Consider a Daniell cell with 1.0 M ion concentrations:

Does the cell voltage increase, decrease, or remain the same when each of the following changes is made? Explain. (a) Write a balanced equation for each cell reaction.

Video transcript

Hello. In this problem we are asked to determine what happens to the cell voltage when each of the following changes made to the daniell cell. Originally in standard conditions were asked to show the balanced equation for each reaction. So under standard conditions it means then that the concentration of all solutions is one moller. When we look at the daniel sell the a note is on the left and the cathode is on the right. So the reaction taking place at the piano. Then you have zinc metal being oxidized forms inclines and at the cathode we have reduction taking place so we have copper irons being reduced to form copper metal overall reaction. Then when we combine these two we have two moles of electrons being lost, canceled with two moles, electrons being gained our overall reaction. Then zinc metal reacting with copper lines form zinc ions in complementary all we can answer the question as to what happens to our cell voltage with each of the changes. If we consider the nurse equation, the nurse equation then we have the cell potential is equal to the standard cell potential -0.05916V divided by the moles of electrons that are transferred times the log of our reaction quotient. Our reaction quotient then is equal to our product concentrations but by reacting concentrations our peer solids do not appear in our reaction quotient expression going to the first change. Then we're told that we have $2.25 to nitrate, zinc nitrate is added to the Inuit compartment. And so in the original daniel cell we were under standard conditions. So the concentration of zinc ions and copper mines was one Moeller. So by increasing the concentration of zinc lines we're gonna increase our reaction quotient this world ain't cause the log of the reaction quotient to be increased. And looking at the first equation that will cause the second term to increase which has a negative sign. This will then result in decreasing our so potential. We consider the next case where we add copper two sulfate 2.25 molar copper two sulfate to the cathode compartment. Looking at our reaction potion, we have zinc mines over copper irons. So we're going to end up increasing in the copper iron concentration from where they were under standard conditions This will result then in Cuba. And was the one which will cause then the log of our reaction portion to be negative and signed going back to our nurse equation. This then will cause the second term to be positive and negative and negative is a positive. This will then cause the cell potential to increase. Considering C then says we have a one molar copper to nitrate solution is added to the third compartment. So the original daniel also had a concentration of one molar. So by adding a one molar solution of copper to nitrate we're not gonna cause any change. And then the last change we're told that We're gonna add sulfuric acid, we're gonna add 2.25 molar sulfuric acid to the cathode compartment. When we add sulfuric acid to the cathode compartment, this is going to change. Then the production half reaction. We're now going to have hydro nines being reduced form hydrogen gas. And the standard reduction potential is equal to 0V. So if we consider our original Danielle cell, then at the an ode we had sink being oxidized form zinc ions. Standard oxidation potential is equal to 0.76V at the cathode we had. Then copper ions being reduced from copper metal. Standard reduction potential 0.34V. So the standard cell potential providing these two works out to 1.10V in our modified cell. The reaction taking place at the anodes has not changed. We still have zinc metal being oxidized to form zinc ions. So the standard oxidation potential remains the same now with the cathode though, we have hydrogen ions being reduced form hydrogen gas. The standard reduction potential Zero volts. So the standard cell potential is equal to 0.76V. So by adding then sulfuric acid to the cathode compartment because the Standard cell potential to be decreased from 1.10V to 0.76V. Thanks for watching. Hope this helped