The voltage of a galvanic cell, E, depends on the concentration of reactants and products. The Nernst equation,

E = E^{o} – (0.0592/n)log(Q)

can be used to calculate the cell potential. E^{o} is the standard cell potential, *n* is the number of moles of electrons exchanged in the reaction, and *Q* is the reaction quotient. Which of the following is true regarding the value of E?

A. E is always less than E^{o}

B. E is always greater than E^{o}

C. E is always equal to E^{o}

D. E can be less than, greater than, or equal to E^{o}

**Explanation**

According to the Nernst equation, the cell potential E will vary based on the logarithm of the reactant quotient *Q*. The reactant quotient follows the same equation as the equilibrium constant *K*, but measures the ratio of the actual concentration of products to concentration of reactants at any time during the reaction. Therefore, when *Q *> 1, log(Q) is positive and E < E^{o}. If *Q *< 1, log(Q) is negative, then E > E^{o}. Lastly, E is equal to E^{o }when *Q *= 1 because the log of one is equal to zero. Thus, E can be less than, greater than, or equal to E^{o} making D the correct answer.

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