If the cell emf of a Zn-Cu cell is 0.80 V
when the concentration of Zn2+ is 2.0M, what is the concentration of Cu2+?
1.5x10^-10M
Well, it seems like Zn2+ is a pretty outgoing ion, always showing up to the party with a concentration of 2.0M. Meanwhile, Cu2+ seems to be a bit of a wallflower. So, to figure out its concentration, let's use a little chemistry humor.
If Zn2+ is rockin' at 2.0M, then Cu2+ must be playin' it cool to maintain that cell emf of 0.80V. So, let's imagine Cu2+ is whispering in a low, sultry voice, "I can't reveal my concentration, it's confidential."
But fear not! We can use the equation for the cell emf to get to the bottom of this. The cell emf equation is:
Ecell = E°cell - (0.0592/n) * log(Q)
In this case, since we're dealing with a Zn-Cu cell, the cell emf is 0.80V. Now, E°cell represents the standard cell potential when all the species are at 1M concentration. To find E°cell, you can look it up in a table or calculate it using standard reduction potentials.
Now, we have Zn2+ at 2.0M and Cu2+ at an unknown concentration. The number 'n' represents the stoichiometric coefficient of electrons in the balanced cell reaction, which in this case is 2.
And Q, my friend, is the reaction quotient, which is the ratio of the concentration of product species (Cu2+) raised to the power of their stoichiometric coefficient, to the concentration of the reactant species (Zn2+) raised to their stoichiometric coefficient.
Now, we can rearrange the equation and solve for the concentration of Cu2+. But remember, we can't rush Cu2+. It's all about the slow jam.
So, while I can't provide an exact answer without more information, I've given you some tools to unlock the answer and hopefully put a smile on your face along the way. Keep calm and keep experimenting!
To determine the concentration of Cu2+ in the Zn-Cu cell, we can use the Nernst equation. The Nernst equation relates cell potential to concentrations of the ions involved in the cell reaction.
The general form of the Nernst equation is:
Ecell = E°cell - (0.0592/n) * log(Q)
where:
Ecell = cell potential
E°cell = standard cell potential
n = number of moles of electrons transferred
Q = reaction quotient
For the Zn-Cu cell, the cell reaction is:
Zn(s) + Cu2+(aq) -> Zn2+(aq) + Cu(s)
The standard cell potential for this reaction is 1.10 V. (Note: The standard cell potential is the potential when all concentrations are at 1.0M.)
Given:
Ecell = 0.80 V
E°cell = 1.10 V
[Zn2+] = 2.0 M
We want to find [Cu2+].
We can rearrange the Nernst equation to solve for Q:
Q = 10^[(E°cell - Ecell) / (0.0592/n)]
For this reaction, n = 2 since 2 moles of electrons are transferred.
Plugging in the values:
Q = 10^[(1.10 - 0.80) / (0.0592/2)]
= 10^[(0.30) / 0.0296]
= 10^(10.14)
To find [Cu2+], we need to know the value of Q. However, we don't have information about the reaction quotient Q. So, without more information, we cannot calculate the concentration of Cu2+.
To find the concentration of Cu2+ in the Zn-Cu cell, you can use the Nernst equation. The Nernst equation relates the cell potential (E) to the concentrations of the reactants and products in the cell.
The Nernst equation is given by:
E = E° - (RT/nF) * ln(Q)
Where:
- E is the cell potential
- E° is the standard cell potential (0.80V in this case)
- R is the ideal gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin
- n is the number of moles of electrons transferred in the balanced cell reaction
- F is Faraday's constant (96485 C/mol)
- Q is the reaction quotient
In this case, the balanced reaction in the Zn-Cu cell is:
Zn(s) + Cu2+(aq) -> Zn2+(aq) + Cu(s)
Since the concentration of Zn2+ is given (2.0 M), we need to find the concentration of Cu2+.
To use the Nernst equation, we need to find Q, the reaction quotient, which is the ratio of the concentrations of the reaction products to the reactants. In this case, Q is given by:
Q = [Zn2+]/[Cu2+]
We can rearrange the Nernst equation to solve for [Cu2+]:
[E - E° = (RT/nF) * ln(Q)]
[E - E° = (RT/nF) * ln([Zn2+]/[Cu2+])]
[E - E° = (RT/nF) * (ln[Zn2+] - ln[Cu2+])]
Now plug in the known values:
[E - 0.80V = (8.314 J/(mol·K)) * T/2 * (ln(2.0M) - ln[Cu2+])]
Using the given values, you can solve this equation for [Cu2+]. The constant temperature (T) is also needed for the calculation.
Zn + Cu^2+ ==> Cu + Zn^2+
Eo cell = Eo oxdn Zn Eo redn Cu = about 1.1 v but you need to look up the numbers and use those in your text.
Then Ecell = Eocell -(0.0592/2)log Q where
Q = (Cu)(Zn^2+)/(Zn)(Cu^2+).
Substitute Zn^2+ and Ecell and solve for (Cu^2+).