A galvanic cell has an iron electrode in contact with 0.10 M FeSO4 and a copper electrode in contact with a CuSO4 solu-tion. If the measured cell potential at 25 °C is 0.67 V, what is the concentration of Cu2+ in the CuSO4 solution?

Video transcript

Hello. In this problem, we are told that we have a galvanic cell that is made up of a chromium electrode dipped in a 0.15 molar chromium three nitrate solution and a lantana electrode dipped in a solution of lantana three nitrate. We wanna know what is the concentration of lantana three in the land and um three nitrate solution. If the cell potential major 25 degrees Celsius is 1.63 volts for given the standard reduction potential for two ions. That's right. The reduction half reaction for two ions. We have chromium three being reduced to form chromium. The standard reduction potential is negative .73V. And we have one phantom three being reduced form one phantom. The standard reduction potential -2.38V. So when we compare the standard reduction potentials, the one that is the least negative has a greater potential to be reduced. So that reaction will take place at the cathode. The other reaction written in reverse will take place at the anodes. So the cathode we will have chromium three being reduced form chromium metal. Standard reduction potential is negative 0. volts at the an ode. Then we will have one phantom being oxidized And the standard oxidation potential is equal to negative of the standard reduction potential. This is equal to 2.38V. We can combine these two half reactions to come up with. Overall reaction, we see that the moles of electrons that are gained is equal to those that are lost. So those cancel what remains then is our overall reaction. So chromium three reacting with land phantom form chromium And land and um three. Our standard cell potential is equal to the standard reduction potential plus the standard oxidation potential. This is native .73V plus 2.38V. This works out to 1.65V. We can then make use of the nurse equation to find our reaction portion. The nurse equation, then we have the cell potential is equal to the standard cell potential -0.05916V divided by the most of electrons that are transferred times the log of our reaction quotient. We can make use of this equation that is simplified because we're at 25°C solving for the log of our reaction quotient, move everything else to the other side. So we are told that the cell potential is 1.63V. We found the standard cell potential to be 1.65V, three moles of electrons are being transferred. Are you know the volts will cancel. And this works out then to one 1.014, Log of our reaction quotient then is equal to log of our product concentrations which is land and um three provided by reacting concentration which is chromium three. The pure souls do not appear in our expression for the reaction quotient, this is equal to 1.0141. When we take the log of ratio is the difference of the logs. So we get then three, We have a log of phantom three minus the log of chromium three Is equal to 1.0141. Move the log of coming 3 to the other side. We have a log of phantom three That is equal to 1.0141 plus log of chromium three. Which we were told this concentration was 0.015 polar. This works out to negative 0.8097. So the concentration of land titanium three we can find by taking the inverse laudable sides. And this works out to then 0.15 Mueller. This concentration of land and um three and corresponds to answer a thanks for watching. Hope this helps