Question

Given a potential of 0.456 V, for the following electrochemical cell, what is the concentration of Ag+?

Cu2+(aq) (1.0 M)|Cu(s)||Ag+(aq)|Ag(s)

Answer 1

The first that we need to define are the reactions involved, and by the nomenclature of the cell we can see that in the anode is taking place the oxidation of the Cu, and in the cathode the reduction of the Ag, if we look in a table of reduction potentials.

now, we need to turn around the equation of the Cu to obtain the oxidation reaction (remember that this will change the sign of the potential) and multiply the Ag reaction by 2, in order to obtain the global reaction of the cell.

if we evaluate the Nernst equation,

from this equation we already know the potential of the cell, E, the estandar potencial, E⁰ , the numbers of moles of electrons, n, and the concentration of Cu⁺² , 1M. We only need to isolate the Ag⁺ concentration.

The concentration of Ag⁺ is 0,855 M.