cu/cu^2+ concentration cell has voltage of .21V at 25 c. the concentration of cu

Question

cu/cu^2+ concentration cell has voltage of .21V at 25 c. the concentration of cu^2+ in one of the other half cell is 1.4E-3M. What is the concentration of{Cu^2 + in the other half-cell?

Explanation / Answer

the Eo for a concentration cell has the voltage for : (oxidation) Cu --> Cu+2 & 2 e- @ - 0.337 volts, cancelled by (reduction) Cu+2 & 2 e- --> Cu @ + 0.337 volts giving a total Eo of zero, @ standard conditions, (1 Molar), for the cell: Cu & Cu+2 Cu & Cu+2 Eo = zero, a dead cell @ equilibrium using the Nernst equation: E = Eo - (0.0592/n) (logQ) E = 0 - (0.0592/n) (logQ) 0.22 volts = 0 - (0.0592 / 2 mole e-) (log Q) 0.22 volts = - (0.0296) (log [products] / [reactants] now to force a concentration cell into working, begs the question ... which concentration goes where... LeChatelier answer the question, when he says that the equilibrium of the cell can be shifted into working as written from left ot right if you increase the concentration of the starting material, &/or decrease the concentration of the products so: Cu & Cu+2 (more) --> --> Cu & Cu+2 (less) since 1.6×10-3 Molar is already quite less than the standard "1 Molar",.... I would like to make it the "lesser porduct" concentration,... and then solve for the "greater starting material" concentration: 0.22 volts = - (0.0296) (log [products] / [reactants] 0.22 volts = - (0.0296) (log [1.6 e-3 ] / [reactants] 0.22 / - (0.0296) = (log [1.6 e-3 ] / [reactants] - 7.4 = log [1.6 e-3 ] / [reactants] we will take it out of logs, by doing a 10^x of both sides 3.695 e-8 = [1.6 e-3 ] / [reactants] [reactants] = [1.6 e-3 ] / 3.695 e-8 [reactants] = [1.6 e-3 ] / 3.695 e-8 reactants = [Cu+2] = 4.3 e4 Molar only the ion concentrations are used, because the solid Cu metal are not soluble,... [...] contain soluble materials only is a 43,000 Molar concentration logical... not really, but neither is a concentration cell with a 0.22 voltage,... that's huge your answer is : 4.3 ×10 +4 Molar

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