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

(a) Using the formula, m = Zit, we get:
$$\begin{gathered} 1\times {10}^{-3}=\frac{\mathrm{Atomic}\mathrm{weight}}{3\times 96500}\times 2\times 5400 \\ \Rightarrow \mathrm{Atomic}\mathrm{weight}=\frac{3\times 96500\times {10}^{-3}}{2\times 5400}=26.8\times {10}^{-3}\mathrm{kg}/\mathrm{mole} \\ \Rightarrow \mathrm{Atomic}\mathrm{weight}=26.8\mathrm{g}/\mathrm{mole} \end{gathered}$$

(b) Using the relation between equivalent mass and mass deposited on plates, we get:
$$\begin{gathered} \frac{E_{\mathit{1}}}{E_{\mathit{2}}}=\frac{m_1}{m_2} \\ \Rightarrow \frac{26.8}{3\times 107.9}=\frac{1}{m_2} \\ \Rightarrow m_2=12.1\mathrm{g} \end{gathered}$$