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

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 = Zit, we get:-

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

\[ \Rightarrow\text{Atomic weight }= \frac{3 \times 96500 \times {10}^{- 3}}{2 \times 5400} =   26 . 8 \times  {10}^{- 3}\text{ kg/mole}\]

\[ \Rightarrow \text{Atomic weight }= 26 . 8\text{ g/mole}\]

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

\[\frac{E_1}{E_2} = \frac{m_1}{m_2}\]

\[ \Rightarrow \frac{26 . 8}{3 \times 107 . 9} = \frac{1}{m_2}\]

\[ \Rightarrow m_2 = 12 . 1 g\]