The cell potentials we have been calculating on the previous pages were all for an electrochemical cell under "standard" conditions. By definition, standard conditions mean that any dissolved species have concentrations of 1 M, any gaseous species have partial pressures of 1 atm and the cell is operating at 25ºC.

In this section, we consider cells in non-standard conditions.

Let’s start by considering an electrochemical cell that is spontaneously pushing electrons along the wire, and ions across the salt bridge.

The voltage of an electrochemical cell is an indication that the system is out of equilibrium. The redox reaction is spontaneously approaching equilibrium, and as the reaction proceeds, electrons flow within the cell. At equilibrium, the voltage drops to zero and the current stops. (A battery at equilibrium is a dead battery.) If we make a change to the system that pushes it further away from equilibrium, the potential goes up. If we make a change that pushes the system towards equilibrium, the potential goes down.

In this example, the spontaneous reaction increases [Zn²⁺] and decreases [Cu²⁺]. If we dilute the Zn half-cell, (lower [Zn²⁺]) we are moving the reaction away from equilibrium and so the cell potential goes up. If we dilute the Cu half-cell (lower [Cu²⁺]), we are moving the reaction towards equilibrium, and so the cell potential goes down.

For cases where we change both [Zn²⁺] and [Cu²⁺], we have to refine what we mean by moving the system closer to or away from equilibrium. The way to do this is to consider the reaction quotient, Q, for the redox reaction. For this reaction Q = [Zn²⁺]/[Cu²⁺]. At standard conditions, all concentrations are 1M and so Q=1. Since the reaction is spontaneous as written, K >> 1. As the cell runs, Q approaches K and the cell potential drops. So in cases where both [Zn²⁺] and [Cu²⁺] are altered, we need to consider the effects on their ratio (i.e. Q=[Zn²⁺]/[Cu²⁺])when deciding the effects on the cell potential.

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