Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Zn2 ] = 0.842 M and [Ni2 ] = 0.0100 M. Standard reduction potentials can be found here.
reaction:
Zn(s)+Ni^2+(aq)--->Zn^2+(aq)+Ni(s)
standard reduction Zn: -.76
standard reduction Ni: 0.26
Eo(cell)= 0.26-(-0.76)=1.02
E(cell)= Eo(cell) -RT/nF (ln(Zn/Ni))
E(cell)= 1.02 - ((8.314*298)/(2)(9.6485))(ln(.842/.01))
E(cell) =0.963
But the answer is wrong and I don't know why. Can someone explain it to me and where I did it wrong.
I worked it using the numbers you provided and came up with the same answer you did of 0.963 so there is nothing wrong with your math. The problem is that the reduction potential of Ni is not 0.26. It is -0.26.
Well, it seems like you've made a slight calculation error in your equation. Let me help you out with it.
First of all, you correctly calculated the standard cell potential (Eo(cell)) as 1.02 V by subtracting the standard reduction potentials of Ni and Zn.
Next, to calculate the actual cell potential (E(cell)), you need to consider the effect of concentration on the cell potential. The equation you used is E(cell) = Eo(cell) - (RT/nF) * ln(Q), where Q is the reaction quotient.
In this case, Q can be calculated using the concentrations of the species involved in the reaction: Q = ([Zn^2+]/[Ni^2+]).
However, you made an error in the right-hand side of the equation. Let's correct it step by step:
E(cell) = Eo(cell) - (RT/nF) * ln(Q)
E(cell) = 1.02 V - (8.314 J/(K*mol) * 298 K / (2 mol e⁻) * (9.6485 C) * ln(0.842/0.01)
E(cell) = 1.02 V - (57.10 / 2) * ln(84.2)
Now, let's do the math:
E(cell) = 1.02 V - 28.55 * ln(84.2)
E(cell) ≈ 1.02 V - 36.06
E(cell) ≈ -35.04 V
So, the correct cell potential for the given reaction at 25.00 °C is approximately -35.04 V, which is indeed quite different from your initial calculation. I hope this clears up the confusion!
It seems like you made a mistake in your calculations. Let's go through the correct calculation step-by-step:
1. Write the half-reactions for the oxidation and reduction processes:
Zn(s) → Zn2+(aq) + 2e- (reduction)
Ni2+(aq) + 2e- → Ni(s) (oxidation)
2. Find the standard cell potential (Eo(cell)) by subtracting the standard reduction potential of the oxidation half-reaction from the reduction half-reaction:
Eo(cell) = Eo(reduction) - Eo(oxidation)
Eo(cell) = 0.26 V - (-0.76 V)
Eo(cell) = 1.02 V
3. Calculate the actual cell potential at 25.00 °C using the Nernst equation:
E(cell) = Eo(cell) - (RT / nF) * ln(Q)
Where:
- R = gas constant (8.314 J/(mol·K))
- T = temperature in Kelvin (25.00 °C + 273.15 = 298.15 K)
- n = number of electrons transferred (2 in this case)
- F = Faraday's constant (96,485 C/mol)
- Q = reaction quotient = [Zn2+]/[Ni2+]
4. Calculate the reaction quotient (Q):
Q = [Zn2+]/[Ni2+]
Q = 0.842 M / 0.01 M
Q = 84.2
5. Substitute the values into the Nernst equation:
E(cell) = 1.02 V - [(8.314 J/(mol·K)) * 298.15 K / (2 * 96,485 C/mol)] * ln(84.2)
E(cell) = 1.02 V - (2.073 * 10^-3 V) * ln(84.2)
E(cell) ≈ 0.83 V
Therefore, the corrected cell potential for the given reaction at 25.00 °C would be approximately 0.83 V, not 0.963 V.
To calculate the cell potential for the given reaction, we can use the Nernst equation:
E(cell) = Eo(cell) - (RT/nF)ln(Q)
Where:
E(cell) is the cell potential
Eo(cell) is the standard cell potential
R is the gas constant (8.314 J/(mol K))
T is the temperature in Kelvin (25.00 + 273.15 = 298 K)
n is the number of moles of electrons transferred in the balanced equation (2 in this case, as both Zn and Ni undergo transfer)
F is the Faraday constant (96500 C/mol)
ln is the natural logarithm
Q is the reaction quotient, which can be calculated using the concentrations of the species involved.
Let's calculate step by step:
1. Calculate Q using the given concentrations:
Q = [Zn^2+]/[Ni^2+]
= 0.842 M / 0.0100 M
= 84.2
2. Plug in the values into the Nernst equation:
E(cell) = 1.02V - ((8.314 J/(mol K))(298 K) / (2)(96500 C/mol)) ln(84.2)
3. Calculate the natural logarithm:
ln(84.2) ≈ 4.4308
4. Plug the value of ln(84.2) into the equation:
E(cell) ≈ 1.02V - ((8.314 J/(mol K))(298 K) / (2)(96500 C/mol)) (4.4308)
5. Perform the calculations in the parentheses:
E(cell) ≈ 1.02V - (0.0096485) (4.4308)
6. Calculate the value inside the parentheses:
0.0096485 x 4.4308 ≈ 0.042757
7. Plug the calculated value back into the equation:
E(cell) ≈ 1.02V - (0.042757)
8. Perform the final calculation:
E(cell) ≈ 1.02V - 0.042757
E(cell) ≈ 0.977243 V
Therefore, the correct cell potential is approximately 0.977 V. It seems that the mistake in your calculation might be related to the calculation of ln(84.2) or the overall arithmetic in step 6, which resulted in a different answer.