An electrolytic cell contains a solution of Ag2SO4 and platinum electrodes. A current is passed unti — Electrochemistry Chemistry Question

💡 Solution & Explanation

To find the mass of silver deposited at the cathode, we can use the concept of chemical equivalence as defined by Faraday's laws of electrolysis. \subsubsection*{Step 1 - Determine the reactions and $n$-factors} At the anode, oxygen gas is liberated during the electrolysis of the aqueous solution [1]. The oxidation half-reaction of water at the anode is: \begin{equation*} \ce{2H2O -> O2 + 4H+ + 4e-} \end{equation*} Since $4$ moles of electrons are transferred per mole of oxygen gas ($\ce{O2}$) liberated, the $n$-factor of $\ce{O2}$ is $4$. At the cathode, silver ions are reduced and deposited as metallic silver [1]: \begin{equation*} \ce{Ag+ + e- -> Ag} \end{equation*} Since $1$ mole of electrons is transferred per mole of silver ($\ce{Ag}$) deposited, the $n$-factor of $\ce{Ag}$ is $1$. \subsubsection*{Step 2 - Calculate the equivalent weights} The equivalent weight ($E$) of a substance is given by: \begin{equation*} E = \frac{\text{Molar Mass}}{n\text{-factor}} \end{equation*} For oxygen ($\ce{O2}$) with a molar mass of $32\text{ g/mol}$: \begin{equation*} E_{\ce{O2}} = \frac{32}{4} = 8\text{ g/eq} \end{equation*} For silver ($\ce{Ag}$) with a molar mass of $108\text{ g/mol}$ [1]: \begin{equation*} E_{\ce{Ag}} = \frac{108}{1} = 108\text{ g/eq} \end{equation*} \subsubsection*{Step 3 - Calculate the chemical equivalents of oxygen liberated} The number of equivalents of a substance is: \begin{equation*} \text{Equivalents} = \frac{\text{Mass}}{\text{Equivalent Weight}} \end{equation*} Substituting the given mass of oxygen ($1.6\text{ g}$) [1]: \begin{equation*} \text{Equivalents of }\ce{O2} = \frac{1.6}{8} = 0.2\text{ eq} \end{equation*} \subsubsection*{Step 4 - Calculate the mass of silver deposited using Faraday's Law} According to Faraday's second law of electrolysis, when the same quantity of electricity is passed through a cell, equal equivalents of products are liberated or deposited at the electrodes: \begin{equation*} \text{Equivalents of }\ce{Ag}\text{ deposited} = \text{Equivalents of }\ce{O2}\text{ liberated} \end{equation*} Thus: \begin{equation*} \text{Equivalents of }\ce{Ag} = 0.2\text{ eq} \end{equation*} Now, we calculate the mass of silver deposited: \begin{equation*} \text{Mass of }\ce{Ag} = \text{Equivalents of }\ce{Ag} \times E_{\ce{Ag}} \end{equation*} \begin{equation*} \text{Mass of }\ce{Ag} = 0.2 \times 108 = 21.6\text{ g} \end{equation*} This matches option (D) [1]. \begin{equation*} \boxed{21.6\text{ g}} \end{equation*}