A galvanic cell's voltage drops to zero at equilibrium because the electrical potential of the two half-cells becomes perfectly equal. The cell produces voltage by allowing a spontaneous redox reaction to occur. At equilibrium, the forward and reverse reactions proceed at the exact same rate, meaning no net work can be done and no electrons can flow through the external circuit. As the cell operates, products are formed and reactants are consumed, driving the reaction quotient Q toward the equilibrium constant K. At chemical equilibrium, Q = K.
Because there is no longer a driving force for the reaction, the Gibbs free energy change Delta G equals zero. Consequently, the cell potential Ecell drops to exactly '0'volts.
Mathematicaly we know that
Delta G = -nFEcell = -(nF)V = - (total charge transfered) x Voltage = work done by the cell = loss in its potential energy.
so Delta G = 0 is possible only when Ecell = 0
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As the reaction approaches equilibrium, the anode and cathode half-cells alter their concentrations:
At the anode: The concentration of oxidized species (products) increases, making it less negative.
At the cathode: The concentration of reduced species (reactants) decreases, making it less positive.
At equilibrium, the reduction potentials of both electrodes become completely equal, making the potential difference (voltage) between them zero.
As the cell runs, reactants are consumed and products are formed, which causes Q to increase. This causes the subtracting term to grow , steadily decreasing the voltage. At equilibrium, the reaction quotient equals the equilibrium constant Q = K, and the equation becomes E cell = 0.
Because as Q approaches K , (RT/nF)ln(Q) becomes = Ecell_standard. Hence Ecell = 0.
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