The mass of a gold atom is 3.27 × 10−25 kg . If 3.6 kg of gold is deposited on the negative electrode of an electrolytic cell in a period of 14.88 h , what is the current in the cell in this period? Assume that each gold ion carries one fundamental unit (1.602 × 10−19 C) of positive charge. Answer in units of A.

Current is 32.8 A

Explanation:

1 mass of Au atom = 3.27×10⁻²⁵kg

Mass of gold = 3.6 kg

We determine the amount of atoms, in that mass of gold therefore we need a rule of three:

3.27×10⁻²⁵kg is the mass for 1 atom

3.6 kg will be the mass for (3.6 . 1)/3.27×10⁻²⁵ = 1.10×10²⁵ atoms

Now we have to find out the total charge of the 3.6 kg of gold so we make another rule of three:

1 atom of Au has a charge of 1.602 × 10⁻¹⁹ C

1.10×10²⁵ atoms of Au will have a charge of (1.10×10²⁵.1.602 × 10⁻¹⁹) / 1 = 1.76×10⁶ C

Formula for the current is: q = i . t

where q is the C, i the value of current and t, time (s)- We make time conversion from h to s

14.88 h . 3600s / 1h = 53568 s  → We replace values:

1.76×10⁶ C  = i . 53568s

1.76×10⁶ C  / 53568s = i → 32.8 A