The following cell was observed to have an emf of 2.25 volts at 25.0oC :
Zn(s) | Zn2+(aq, 0.143 M) || Cl2(g, 0.548 atm) | Cl-(aq, ?M) | Pt(s) .
Use tables of standard reduction potentials in Petrucci to calculate the concentration of chloride ion in the cathode compartment, [Cl-] / M . Your answer should be dimensionless (no units) and be given to 2
i got 62
To calculate the concentration of chloride ions in the cathode compartment, we need to use the Nernst equation, which relates the reduction potential, half-cell reaction, and concentrations of the species involved.
The Nernst equation for a half-cell reaction is as follows:
E = E° - (0.0592/n) * log(Q)
Where:
E = cell potential at non-standard conditions
E° = standard cell potential
n = number of electrons transferred in the half-cell reaction
Q = reaction quotient
In this case, we have the half-cell reactions:
Zn(s) → Zn2+(aq) + 2e- (reduction half-reaction)
Cl2(g) + 2e- → 2Cl-(aq) (oxidation half-reaction)
We can check the standard reduction potentials (E°) of these half-cell reactions from the tables in Petrucci. Assuming we're considering standard conditions, we find:
E°(Zn2+(aq) + 2e- → Zn(s)) = -0.763 V
E°(Cl2(g) + 2e- → 2Cl-(aq)) = +1.358 V
Now, let's substitute these values into the Nernst equation:
2.25 V = -0.763 V - (0.0592/2) * log(Q)
Simplifying:
2.25 V = -0.763 V - 0.0296 * log(Q)
2.25 V + 0.763 V = -0.0296 * log(Q)
3.013 V = -0.0296 * log(Q)
Now, let's solve for log(Q):
log(Q) = 3.013 V / -0.0296
log(Q) ≈ -101.96
Finally, we can calculate the value of Q and then use it to determine the concentration of chloride ions using the following equation:
Q = [Cl-]^2 / [Cl2(g)]
Since we want to find [Cl-], we need to rearrange the equation:
[Cl-]^2 = Q * [Cl2(g)]
Using the value of Q ≈ 10^(-101.96), and assuming [Cl2(g)] = 0.548 atm, we can calculate [Cl-]:
[Cl-]^2 ≈ 10^(-101.96) * 0.548
[Cl-]^2 ≈ 5.65 x 10^(-101)
Taking the square root:
[Cl-] ≈ √(5.65 x 10^(-101))
Finally, we can round the value of [Cl-] to 2 significant figures:
[Cl-] ≈ 7.5 x 10^(-51) M
Therefore, the concentration of chloride ions in the cathode compartment is approximately 7.5 x 10^(-51) M.