Question

Question:Three electrochemical cells were connected in series so that the same quantity of electrical current passes through all three cells. In the first cell, 1.15 g of chromium metal was deposited from chromium (III) nitrate solution. In the second cell, 3.15 g of osmium was deposited from a solution made of Osⁿ⁺ and nitrate ions. What is the name of the salt? In the third cell, the electrical charge passed through a solution containing X²⁺ ions caused deposition of 2.11 g of metallic X. Identify X.

Short answer

The salt is Osmium nitrate $\left(\mathrm{Os}{\left({\mathrm{NO}}_3\right)}_4\right)$. X is copper$\left(\mathrm{Cu}\right)$.

Step-by-step solution

Determine the salt

Chromium nitrate has +3 oxidation state. Therefore, charge of chromium (Cr) in the sample will be

$$\begin{gathered} =1.15\mathrm{g}\mathrm{Cr}\times \frac{1\mathrm{mol}\mathrm{Cr}}{52\mathrm{g}}\times \frac{3\mathrm{mol}{\mathrm{e}}^{-1}}{1\mathrm{mol}\mathrm{Cr}}\times \frac{96485\mathrm{C}}{{\mathrm{mole}}^{-1}} \\ =6.4\times {10}^3\mathrm{C} \end{gathered}$$

For Os cell also $6.4\times {10}^3$charge passes.

$\begin{array}{rcl}3.5\mathrm{g}\mathrm{Os}\times \frac{1\mathrm{mol}\mathrm{Os}}{190.2\mathrm{g}}& =& 0.0166\mathrm{mol}\\ 6.4\times {10}^3\mathrm{C}\times \frac{1\mathrm{mol}{\mathrm{e}}^{-1}}{96485\mathrm{C}}& =& 0.0663\mathrm{mol}{\mathrm{e}}^{-1}\\ \frac{\mathrm{mol}{\mathrm{e}}^{-1}}{\mathrm{mol}\mathrm{Os}}& =& \frac{0.0663}{0.0166}\approx 4\end{array}$

Therefore, the salt is comprised of ion${\mathbf{Os}}^{\mathbf{4}\mathbf{+}}$and . The compound will be${{\mathbf{NO}}_{\mathbf{3}}}^{\mathbf{-}}$ Osmium nitrate$\left(\mathrm{Os}{\left({\mathrm{NO}}_3\right)}_4\right)$.

Determine X

For the third cell, identify X by determining its molar mass. Two moles of electrons are transferred when X²⁺ is reduced to X.

$\mathrm{Molar}\mathrm{mass}=\frac{2.11\mathrm{g}\mathrm{X}}{6.40\times {10}^3\mathrm{C}\times {\displaystyle \frac{1\mathrm{mol}{\mathrm{e}}^{-1}}{96485\mathrm{C}}}\times {\displaystyle \frac{1\mathrm{mol}\mathrm{X}}{2\mathrm{mol}{\mathrm{e}}^{-}}}}=63.3\mathrm{g}/\mathrm{mol}$

This is copper (Cu).