Element A (atomic mass = 112) and element B (atomic mass = 27) form chlorides. Solutions of these ch — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand Faraday's Second Law of Electrolysis According to **Faraday's Second Law of Electrolysis**, when the same quantity of electricity is passed through different electrolytes, the masses ($W$) of the substances deposited or liberated at the respective electrodes are directly proportional to their chemical equivalent weights ($E$): $$\frac{W_A}{W_B} = \frac{E_A}{E_B}$$ Where: * $W_A$ and $W_B$ are the masses of elements $A$ and $B$ deposited. * $E_A$ and $E_B$ are the chemical equivalent weights of elements $A$ and $B$. The chemical equivalent weight ($E$) of an element is defined as its atomic mass ($M$) divided by its valency or oxidation state ($z$): $$E = \frac{M}{z}$$ At the cathodes, the reduction half-reactions of the metal ions from their respective chloride solutions can be represented as: $$\ce{A^{z_A+} + z_A e^- -> A(s)}$$ $$\ce{B^{z_B+} + z_B e^- -> B(s)}$$ Step 2 - Calculate the Chemical Equivalent Weight of Element B We are given the following values for element $B$: * Atomic mass of $B$ ($M_B$) = $27\text{ g/mol}$ * Valency of $B$ ($z_B$) = $3$ * Mass of $B$ deposited ($W_B$) = $0.9\text{ g}$ Using the equivalent weight formula: $$E_B = \frac{M_B}{z_B}$$ Substitute the given values into the formula: $$E_B = \frac{27\text{ g/mol}}{3} = 9\text{ g/eq}$$ Step 3 - Set up the Ratio to Find the Valency of Element A We are given the following values for element $A$: * Atomic mass of $A$ ($M_A$) = $112\text{ g/mol}$ * Mass of $A$ deposited ($W_A$) = $5.6\text{ g}$ * Valency of $A$ ($z_A$) = ? The equivalent weight of element $A$ is expressed as: $$E_A = \frac{M_A}{z_A} = \frac{112}{z_A}\text{ g/eq}$$ Applying Faraday's Second Law of Electrolysis: $$\frac{W_A}{W_B} = \frac{E_A}{E_B}$$ Substitute the values and units into the formula: $$\frac{5.6\text{ g}}{0.9\text{ g}} = \frac{\frac{112}{z_A}\text{ g/eq}}{9\text{ g/eq}}$$ Simplify the expression by dividing both sides of the equation: $$\frac{5.6}{0.9} = \frac{112}{9 \times z_A}$$ Multiply both sides of the equation by $9$ to isolate the variable term: $$9 \times \frac{5.6}{0.9} = \frac{112}{z_A}$$ $$10 \times 5.6 = \frac{112}{z_A}$$ $$56 = \frac{112}{z_A}$$ Solving for the valency $z_A$: $$z_A = \frac{112}{56}$$ $$z_A = \boxed{2}$$ Thus, the valency of element $A$ is **2**. Step 4 - Evaluate and Explain the Options * **Option (A) is incorrect:** If the valency of $A$ were $1$, its equivalent weight would be $E_A = 112\text{ g/eq}$. The deposited mass of $A$ would then be $W_A = 0.9\text{ g} \times \frac{112}{9} = 11.2\text{ g}$, which is twice the actual value. * **Option (B) is correct:** As mathematically calculated, a valency of $2$ perfectly satisfies Faraday's second law, yielding exactly $5.6\text{ g}$ of deposited element $A$. * **Option (C) is incorrect:** If the valency of $A$ were $3$, its equivalent weight would be $E_A = \frac{112}{3} \approx 37.33\text{ g/eq}$, resulting in a deposited mass of $W_A = 0.9\text{ g} \times \frac{37.33}{9} \approx 3.73\text{ g}$, which does not match the experimental data. * **Option (D) is incorrect:** If the valency of $A$ were $4$, its equivalent weight would be $E_A = \frac{112}{4} = 28\text{ g/eq}$, resulting in a deposited mass of $W_A = 0.9\text{ g} \times \frac{28}{9} \approx 2.8\text{ g}$, which is half the actual value. $$\text{Correct Option: } \boxed{\text{B}}$$