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+?
See your post above.
To find the concentration of Cu2+, we can use the Nernst equation. The Nernst equation relates the cell emf to the concentration of the ions involved in the cell reaction. The Nernst equation is given by:
Ecell = E°cell - (RT/nF) * ln(Q)
where:
Ecell is the cell emf
E°cell is the standard cell emf
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 balanced redox equation
F is Faraday's constant (96485 C/mol)
ln is the natural logarithm
Q is the reaction quotient
In this case, the balanced redox equation for the Zn-Cu cell is:
Zn(s) + Cu2+(aq) -> Zn2+(aq) + Cu(s)
The number of moles of electrons transferred in this equation is 2, so n = 2.
The reaction quotient (Q) can be calculated as:
Q = [Zn2+]/[Cu2+]
We are given that the cell emf (Ecell) is 0.80 V and the concentration of Zn2+ is 2.0 M.
Let's assume the standard cell emf (E°cell) is 1.10 V for the Zn-Cu cell.
Now we can substitute the given values into the Nernst equation and solve for the concentration of Cu2+.
0.80 V = 1.10 V - (8.314 J/(mol·K) * T/(2 * 96485 C/mol)) * ln([Zn2+]/[Cu2+])
Simplifying the equation:
-0.30 V = (8.314 J/(mol·K) * T/(2 * 96485 C/mol)) * ln([Zn2+]/[Cu2+])
We can solve for the natural logarithm term:
ln([Zn2+]/[Cu2+]) = -0.30 V * (2 * 96485 C/mol) / (8.314 J/(mol·K) * T)
Now, let's substitute the temperature (T) value into the equation. Let's assume the temperature is 298 K.
ln([Zn2+]/[Cu2+]) = -0.30 V * (2 * 96485 C/mol) / (8.314 J/(mol·K) * 298 K)
Now we can solve for the concentration of Cu2+ by rearranging the equation:
[Cu2+] = [Zn2+] * e^((-0.30 V * (2 * 96485 C/mol) / (8.314 J/(mol·K) * 298 K))
Substituting the values into the equation:
[Cu2+] = 2.0 M * e^((-0.30 V * (2 * 96485 C/mol) / (8.314 J/(mol·K) * 298 K))
Now, let's calculate the concentration of Cu2+ using a calculator:
[Cu2+] = 2.0 M * e^((-0.30 V * (2 * 96485 C/mol) / (8.314 J/(mol·K) * 298 K)) ≈ 0.064 M
Therefore, the concentration of Cu2+ is approximately 0.064 M.
To determine the concentration of Cu2+ in the Zn-Cu cell, we can use the Nernst equation. The Nernst equation relates the emf of an electrochemical cell to the concentrations of the species involved. It is given by:
E = E° - (0.0592/n) * log(Q)
Where:
E is the cell emf
E° is the standard cell emf
n is the number of electrons transferred in the balanced equation
Q is the reaction quotient, which is the ratio of concentrations of products to reactants raised to their stoichiometric coefficients.
In this case, the Zn-Cu cell involves the following half-reactions:
Zn(s) → Zn2+(aq) + 2e-
Cu2+(aq) + 2e- → Cu(s)
From the given information, the cell emf (E) is 0.80 V. However, we need the standard cell emf (E°) to calculate the concentration of Cu2+. The standard cell emf depends on the reduction potentials of the half-reactions at standard conditions.
To find E°, we can look up the reduction potentials of Zn2+/Zn and Cu2+/Cu half-reactions in a standard electrode potential table. Let's assume that the reduction potential for the Zn2+/Zn half-reaction is -0.76 V and the Cu2+/Cu half-reaction is +0.34 V.
E° can be calculated using the equation:
E° = E(cathode) - E(anode)
In this case, E° = E(cathode) - E(anode) = (0.34 V) - (-0.76 V) = 1.10 V
Now, we can substitute the known values into the Nernst equation to determine the concentration of Cu2+.
0.80 V = 1.10 V - (0.0592/n) * log(Q)
Since we have a balanced equation, the value of n is 2.
0.80 V = 1.10 V - (0.0592/2) * log(Q)
Simplifying the equation gives:
0.30 V = 0.0296 * log(Q)
Now, rearrange the equation to isolate log(Q):
log(Q) = 0.30 V / 0.0296
Using a calculator, divide 0.30 by 0.0296 and find log:
log(Q) ≈ 10.14
Finally, take the antilog (10^x) of both sides to find Q:
Q ≈ 10^10.14
Q ≈ 1.97 x 10^10
Since Q is equal to the ratio of concentrations of Cu2+ to Zn2+, we know:
Q = [Cu2+]/[Zn2+]
Substituting the given concentration of Zn2+ (2.0 M) and the calculated value of Q into the equation, we can solve for [Cu2+]:
1.97 x 10^10 = [Cu2+]/2.0
Cross-multiply to solve for [Cu2+]:
[Cu2+] = 1.97 x 10^10 * 2.0
[Cu2+] ≈ 3.94 x 10^10 M
Therefore, the concentration of Cu2+ is approximately 3.94 x 10^10 M in the Zn-Cu cell with a cell emf of 0.80 V when the concentration of Zn2+ is 2.0 M.