The same current was passed successively through solution of zinc-ammonium sulphate and nickel-ammon — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - State Faraday's Second Law of Electrolysis When the same electric current is passed successively through different electrolytic cells connected in series for the same duration of time, the total quantity of electric charge ($Q = I \cdot t$) passing through them is identical. According to **Faraday's Second Law of Electrolysis**, the masses ($W$) of different substances deposited or liberated at the respective electrodes by the passage of the same quantity of electricity are directly proportional to their chemical equivalent weights ($E$): $$W \propto E$$ For a system containing zinc ($\ce{Zn}$) and nickel ($\ce{Ni}$) cells connected in series, we can express this relationship as: $$\frac{W_{\ce{Zn}}}{W_{\ce{Ni}}} = \frac{E_{\ce{Zn}}}{E_{\ce{Ni}}}$$ Rearranging this formula to solve for the chemical equivalent weight of nickel ($E_{\ce{Ni}}$): $$E_{\ce{Ni}} = E_{\ce{Zn}} \times \frac{W_{\ce{Ni}}}{W_{\ce{Zn}}}$$ Step 2 - Identify the Given Experimental Parameters From the question, we have the following values: * Mass of zinc deposited ($W_{\ce{Zn}}$) = $22.89\text{ g}$ * Mass of nickel deposited ($W_{\ce{Ni}}$) = $20.55\text{ g}$ * Chemical equivalent weight of zinc ($E_{\ce{Zn}}$) = $32.7\text{ g eq}^{-1}$ Step 3 - Calculate the Chemical Equivalent Weight of Nickel ($E_{\ce{Ni}}$) First, let us find the number of equivalents of zinc deposited by dividing its mass by its equivalent weight: $$\text{Equivalents of Zn} = \frac{W_{\ce{Zn}}}{E_{\ce{Zn}}} = \frac{22.89\text{ g}}{32.7\text{ g eq}^{-1}}$$ $$\text{Equivalents of Zn} = 0.70\text{ eq}$$ According to Faraday's second law, the number of equivalents of nickel deposited must be exactly equal to the equivalents of zinc deposited because they received the same amount of charge: $$\text{Equivalents of Ni} = \frac{W_{\ce{Ni}}}{E_{\ce{Ni}}} = 0.70\text{ eq}$$ Now, substitute the values to calculate $E_{\ce{Ni}}$: $$0.70\text{ eq} = \frac{20.55\text{ g}}{E_{\ce{Ni}}}$$ $$E_{\ce{Ni}} = \frac{20.55\text{ g}}{0.70\text{ eq}}$$ $$E_{\ce{Ni}} = \mathbf{29.3571\text{ g eq}^{-1}} \approx \mathbf{29.36\text{ g eq}^{-1}}$$ Thus, the chemical equivalent weight of nickel is approximately $29.36\text{ g eq}^{-1}$. Step 4 - Understand the Underlying Chemistry In their respective alkaline ammonia solutions, zinc and nickel form stable ammine coordination complexes: * Zinc exists as the tetraamminezinc(II) complex: $\ce{[Zn(NH3)4]^2+}$ * Nickel exists as the hexaamminenickel(II) complex: $\ce{[Ni(NH3)6]^2+}$ During electrolysis, both divalent metal cations undergo a $2\text{-electron}$ reduction at their respective cathodes to deposit as solid metals: $$\ce{[Zn(NH3)4]^2+(aq) + 2e^- -> Zn(s) + 4NH3(aq)}$$ $$\ce{[Ni(NH3)6]^2+(aq) + 2e^- -> Ni(s) + 6NH3(aq)}$$ Since the valency factor ($n$-factor) is $2$ for both metals, their chemical equivalent weights are half of their respective atomic masses ($E = \frac{\text{Atomic Mass}}{2}$): * For Zinc: $\text{Atomic Mass} \approx 65.4\text{ g/mol} \implies E_{\ce{Zn}} = \frac{65.4}{2} = 32.7\text{ g eq}^{-1}$ * For Nickel: $\text{Atomic Mass} \approx 58.7\text{ g/mol} \implies E_{\ce{Ni}} = \frac{58.7}{2} = 29.35\text{ g eq}^{-1}$ This chemical analysis beautifully confirms our mathematically calculated value of $29.36$. Step 5 - Evaluate and Explain the Options * **Option (A) is incorrect:** The value $58.71$ is approximately the atomic mass of nickel ($\ce{Ni}$). This would only represent the equivalent weight if nickel were reduced from a $+1$ oxidation state ($n = 1$), which is not the case for nickel-ammonium sulphate ($+2$ state). * **Option (B) is correct:** As mathematically shown using Faraday's second law, the chemical equivalent weight of nickel is $29.36$. * **Option (C) is incorrect:** The value $14.39$ is approximately half of the correct equivalent weight, which would only occur if nickel had an incorrect valency factor of $4$. * **Option (D) is incorrect:** The value $36.42$ is mathematically incorrect and does not satisfy the proportional ratio established by Faraday's second law. $$\text{Correct Option: } \boxed{\text{B}}$$