Consider a galvanic cell that uses the reaction Calculate the potential at 25 °C for a cell that has the following ion concentrations: [Ag+] = 0.010M, [Ni2+] = 0.100M.

Video transcript

Hello. In this problem we are asked to consider a galvanic cell with the following reaction. We want to know what is potential to sell at 25 degrees Celsius at the concentration of lead to and gold three is 30.250 moller and 0.150 molar. Respectfully were given the standard reduction potential for lead to and gold three. Looking at the overall reaction that were provided, we have gold three being reduced. Warm gold standard reduction potential, It's equal to 1.50V. And then we have led being oxidized from lead to. We have the standard oxidation potential which is the negative of the standard reduction potential. This works out 2.13V. So to combine these two reactions to get our overall reaction moles of electrons being gained has to equal those being lost. And so we'll take and multiply the first reaction by two and the second reaction by three. So we then have two times three is six moles of electrons being gained and three times two is six being lost. So those will cancel what's left over then gets us to the overall reaction that we were provided in the problem statement. And our standard cell potential then is our standard production potential plus our standard oxidation potential 1.50 volts plus 0.13 volts. So 1.63 volts. You can then find the potential of the cell using the equation. So we have we sell is equal to the standard cell potential -0.05916V divided by the moles of electrons that are transferred times log of our reaction portion. So we found the standard cell potential to be 1.63V. The moles of electrons that are being transferred is six. And the log of our reaction quotient comes from our balanced reaction. So we have our product concentrations, the concentration of lead to that's cubed divided by the concentration of our reactant that's called three and that squared. So the parasol is don't appear in our reaction quotient expression, Then have 1.63V -0.05916V divided by six. And we're told that the concentration of lead two is 0.250 Mueller. So that's cubed and the concentration of gold three is 0.150. And that will be squared. So our self potential then works out to 1.61V. This then corresponds to answer D Thanks for watching. Hope this out