I'm having trouble with the following question, any help would be greatly appreciated.
A voltaic cell consists of a Zn/Zn^2+ half-cell and a Ni/Ni^2+ half-cell at 25 C . The initial concentrations of Ni^2+ and Zn^2+ are 1.30 M and 0.100 M , respectively.
What is the cell potential when the concentration of Ni^2+ has fallen to 0.500 M?
What is the concentrations of Ni^2+ when the cell potential falls to 0.45 M?
What is the concentration of Zn^2+ when the cell potential falls to 0.45 M?
I haven't tried to work on this problem at all because I don't understand how Ni2+ ion can start at 0.100M, use some of it and END UP FALLING to 0.500M. It appears to me that Ni2+ can become smaller but not larger.
The second part of this question does not make sense cell potential has to be in V not M...M is for concentration
I got the correct answer by making a simple Initial/Final chart. the initial [Ni^2+] = 1.30, the final [Ni^2+] = 0.500. The change is 0.80 M. This means you would add 0.80 to the initial [Zn^2+] (which is .100 M) and you would get a final [Zn^2+] of 0.90 M.
The next step is to find the new Q which is [Zn^2+]/[Ni^2+] = (0.90)/(0.500). Plug this in the Nernst equation and you will find the cell potential.
To find the cell potential when the concentration of Ni^2+ has fallen to 0.500 M, you can use the Nernst equation. The Nernst equation relates the cell potential (Ecell) to the standard cell potential (E°cell), the gas constant (R), the temperature in Kelvin (T), the Faraday constant (F), and the concentrations of the species involved in the reaction.
The Nernst equation is given by:
Ecell = E°cell - (RT/nF) * ln(Q)
Where:
- Ecell is the cell potential
- E°cell is the standard cell potential
- R is the gas constant (8.314 J/mol·K)
- T is the temperature in Kelvin
- n is the number of moles of electrons transferred in the cell reaction
- F is the Faraday constant (96,485 C/mol)
- Q is the reaction quotient, which is calculated using the concentrations of the species involved in the reaction
In this case, the cell reaction is:
Zn(s) + Ni^2+(aq) -> Zn^2+(aq) + Ni(s)
The standard cell potential (E°cell) for this reaction is provided elsewhere or can be looked up in a table.
To solve the problem, you need to calculate the reaction quotient (Q) using the concentrations of Ni^2+ and Zn^2+ and then substitute the values into the Nernst equation to find Ecell.
For the second part of the question, to find the concentration of Ni^2+ when the cell potential falls to 0.45 V, you will need to rearrange the Nernst equation and solve for the concentration. The steps to solve the equation are as follows:
1. Rearrange the Nernst equation to solve for the concentration of Ni^2+ :
Ecell = E°cell - (RT/nF) * ln(Q)
Ecell - E°cell = -(RT/nF) * ln(Q)
ln(Q) = -((Ecell - E°cell) * nF) / (RT)
Q = e^-((Ecell - E°cell) * nF) / (RT)
2. Substitute the given values into the formula: Ecell = 0.45 V, E°cell (from table or given), R = 8.314 J/mol·K, T = 25+273 K, n = number of electrons transferred in the cell reaction, F = 96,485 C/mol.
3. Calculate Q using the formula from step 1.
4. Set Q equal to the reaction quotient of the cell reaction:
Zn(s) + Ni^2+(aq) -> Zn^2+(aq) + Ni(s)
The reaction quotient (Q) in this case is equal to the concentration of Zn^2+ divided by the concentration of Ni^2+. Since the concentration of Ni^2+ is unknown, you can assign a variable (x) to represent it.
So, Q = [Zn^2+]/[Ni^2+] = [0.100 M]/[x]
5. Solve the equation Q = e^-((Ecell - E°cell) * nF) / (RT) for x.
For the last part of the question, to find the concentration of Zn^2+ when the cell potential falls to 0.45 V, you can follow similar steps as in the previous part, but this time the reaction quotient (Q) will be [Ni^2+]/[Zn^2+]. Again, you can assign a variable (y) to represent the concentration of Zn^2+.
So, Q = [Ni^2+]/[Zn^2+] = [0.500 M]/[y]
Substitute the given values into the Nernst equation of the cell reaction and solve for y.