Question

Calculate E° for the following reaction at 298K:

2Al(s) + 3 Cu²⁺ (0.01M) → 2Al³⁺ (0.01M) + 3Cu(s)
Given E_cell = 1.98 , V.

A. 1.98 V
B. 0.99 V
C. 2.97 V
D. 2.48 V

Answer 1

The E° for the reaction 2Al(s) + 3 Cu²⁺ (0.01M) → 2Al³⁺ (0.01M) + 3Cu(s) at 298K is 1.98 V (Option A).

Step-by-step explanation:

In order to calculate E° for the given reaction, we can use the formula:

E° = E°cell - (0.0592/n) * logQ

Where E° is the standard cell potential, E°cell is the given value of 1.98 V, n is the number of electrons transferred in the reaction (in this case 6), and Q is the reaction quotient.

First, we need to calculate the reaction quotient, Q. The concentration of Al³⁺ is given as 0.01M and the concentration of Cu²⁺ is also given as 0.01M, so

Q = [Al³⁺]²/[Cu²⁺]³ = (0.01)²/(0.01)³ = 1.

Next, we substitute these values into the formula and calculate E°:

E° = 1.98 - (0.0592/6) * log(1)

= 1.98 - (0.00987) * 0 = 1.98 V.

Therefore, the correct answer is (A) 1.98 V.