If a current of 1.0 A is drawn from the Daniel cell for 96.5 min, the cathode will gain in weight by — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the Electrodes and the Cathodic Reaction of a Daniell Cell A Daniell cell is a classic electrochemical system consisting of a zinc anode and a copper cathode: $$\ce{Zn(s) | Zn^2+(aq, 1 M) || Cu^2+(aq, 1 M) | Cu(s)}$$ * **At the Anode (Negative Electrode - Oxidation):** Metallic zinc dissolves into the solution as zinc ions: $$\ce{Zn(s) -> Zn^2+(aq) + 2e^-}$$ * **At the Cathode (Positive Electrode - Reduction):** Copper ions in the aqueous electrolyte migrate to the cathode, gain electrons, and deposit as metallic copper: $$\ce{Cu^2+(aq) + 2e^- -> Cu(s)}$$ Since reduction occurs at the cathode, the deposition of solid copper ($\ce{Cu}$) causes the cathode to gain weight. The number of moles of electrons transferred per mole of copper deposited is: $$n = 2$$ Step 2 - Determine the Chemical Equivalent Weight ($E$) of Copper The chemical equivalent weight ($E$) of an element is its atomic mass divided by its valency factor ($n$-factor): $$E = \frac{\text{Atomic Mass}}{n}$$ Given that the atomic mass of copper ($\ce{Cu}$) is $63.5\text{ g/mol}$: $$E_{\ce{Cu}} = \frac{63.5\text{ g/mol}}{2} = 31.75\text{ g/eq}$$ Step 3 - Convert the Time into Seconds The duration of the electrolysis is given in minutes: $$t = 96.5\text{ min}$$ Converting minutes to seconds ($1\text{ min} = 60\text{ s}$): $$t = 96.5 \times 60\text{ s} = 5790\text{ s}$$ Step 4 - Calculate the Mass ($w$) of Copper Deposited using Faraday's First Law Faraday's First Law of Electrolysis states that the mass ($w$) of a substance deposited or liberated at an electrode is directly proportional to the quantity of electricity passed: $$w = \frac{E \cdot I \cdot t}{F}$$ Where: * $E_{\ce{Cu}}$ is the equivalent weight of copper = $31.75\text{ g/eq}$ * $I$ is the electric current = $1.0\text{ A}$ * $t$ is the time in seconds = $5790\text{ s}$ * $F$ is Faraday's constant = $96,500\text{ C/eq}$ Substituting these values with their respective units: $$w = \frac{31.75\text{ g/eq} \times 1.0\text{ A} \times 5790\text{ s}}{96500\text{ C/eq}}$$ $$w = 31.75 \times \left( \frac{5790}{96500} \right)\text{ g}$$ $$w = 31.75 \times 0.06\text{ g} = \mathbf{1.905\text{ g}}$$ Thus, the copper cathode gains exactly $1.905\text{ g}$ in weight. Step 5 - Evaluate and Explain the Options * **Option (A) is correct:** As calculated, the mass of copper deposited at the cathode is exactly $1.905\text{ g}$. * **Option (B) is incorrect:** This value ($1.962\text{ g}$) represents the mass of zinc that dissolves from the anode. Using the equivalent weight of zinc ($E_{\ce{Zn}} = \frac{65.4}{2} = 32.7\text{ g/eq}$): $$w_{\ce{Zn}} = \frac{32.7 \times 1.0 \times 5790}{96500} = 1.962\text{ g}$$ This is the weight *lost* by the anode, not the weight gained by the cathode. * **Option (C) is incorrect:** This value ($3.81\text{ g}$) represents the mass of copper deposited if one incorrectly assumes a $1:1$ electron-to-ion stoichiometry ($n = 1$ instead of $2$). * **Option (D) is incorrect:** This value ($3.924\text{ g}$) represents twice the mass of zinc dissolved, which is a combined stoichiometry and electrode-assignment error. $$\text{Correct Option: } \boxed{\text{A}}$$