The left hand half-cell containing the metal M and its ion Mz+ in solution is in equilibrium with electrons in the circuit to the left of the potentiometer. The half-cell reaction describing this equilibrium is:
The right hand compartment comprises a hydrogen half-cell. Gaseous H2 saturates the solution with molecular hydrogen, which, by virtue of rapid reaction on the inert metal electrode, maintains equilibrium with hydrogen ions in solution and electrons in the right side of the external circuit. This half-cell reaction is:
The overall cell reaction is the difference between the two half-cell reactions (with Eq (10.20) multiplied by z/2 in order to cancel the electrons):
Digression: If the cell is short-circuited, the reactions in the half-cells correspond to chemical attack of the metal by acid. The overall cell reaction of (10.21) proceeds from left to right. The metal half-cell is the anode, where oxidation (in the sense of increasing the valence of M) occurs and reaction (10.19) proceeds from left to right. The hydrogen half-cell is the cathode, where protons are reduced to elemental hydrogen, or reaction (10.20) goes from right to left.
The introduction of metal ions into the anode solution and removal of hydrogen ions from the cathode solution upsets electrical neutrality of these solutions. This charge imbalance is rectified by flow of anions (e.g., Cl-), from right to left. This compensating charge flow must pass through the bridge in Fig. 10.5. The bridge is a porous structure or a gel that serves to prevent gross mixing of the solutions in the two half-cells while allowing for transport of negative ions to maintain electrical neutrality during cell operation.
In the equilibrium mode, the cell EMF is the result of the difference of the electric potentials of the electrons in the left-hand and right-hand electrodes. To determine how this EMF relates to the chemical potentials of the constituents of overall reaction (10.21), the chemical potential balance is written including the electric potentials of the electrodes. For the left-hand and right-hand cells in Fig. 10.5, the total potentials are:
where <L and <R are the electric potentials (in Volts) in the left-hand and right-hand electrodes, respectively. They are multiplied by the Faraday, F = 96.5 kJ/mole-Volt in order to convert the electric potentials to the same units as the chemical potentials. The z/2 factor in Eq (10.22R) is a consequence of the stoichiometry of overall reaction (10.21)
Equilibium is expressed by: ^ L R (10.23)
Substituting Eqs (10.22L) and (10.22R) into Eq (10.23) yields
The term on the left is A^, the chemical-potential difference of overall reaction (10.21). It is the aqueous equivalent of the free-energy difference AG used in describing nonaqueous cells. The electric potential difference on the right is the cell EMF, s, so the equation is:
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