Voltaic cell. The initial concentrations of Ni2+ and Zn2+ are 1.50 M and 0.100 M. Initial cell potential=.56 V. What are the concentrations of Ni2+ and Zn2+ when the cell potential falls to 0.45 V?
To determine the concentrations of Ni2+ and Zn2+ when the cell potential falls to 0.45 V, we can use the Nernst equation:
Ecell = E°cell - (0.0592/n) * log(Q)
Where:
- Ecell is the cell potential at a given time
- E°cell is the standard cell potential
- n is the number of electrons transferred in the cell reaction
- Q is the reaction quotient, which is equal to [Ni2+]/[Zn2+]
Let's assume the cell reaction is:
Ni(s) + Zn2+(aq) -> Ni2+(aq) + Zn(s)
The balanced equation shows that 2 moles of electrons are transferred for each mole of Zn2+. Thus, n = 2.
Now, let's calculate the reaction quotient Q at the initial conditions:
Q = [Ni2+]/[Zn2+] = (1.50 M)/(0.100 M) = 15
By rearranging the Nernst equation, we have:
log(Q) = (n/0.0592) * (E°cell - Ecell)
Plugging in the given values:
log(15) = (2/0.0592) * (0.56 V - 0.45 V)
Now, let's solve for Ecell:
0.034 = (2/0.0592) * (0.11 V)
0.034 * 0.0592 = 2 * 0.11 * Ecell
0.00200608 = 0.22 * Ecell
Ecell = 0.00200608 / 0.22
Ecell ≈ 0.00912 V
Now, we can use this new Ecell value to calculate Q at the new conditions. Let's assume the new Q value is Q':
Q' = [Ni2+]/[Zn2+]
Rearranging the Nernst equation again, we have:
log(Q') = (n/0.0592) * (E°cell - Ecell)
Plugging in the known values:
log(Q') = (2/0.0592) * (0.56 V - 0.00912 V)
Now we can solve for Q':
log(Q') = (2/0.0592) * (0.55088 V)
log(Q') = 1.859 + C
Here, C is the constant obtained from the calculation above.
Next, we can use logarithmic properties to solve for Q':
Q'= 10^(1.859 + C)
Q' = 10^(1.859) * 10^(C)
Finally, we can calculate the new concentrations of Ni2+ and Zn2+ using the new Q' value. Since Q' = [Ni2+]/[Zn2+], we can find the concentration of Ni2+ as follows:
[Ni2+] = Q' * [Zn2+]
Substituting the known values:
[Ni2+] = (Q') * (0.100 M)
And similarly, we can find the concentration of Zn2+:
[Zn2+] = (1/Q') * (1.50 M)
By substituting Q' from the previous calculation, we can obtain the values of [Ni2+] and [Zn2+].
To determine the concentrations of Ni2+ and Zn2+ when the cell potential falls to 0.45 V, we can use the Nernst equation. The Nernst equation relates the cell potential to the concentrations of the reactants and products involved in the cell reaction.
The Nernst equation is given as:
E = E° - (RT / nF) * ln(Q)
Where:
E is the cell potential
E° 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 electrons transferred in the balanced cell reaction
F is the Faraday constant (96485 C/mol)
Q is the reaction quotient, which is the ratio of product concentrations to reactant concentrations (raised to the power of their stoichiometric coefficients) at any point during the reaction.
In this case, the balanced cell reaction is:
Ni2+(aq) + Zn(s) → Ni(s) + Zn2+(aq)
The number of electrons transferred (n) in this reaction will be 2, as each Ni2+ ion gains two electrons and each Zn atom loses two electrons.
Given that the initial cell potential is 0.56 V and the initial concentrations of Ni2+ and Zn2+ are 1.50 M and 0.100 M, respectively, we can determine the values of E° and Q.
First, we need to calculate Q using the concentrations of the reactants and products based on the reaction quotient:
Q = [Ni2+]/[Zn2+]
We are given the initial concentrations of Ni2+ and Zn2+ as 1.50 M and 0.100 M, respectively. So, initially:
Q = (1.50 M)/(0.100 M)
= 15
Now, we can plug in the given values into the Nernst equation and solve for the cell potential (E) when the cell potential falls to 0.45 V:
0.45 V = E° - [(8.314 J/(mol·K)) * (T in K)/(2 * 96485 C/mol) * ln(15)]
To solve for the concentrations of Ni2+ and Zn2+, we rearrange the Nernst equation to:
ln(Q) = (E° - E) * (nF / (RT))
Then, we substitute the values of E°, E, n, F, R, and T (make sure to use the same temperature unit as used in the gas constant) into the equation.
Finally, we can calculate the concentrations of Ni2+ and Zn2+ by using the reaction quotient equation:
Q = [Ni2+]/[Zn2+]
By substituting the values of Q (calculated using the Nernst equation) and one of the concentrations, we can solve for the other concentration.