Electrochemical cells: voltage, conditions and controls

What you’ll learn

• Calculate a cell potential from reduction potentials.
• Explain the roles of electrodes, a salt bridge and a high-resistance voltmeter.

Understand the idea

Core practical 12 relates measured cell voltages to two half-cells. Tabulated standard electrode potentials are reduction potentials. For a feasible standard cell direction, the more positive reduction occurs at the cathode and the other half-reaction is reversed at the anode. Subtract the anode reduction potential from the cathode reduction potential. Multiplying a half-equation to balance electrons does not multiply its electrode potential.

The external connection allows electron transfer, while a salt bridge provides ionic conduction and prevents charge build-up in the solutions. A high-resistance voltmeter draws very little current, keeping the reading close to the cell's open-circuit potential. A clean conducting electrode provides contact with its redox system; an inert electrode is needed when the couple contains no suitable solid conductor. Salt-bridge ions must not react with the half-cell components.

Standard comparisons refer to the specified temperature, gas pressure and solution concentrations: conventionally $298\ \mathrm{K}$, $100\ \mathrm{kPa}$ and $1.00\ \mathrm{mol\,dm^{-3}}$ for the relevant dissolved species in this course. Different concentrations, contaminated electrodes or poor connections can change a measured result. A positive standard cell potential suggests thermodynamic feasibility under standard conditions; it does not promise a fast reaction or describe every non-standard mixture.

Key formulae

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

$$\mathrm{Zn(s)}\rightarrow\mathrm{Zn^{2+}(aq)}+2e^-$$

$$\mathrm{Cu^{2+}(aq)}+2e^-\rightarrow\mathrm{Cu(s)}$$

$$\mathrm{Zn(s)}+\mathrm{Cu^{2+}(aq)}\rightarrow\mathrm{Zn^{2+}(aq)}+\mathrm{Cu(s)}$$

Worked example

Standard reduction potentials are $E^{\circ}(\mathrm{Zn^{2+}/Zn})=-0.76\ \mathrm{V}$ and $E^{\circ}(\mathrm{Cu^{2+}/Cu})=+0.34\ \mathrm{V}$. Find the standard cell potential for the spontaneous cell and state the electron-flow direction in the wire.

1. Copper has the more positive reduction potential, so copper ions are reduced at the cathode.
2. Zinc is oxidised at the anode. The potential difference is $0.34-(-0.76)=1.10\ \mathrm{V}$.
3. Electrons released by zinc travel through the wire from the zinc electrode to the copper electrode.

Answer: $1.10\ \mathrm{V}$; electrons flow from zinc to copper in the external wire.