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

vanderwaal's forces of attraction are the electrostatic intermolecular forces that exist in a molecule due to the temporary or permanent dipoles that are developed in a molecule. in , the major intermolecular forces are dipole-dipole interactions as the molecule is polar (bent structure). so the vanderwaal's forces of interaction in between molecules are the permanent dipole-dipole interactions that develop due to the dipole that exists in between the polar h - s bond. the positively charged hydrogen atom is attracted towards the negative end of the dipole that is sulfur of the adjacent molecule. therefore, these forces of attraction are relatively stronger than that of london dispersion forces that exist in non polar molecules.