Electrode potentials, cells and fuel cells

What you’ll learn

• Calculate cell emf from standard reduction potentials.
• Interpret cell diagrams and the reference electrode.
• Explain fuel-cell reactions and the limits of feasibility predictions.

Understand the idea

A standard electrode potential compares a half-cell with the standard hydrogen electrode, assigned $0.00\,\mathrm V$. Standard measurements use $298\,\mathrm K$, gases at $100\,\mathrm{kPa}$ and aqueous ions at $1.00\,\mathrm{mol\,dm^{-3}}$. Tables write half-equations as reductions: the more positive potential has the greater tendency to proceed in that direction. An inert platinum electrode supplies an electrical contact when a half-cell has no conducting solid species.

For a spontaneous cell as written, oxidation occurs at the anode and reduction at the cathode. A single line in a cell diagram separates phases and a double line represents the salt bridge, which allows ion movement to maintain electrical neutrality. A positive standard cell emf predicts a thermodynamically favourable reaction under standard conditions, but a slow electron-transfer process, an activation barrier or non-standard concentrations can prevent the expected observable reaction.

A fuel cell converts a continuously supplied fuel and oxidant into electrical energy. In an acidic hydrogen–oxygen cell, hydrogen forms hydrogen ions and electrons at the anode, while oxygen consumes them to form water at the cathode. In an alkaline cell, the corresponding half-equations are $\mathrm{2H_2+4OH^-\rightarrow4H_2O+4e^-}$ and $\mathrm{O_2+2H_2O+4e^-\rightarrow4OH^-}$. Both give the same overall water-forming reaction. Hydrogen-rich fuels such as methanol can also be used in suitable fuel-cell designs.

Key formulae

$$E^\circ_{\mathrm{cell}}=E^\circ_{\mathrm{cathode}}-E^\circ_{\mathrm{anode}}\quad\text{(both tabulated as reductions)}$$

$$\mathrm{H_2\longrightarrow2H^++2e^-}$$

$$\mathrm{O_2+4H^++4e^-\longrightarrow2H_2O}$$

$$\mathrm{2H_2+O_2\longrightarrow2H_2O}$$

Worked example

The standard reduction potentials are $E^\circ(\mathrm{Cu^{2+}/Cu})=+0.34\,\mathrm V$ and $E^\circ(\mathrm{Zn^{2+}/Zn})=-0.76\,\mathrm V$. Find the spontaneous cell reaction and standard emf.

1. The more positive copper half-cell undergoes reduction: $\mathrm{Cu^{2+}+2e^-\rightarrow Cu}$.
2. Reverse the zinc reduction half-equation to give $\mathrm{Zn\rightarrow Zn^{2+}+2e^-}$.
3. Add the equations and calculate $E^\circ_{\mathrm{cell}}=0.34-(-0.76)=1.10\,\mathrm V$.

Answer: $\mathrm{Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu}$, with $E^\circ_{\mathrm{cell}}=1.10\,\mathrm V$.