Error Analysis: The error analysis discussion refers to Part B Does an error of

Question

Error Analysis: The error analysis discussion refers to Part B Does an error of one or two hundredths of a volt in your measured cell potential cause a discrepancy between your value for K_d and the value found in literature for K_d? Will this affect K_d a little or a lot? Explain, (b) If you accidentally added 21 drops of O.IOM copper (II) sulfate solution instead of 20 drops and you did not recognize this error, what effect would it have on your voltaic cell potential? Will this affect E_cell a little or a lot? Explain.

Explanation / Answer

a) We know that

G0 = -nFEcell

G0 = -RTlnK

Ecell is the measured cell potential and K is the equilibrium constant. Therefore,

-nFEcell = -RTlnK

===> Ecell = (RT/nF)ln K

We are given, Ecell = 1/100 V so that

E’cell = (RT/nF)ln Kd’ (Kd’ is the theoretical value)

E”cell = (RT/nF)ln Kd” (Kd” is the actual value)

Ecell = E’cell – E”cell = (RT/nF)[ln Kd’ – ln Kd”]

===> (1/100 V) = (RT/nF) ln (Kd’/Kd”) ….(1)

Since we are dealing with the reaction,

Cu2+ (aq) + 2 e- -----> Cu (s),

n = 2 moles and F = 96485 C = 96485 J/V. Plug in values in 1 (assume room temperature = 298 K) and write,

0.01 V = [(8.314 J/mol.K)*(298 K)/(2 mol).(96485 J/V)]ln (Kd’/Kd”)

===> 0.01 V = (0.013 V)ln (Kd’/Kd”)

===> 0.769 = ln (Kd’/Kd”)

===> Kd’/Kd” = 2.15 2.0

===> Kd’ = 2*Kd”

There will be a huge error between the theoretical and actually obtained Kd values if there is an error of 1/100 V between the actual and the recorded cell potential. When the error is 1/200 V, the actually obtained Kd is almost 1.5 times less than the theoretical value.

Please note that I have assumed measured Kd to be less than theoretical Kd, but the same results hold if the measured Kd is greater than the theoretical value.

Related Questions

Navigate

• Browse (All)
• Browse E
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.