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So, my question is regarding how reactant concentration affects voltage of a galvanic cell. If the concentration is decreased significantly for instance, does voltage increase because, according to the Nernst equation, Q (reactant/products) is reduced and log of a fraction is a negative number so the right side of the equation becomes positive since a negative x a negative = a positive (sorry that was a confusing sentence)?
Nernst Equation: E=Estandard -(2.30RT/nF)log[reactant]/[product]
Ooooor, does increasing the reactant concentration increase voltage because the reaction moves more quickly to the right, meaning more electrons released at the anode and more electrons deposited on the cathode = higher voltage since since emf depends on number of electrons and rate of redox reaction? Or does this not apply because the cell is not at equilibrium? Am I confusing concepts?
Thanks for the help!
Nernst Equation: E=Estandard -(2.30RT/nF)log[reactant]/[product]
Ooooor, does increasing the reactant concentration increase voltage because the reaction moves more quickly to the right, meaning more electrons released at the anode and more electrons deposited on the cathode = higher voltage since since emf depends on number of electrons and rate of redox reaction? Or does this not apply because the cell is not at equilibrium? Am I confusing concepts?
Thanks for the help!
