Electrochemical equivalent is more for — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand Faraday's First Law and Electrochemical Equivalent ($Z$) According to Faraday's First Law of Electrolysis, the mass ($w$) of a substance deposited or liberated at any electrode during electrolysis is directly proportional to the quantity of electricity ($Q$) passed through the electrolyte: $$w \propto Q$$ Since the electric charge is the product of current ($I$) and time ($t$): $$w = Z \cdot Q = Z \cdot I \cdot t$$ The proportionality constant $Z$ is called the **electrochemical equivalent** of the substance. It represents the mass of the substance deposited or liberated by passing exactly $1\text{ Coulomb}$ of electricity (i.e., a current of $1\text{ Ampere}$ for $1\text{ second}$). Step 2 - Relate Electrochemical Equivalent ($Z$) to Chemical Equivalent Weight ($E$) By definition, passing $1\text{ Faraday}$ of charge ($F \approx 96,500\text{ C}$) deposits or liberates exactly $1\text{ equivalent weight}$ ($E$) of a substance. Substituting $Q = F$ and $w = E$ into Faraday's equation: $$E = Z \cdot F \implies Z = \frac{E}{F}$$ Since Faraday's constant ($F$) is a universal electrochemical constant: $$Z \propto E$$ The electrochemical equivalent ($Z$) of any substance is directly proportional to its chemical equivalent weight ($E$). Therefore, the substance with the highest equivalent weight will have the largest electrochemical equivalent. Step 3 - Calculate the Equivalent Weights ($E$) of the Given Species The equivalent weight ($E$) of an element is calculated using the formula: $$E = \frac{\text{Atomic Mass}}{\text{Valency factor } (z)}$$ Let us calculate the equivalent weight for each of the given elements based on their common cathodic reduction half-reactions during electrolysis: 1. **Hydrogen ($\ce{H}$):** Reduction half-reaction: $\ce{H^+(aq) + e^- -> \frac{1}{2} H2(g)} \quad \implies z = 1$ $$E_{\ce{H}} = \frac{1\text{ g/mol}}{1} = 1\text{ g/eq}$$ 2. **Silver ($\ce{Ag}$):** Reduction half-reaction: $\ce{Ag^+(aq) + e^- -> Ag(s)} \quad \implies z = 1$ $$E_{\ce{Ag}} = \frac{108\text{ g/mol}}{1} = 108\text{ g/eq}$$ 3. **Copper ($\ce{Cu}$):** Reduction half-reaction: $\ce{Cu^2+(aq) + 2e^- -> Cu(s)} \quad \implies z = 2$ $$E_{\ce{Cu}} = \frac{63.5\text{ g/mol}}{2} = 31.75\text{ g/eq}$$ *(Note: Even in the case of cuprous ion reduction $\ce{Cu^+(aq) + e^- -> Cu(s)}$ where $z=1$, the equivalent weight is $63.5\text{ g/eq}$, which is still significantly smaller than that of silver).* 4. **Zinc ($\ce{Zn}$):** Reduction half-reaction: $\ce{Zn^2+(aq) + 2e^- -> Zn(s)} \quad \implies z = 2$ $$E_{\ce{Zn}} = \frac{65.4\text{ g/mol}}{2} = 32.7\text{ g/eq}$$ Step 4 - Compare Equivalent Weights and Determine the Correct Option Comparing the calculated equivalent weights of the given elements: $$E_{\ce{Ag}}\ (108\text{ g/eq}) > E_{\ce{Zn}}\ (32.7\text{ g/eq}) > E_{\ce{Cu}}\ (31.75\text{ g/eq}) > E_{\ce{H}}\ (1\text{ g/eq})$$ Since silver ($\ce{Ag}$) has the highest chemical equivalent weight, it has the largest electrochemical equivalent ($Z$). Step 5 - Explain Each Option * **Option (A) is incorrect:** Hydrogen has the lowest atomic mass and an equivalent weight of only $1\text{ g/eq}$, resulting in the smallest electrochemical equivalent among all the choices. * **Option (B) is correct:** Silver has the largest chemical equivalent weight ($108\text{ g/eq}$) and therefore has the highest electrochemical equivalent. * **Option (C) is incorrect:** Copper has a common equivalent weight of $31.75\text{ g/eq}$ (or $63.5\text{ g/eq}$ for cuprous systems), which is much lower than that of silver. * **Option (D) is incorrect:** Zinc has an equivalent weight of $32.7\text{ g/eq}$, which is lower than that of silver. $$\text{Correct Option: } \boxed{\text{B}}$$