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

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° =

thus,

0.34 =

here, n is the number of electrons exchanged

n = 6

1.08 =

or

㏒ k = 109.44

taking anti-log both sides

we get

k =  3 × 10¹⁰⁹

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

a electronegativity decreases beacuse the size of nucleus is bigger.