When 12,000 coulombs of electricity is passed through the electrolyte, 3.0 g of a metal of atomic ma — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Conceptual Understanding of Faraday's First Law of Electrolysis According to Faraday's first law of electrolysis, the mass of a substance ($W$) deposited or liberated at any electrode during electrolysis is directly proportional to the quantity of electricity ($Q$) passed through the electrolyte: $$W = Z \cdot Q$$ Where: * $W$ is the mass of the metal deposited (in grams). * $Q$ is the quantity of electricity passed (in coulombs). * $Z$ is the electrochemical equivalent of the substance, which is defined as: $$Z = \frac{E}{F}$$ Here, $E$ is the equivalent mass of the metal, and $F$ is Faraday's constant ($96500\text{ C/mol}$). The equivalent mass ($E$) of a metal is related to its atomic mass ($M$) and its electro-valency ($z$, which represents the charge of the metal cation) by the relation: $$E = \frac{M}{z}$$ Combining these equations, we get the complete formula for the deposited mass: $$W = \frac{M}{z \cdot F} \cdot Q$$ Step 2 - Substitution of Given Values We are given the following values from the problem: * Mass of metal deposited, $W = 3.0\text{ g}$ * Atomic mass of the metal, $M = 96.5\text{ g/mol}$ * Quantity of electricity, $Q = 12000\text{ C}$ * Faraday's constant, $F = 96500\text{ C/mol}$ Substitute these values into the unified Faraday's law equation: $$3.0\text{ g} = \frac{96.5\text{ g/mol}}{z \times 96500\text{ C/mol}} \times 12000\text{ C}$$ Step 3 - Calculate the Electro-valency ($z$) Simplify the pre-fraction terms: $$3.0 = \frac{96.5}{96500} \times \frac{12000}{z}$$ Since $\frac{96.5}{96500} = \frac{1}{1000}$: $$3.0 = \frac{1}{1000} \times \frac{12000}{z}$$ $$3.0 = \frac{12}{z}$$ Now, solve for the electro-valency ($z$): $$z = \frac{12}{3.0}$$ $$z = 4$$ Since the metal forms a cation ($\ce{M^{z+}}$) by losing electrons, its electro-valency must be $+4$. Step 4 - Evaluation of Options * **Option (A)** represents $+4$, which is the correct electro-valency of the metal cation. * **Option (B)** represents $+3$, which is incorrect. A valency of $+3$ would deposit $\frac{96.5}{3 \times 96500} \times 12000 = 4.0\text{ g}$ of metal. * **Option (C)** represents $+2$, which is incorrect. A valency of $+2$ would deposit $\frac{96.5}{2 \times 96500} \times 12000 = 6.0\text{ g}$ of metal. * **Option (D)** represents $-4$, which is incorrect because metal cations carry a positive charge due to the loss of electrons. A negative valency is physically impossible for a metal cation. $$\text{Correct Option: } \boxed{\text{A}}$$