A chemistry student wants to determine the concentration of copper(II) ions in a solution using electrochemical methods. The student prepares a copper electrode and a reference electrode, and sets up a cell with the copper electrode immersed in the solution and the reference electrode in the salt bridge. The standard cell potential is measured to be +0.34 V. If the student knows that the reduction of copper(II) to copper occurs at the copper electrode, what is the concentration of copper(II) ions in the solution?

Question

A chemistry student wants to determine the concentration of copper(II) ions in a solution using electrochemical methods. The student prepares a copper electrode and a reference electrode, and sets up a cell with the copper electrode immersed in the solution and the reference electrode in the salt bridge. The standard cell potential is measured to be +0.34 V. If the st

Answer 1

To determine the concentration of copper II  ions in the solution, we can use the Nernst equation, which relates the cell potential to the concentrations of the species involved in the redox reaction. The Nernst equation is given by:E = E -  RT/nF  * ln Q where:E = cell potential  +0.34 V E = standard cell potential for the reduction of Cu to Cu  0.34 V R = gas constant  8.314 J/molK T = temperature  assuming 298 K, room temperature n = number of electrons transferred in the redox reaction  2 for Cu to Cu F = Faraday's constant  96485 C/mol Q = reaction quotient, which is the ratio of the concentrations of the products to the reactantsFor the reduction of Cu to Cu, the balanced half-reaction is:Cu + 2e  CuThe reaction quotient, Q, for this half-reaction is:Q = [Cu] / [Cu]Since the concentration of solid copper [Cu] is constant, we can simplify the Nernst equation:E = E -  RT/nF  * ln [Cu] / [Cu] We can rearrange the equation to solve for the concentration of copper II  ions, [Cu]:ln [Cu]  =  E - E  *  nF / RT Now, we can plug in the values:ln [Cu]  =  0.34 V - 0.34 V  *  2 * 96485 C/mol  /  8.314 J/molK * 298 K ln [Cu]  = 0Taking the exponent of both sides:[Cu] = e^0 = 1 MSo, the concentration of copper II  ions in the solution is 1 M  1 mol/L .