A concentration cell has the same half-reactions at the anode and cathode, but a voltage results from different concentrations in the two electrode compartments. (b) A similar cell has 0.10 M Cu2+ in both compartments. When a stoichiometric amount of ethylenediamine (NH2CH2CH2NH2) is added to one compartment, the measured cell potential is 0.179 V. Calculate the formation constant Kf for the complex ion Cu(NH2CH2CH2CH2)22+. Assume there is no volume change.

Video transcript

Hello this problem we are told a cell with the same half reactions at the animal. And the cathode was constructed with 0.125 Mueller silver in both electoral compartments. A stoke geometric amount of cyanide ion generating solution was carefully added to one of the two compartments. Henley solution very carefully. The cell potential was measured and recorded for future reference. The major self potential is 0.380V assuming there is no change of volume. Were asked to calculate the formation constant for our complex. So sketching out a portion of our cells. We have then silver ions at the the electrodes in both compartments and we have silver ions in solution. So when we have the same half reactions in both the old and the cathode, this is referred to as a concentration cell. So the driving force then is a difference in the concentration. So to one of these compartments, then we're gonna add cyanide. When we add cyanide, silver ryans then will react with the cyanide to form our complex on. By forming the complex sign. Then we remove silver mines from solution and so the driving force then will be to put these eyes back in solution. So that means then that the A note is on the left hand side where the cyanide is being added. So at the and then we have oxidation taking place. So we will have silver than being oxidized to form silver ryans to replace those that are being removed by the formation of the complex sign. And then at the cathode we will have reduction taking place. So we will have silver Ryan's and being reduced to form a silver medal. Our overall reaction then the number of electrons lost as you go into that game. So those cancel. And then we're left with silver medal. In the modified compartment, acting with silver Ryan's perform silver ions. The modified compartment cause silver medal. So we can make use of the nerds equation to find our reaction quotient. So the first equation we have the cell potential that is equal to the standard cell potential -0.05916V provided by N. times log of our reaction quotient. So we're told the cell potential is 0.380V for a concentration sell the standard cell potential is zero. Since we have the same reaction is taking place in both the animal and the cathode compartment. The number of electrons that are being transferred is one. And then we have the log of our reaction quotient. So the log of our reaction quotient then will be equal to 0.380V divided by a negative 0.059 16 volts. This works out to the negative 6.4233. To find a reaction quotient by taking the inverse log of both sides. This works out to 3.773 Times 10 to the -7. So our reaction quotient then based on our overall reaction is equal to the concentration of our products, which is the modified silver ion concentration all over. The concentration of our silver Ryan's in the unmodified compartment. The pure solids don't appear in our reaction quotient expression. We're trying to find the concentration of silver ions in our modified compartment. So we'll call this concentration X. And we know that the concentration of silver Ryan's and the other compartment is 0. Mueller. So then this works out to 3.77, 3 times 10 - -7. So solving for X, then Get that X is equal to 4. And sent the -8. Again, this is equal to then our concentration of silver ions in the modified compartment. You then can make a table to find our formation constant. So we'll have the reaction between our silver, bronze and cyanide ions to form the complex sign. We have initial change and final. So initially we have 0.125 Mueller silver ions were told that we had a stroke in metric amount of cyanide. So we would have two moles of cyanide for everyone. Gold silver Ryan's. We'd have 20.250 moles per liter of cyanide and initially we have none of the complex sign. So we're gonna assume that all of our ions react. So that would be -1.25. And for every one mole of silver that reacts we have two modes of cyanide, that would be -1250. And then we'd have plus 0.125 for a complex sign. Final 0, 0 and 0.125. We'll assume that we have a small back reaction. So this would then be plus X plus two X and minus X. So then our equilibrium will be our final plus our small back reaction. So this will be X two X and 0. minus sex. Writing our equilibrium constant expression we have then the concentration of our product all over. The concentration of our reactant, silver, Ryan's and cyanide. I am squared. This then works out to 0.125. My sex using the information from the table above. I'm sex times two X squared. This then works out to 0.125 -1. All over for X cubed. So we found X. Previously. This then works out to 0.125 minus 4. times 10 today, eight And the - All over four times 4.716, eight times 10 to the -8 cube. This then works out to 2.98 times 10 to the 20. So this is our formation constant and this corresponds into answer B. Thanks for watching. Hope to help