What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.370 V?
Given: A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25degrees C . The initial concentrations of Pb2+ and Cu2+ are 5.30×10−2 M and 1.60 M, respectively.
The Einitial cell that I calculated was 0.47. I know how to figure out the Q, but I don't know how to figure out the specific concentrations of Pb2+ and Cu2+ given only this information. Please help me! Thank you :)
Try this. Q = (Pb^2+)/(Cu^2+)= ?
You know Pb goes into soln so it will be 0.053+x and Cu^2+ comes out of soln and it will be 1.6-x.
It worked! Thank you so much :)
pb2+ +.....=pb
Well, well, well, looks like we have a voltage-drop detective here! Let's see if Clown Bot can crack this electrifying case for you.
To find the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.370 V, we need to use the Nernst equation. This equation allows us to calculate the cell potential at any given concentration.
The Nernst equation is:
Ecell = E°cell - (0.0592/n) * log(Q)
Where:
- Ecell is the cell potential
- E°cell is the standard cell potential
- n is the number of electrons transferred in the balanced equation
- Q is the reaction quotient
In this case, we are given the initial cell potential (Einitial) as 0.47 V, and we want to know the concentrations of Pb2+ and Cu2+ when the cell potential drops to 0.370 V.
Now, the reaction quotient (Q) is determined by the concentrations of the species involved in the reaction. In our case, we have:
Q = [Pb2+]/[Cu2+]
Uh-oh, seems like we're missing something here. We need to know the number of electrons transferred (n) in the balanced equation. Could you kindly provide the balanced equation for this electrochemical cell? Then, Clown Bot will be able to calculate this for you and help you find the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.370 V. Let's clear this electrified fog together!
To determine the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.370 V, we can use the Nernst equation. This equation relates the cell potential to the concentrations of the species involved in the redox reaction.
The Nernst equation is given by:
Ecell = E°cell - (0.0592/n) * log(Q)
Where:
Ecell is the cell potential
E°cell is the standard cell potential
n is the number of electrons transferred in the balanced redox equation
Q is the reaction quotient
In this case, the initial cell potential (Einitial) is 0.47 V. To find the concentrations of Pb2+ and Cu2+ at a cell potential of 0.370 V, we need to rearrange the Nernst equation and solve it for Q:
Ecell = E°cell - (0.0592/n) * log(Q)
0.370 V = 0.47 V - (0.0592/n) * log(Q)
Now, let's plug in the known values:
E°cell = 0.47 V (given)
n = number of electrons transferred (which can be determined from the balanced redox equation)
Q = (concentration of products) / (concentration of reactants)
In this case, the balanced redox equation is:
Pb + Cu2+ -> Pb2+ + Cu
From the equation, we can see that n = 2, as two electrons are transferred.
Now, we need to find the concentrations of Pb2+ and Cu2+ at the new cell potential. Rearranging the equation above, we have:
Q = 10^((E°cell - Ecell) * n / 0.0592)
Plugging in the values:
Q = 10^((0.47 V - 0.370 V) * 2 / 0.0592)
Calculating the value of Q:
Q = 10^(0.1 * 2 / 0.0592)
Q = 10^(3.37)
Now, we can use the calculated value of Q to find the concentrations of Pb2+ and Cu2+ using the balanced redox equation:
Q = [Pb2+]/[Cu2+]
3.37 = [Pb2+]/1.60
Solving for [Pb2+]:
[Pb2+] = 3.37 * 1.60
[Pb2+] = 5.392 M
Therefore, at a cell potential of 0.370 V, the concentration of Pb2+ is 5.392 M.
To find the concentration of Cu2+, we can use the relationship Q = [Pb2+]/[Cu2+] and the concentration we just calculated for Pb2+:
3.37 = 5.392/[Cu2+]
Solving for [Cu2+]:
[Cu2+] = 5.392 / 3.37
[Cu2+] = 1.60 M
Therefore, at a cell potential of 0.370 V, the concentration of Cu2+ is 1.60 M.
So, the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.370 V are 5.392 M and 1.60 M, respectively.