SHE), assigned E°=0.000E° = 0.000 V.

Standard electrode potential E°E° is measured against SHE at 298 K, 1 M ion activity, 1 bar gas pressure. By IUPAC convention it is always quoted as a reduction potential.

E°cell=E°cathode−E°anodeE°_{cell} = E°_{cathode} - E°_{anode}

For the Daniell cell:

E°cell=0.34−(−0.76)=1.10 VE°_{cell} = 0.34 - (-0.76) = 1.10\ \text{V}

A positive E°cellE°_{cell} means the reaction as written is spontaneous.

The electrochemical series

Elements arranged by standard reduction potential, most negative first.

  • More negative — stronger reducing agent, more readily oxidised, more reactive metal. Li, K, Ca, Na, Mg, Al, Zn, Fe.
  • More positive — stronger oxidising agent, more readily reduced, more noble metal. Cu, Ag, Pt, Au.

Applications:

  1. Predicting displacement. A metal displaces any metal below it from solution. Zinc displaces copper because zinc is more negative.
  2. Predicting cell EMF, as above.
  3. Predicting corrosion. When two metals are in contact, the more negative one corrodes. This single fact drives the whole of the corrosion topic.
  4. Predicting reaction with acid. Metals above hydrogen liberate H₂ from acids; those below (Cu, Ag, Au) do not.

The Nernst equation

Standard potentials assume 1 M concentrations. Real cells are rarely standard, and the potential shifts as concentrations change — indeed that shift is what makes a battery go flat.

For a general reduction Mn++ne−→M\text{M}^{n+} + ne^- \rightarrow \text{M}:

E=E°−2.303RTnFlog⁡[reduced][oxidised]E = E° - \frac{2.303RT}{nF}\log\frac{[\text{reduced}]}{[\text{oxidised}]}

At 298 K the constant evaluates to

2.303RTF=2.303×8.314×29896485=0.0591 V\frac{2.303RT}{F} = \frac{2.303 \times 8.314 \times 298}{96485} = 0.0591\ \text{V}

so the working form is

 E=E°−0.0591nlog⁡[red][ox] \boxed{\,E = E° - \frac{0.0591}{n}\log\frac{[\text{red}]}{[\text{ox}]}\,}

For a single electrode, the reduced form is the solid metal with activity 1, so it simplifies to

E=E°+0.0591nlog⁡[Mn+]E = E° + \frac{0.0591}{n}\log[\text{M}^{n+}]

For a complete cell:

Ecell=E°cell−0.0591nlog⁡[products][reactants]E_{cell} = E°_{cell} - \frac{0.0591}{n}\log\frac{[\text{products}]}{[\text{reactants}]}

Reading it physically. Raising the concentration of the oxidised species makes reduction more favourable, so EE rises. Reducing it makes EE fall. As a cell discharges, reactants are consumed and products accumulate, so the log term grows and the voltage drops — until at equilibrium E=0E = 0 and the battery is flat. A flat battery is a cell at equilibrium, which is a more useful way to think of it than "empty".