For the following cell with metal X electrodes, X |Xn (0.5M) || Xn (0.05)| X Ecell = -0.028 V at 290 K, if there is no liquid junction potential, valency of X is?

Question

For the following cell with metal X electrodes, X |Xn (0.5M) || Xn  (0.05)| X Ecell = -0.028 V at 290 K, if there is no liquid junction potential, valency of X is?

Answer

Valency of Metal X

Given Information:

Electrodes: X |Xn (0.5M) || Xn (0.05)| X

Ecell = -0.028 V at 290 K

No liquid junction potential

Explanation:

The given cell notation represents a redox reaction where metal X is oxidized at the anode and Xn is reduced at the cathode. The concentration of Xn at the cathode is 10 times lower than that of X at the anode. The negative value of Ecell indicates that the reaction is not spontaneous and requires an external voltage source to run.

Using the Nernst equation, we can calculate the standard reduction potential of Xn:

E°cell = E°cathode - E°anode = E°Xn - E°X

E°Xn = E°cell + E°X = 0 + (-0.028) = -0.028 V

We can then use the standard reduction potential of Xn to calculate the equilibrium constant (K) of the reaction:

E°Xn = - (RT/nF) lnK

lnK = - nF E°Xn / RT

K = e^(-nF E°Xn / RT)

At equilibrium, the concentration of products divided by the concentration of reactants equals K:

K = [Xn] / [X]^(n)

Substituting the values, we get:

e^(-nF E°Xn / RT) = [Xn] / [X]^(n)

Since we know that the concentration of Xn is 10 times lower than that of X:

e^(-nF E°Xn / RT) = 0.1 / [X]^(n)

We can simplify this equation by taking the natural logarithm:

-nF E°Xn / RT = ln(0.1) - n ln([X])

ln([X]) = (nF E°Xn / RT) + ln(0.1) / n

If we plot ln([X]) vs. E°Xn, the slope of the line will be (nF/RT) and the intercept will be ln(0.1)/n. We can then use the slope to determine the valency of X.

Valency Calculation:

Assuming that F = 96500 C/mol and R = 8.314 J/mol·K:

slope = (nF/RT) = (n x 96500 / (8.314 x 290)) = 38.9n

From the graph of ln([X]) vs. E°Xn, the slope is found to be -77.8.

Substituting this value in the equation, we get:

-77.8 = 38.

Answer

2

Similar Class 12 Doubts

The emf of a cell containing sodium/ copper electrodes is 3.05 V, if the electrode potential copper electrode is +0.34 V, the electrode potential of sodium is

One or More than One Options Correct TypeThis section contains 4multiple type questions. Each question has 4 choices (

Which of the following is true in case of an unbiased p-n junction diode? (

Top Courses for Class 12

Suggested Free Tests

Related Content

Schools

Competitive Examinations

Quick Access Links

Technical Exams