Question

What is the value of the equilibrium constant for the reaction between Cr(s) and Cu^2+ (aq) at 25 degree C? Cr(s) + Cu^2+ (q)rightarrow Cr^3+ (aq) + Cu(s) If your calculator has overflow issues with this problem, you can use this identity to solve for K: if log(K) = x + y then K = 10^x times 10^y 3 times 10^109 -6 times 10^5 2 times 10^18 1 times 10^2 3 times 10^40

Answer 1

Answer:

3 × 10¹⁰⁹

Explanation:

2Cr (s) → 2Cr³⁺ + 6e⁻        

and,

Ec° for the above reaction = 0.74 V

also,

3Cu²⁺ (aq) + 6e⁻ →   3Cu(s)

and,

Ea° for the above reaction = 0.34 V

Now,

The cell potential, E° = 0.34 + 0.74 = 1.08

E° = $\frac{\textup{0.059}}{\textup{n}}\log k$

thus,

0.34 = $\frac{\textup{0.059}}{\textup{n}}\log k$

here, n is the number of electrons exchanged

n = 6

1.08 = $\frac{\textup{0.05921}}{\textup{6}}\log k$

or

㏒ k = 109.44

taking anti-log both sides

we get

k =  3 × 10¹⁰⁹

Hence, the answer is option (A) 3 × 10¹⁰⁹