Question

Two voltameters, one with a solution of silver salt and the other with a trivalent-metal salt, are connected in series and a current of 2 A is maintained for 1.50 hours. It is found that 1.00 g of the trivalent metal is deposited. (a) What is the atomic weight of the trivalent metal?
(b) How much silver is deposited during this period? Atomic weight of silver is 107.9 g mol−1.

Answer 1

(a) The atomic weight of the trivalent metal is 26 g/mole .

(b) 12.1 g of silver is deposited during this period.

Explanation:

Given:-

Mass of salt deposited, m = 1 g

Current, i = 2 A

Time,t = 1.5 hours = 5400 s

For the trivalent metal salt:-

Equivalent mass $=\frac{1}{3}$ atomic weight

The E.C.E of the salt,

$Z=\frac{\text { Equivalent mass }}{96500}=\frac{\text { Atomic weight }}{3 \times 96500}$

(a) Using the formula, m=Z i t, we get:-

          $1 \times 10^{-3}=\frac{\text { Atomic weight }}{3 \times 96500} \times 2 \times 5400$

$\text { Atomic weight }=\frac{3 \times 96500 \times 10^{-3}}{2 \times 5400}$

                         $=26.8 \times 10^{-3} \mathrm{kg} / \mathrm{mole}$

$\text { Atomic weight }=26.8 \mathrm{g} / \mathrm{mole}$

(b) Using the relation between equivalent mass and mass deposited on plates, we get:-

$\frac{E_{1}}{E_{2}}=\frac{m_{1}}{m_{2}}$

$\frac{26.8}{3 \times 107.9}=\frac{1}{m_{2}}$

$m_{2}=12.1 g$