A galvanic cell is set up from a zinc bar weighing 100 g and 1.0 L of 1.0 M copper sulphate solution — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the Balanced Cell Reaction In the given galvanic cell (a classic Daniell-type cell), oxidation occurs at the zinc anode and reduction occurs at the copper cathode. The half-reactions are: * **Anode (Oxidation):** $$\ce{Zn(s) -> Zn^{2+}(aq) + 2e^-}$$ * **Cathode (Reduction):** $$\ce{Cu^{2+}(aq) + 2e^- -> Cu(s)}$$ By combining these two half-reactions, we obtain the net spontaneous cell reaction: $$\ce{Zn(s) + Cu^{2+}(aq) -> Zn^{2+}(aq) + Cu(s)}$$ From this balanced equation, we see that $1\text{ mole}$ of solid zinc reacts with $1\text{ mole}$ of copper(II) ions in a $1:1$ stoichiometric ratio, transferring $2\text{ moles}$ of electrons ($n = 2$) in the process. Step 2 - Determine the Limiting Reactant To find out how long the cell can run, we must calculate the initial moles of both reactants and determine which one is the limiting reactant: 1. **Moles of Zinc ($\ce{Zn}$):** Given mass of zinc bar = $100\text{ g}$ Atomic mass of zinc ($M_{\ce{Zn}}$) = $65.4\text{ g/mol}$ $$\text{Initial moles of Zn} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{100\text{ g}}{65.4\text{ g/mol}} \approx 1.53\text{ mol}$$ 2. **Moles of Copper(II) ions ($\ce{Cu^{2+}}$):** Given volume of copper sulphate solution ($V$) = $1.0\text{ L}$ Molar concentration ($C$) = $1.0\text{ M}$ $$\text{Initial moles of }\ce{Cu^{2+}} = C \times V = 1.0\text{ mol/L} \times 1.0\text{ L} = 1.0\text{ mol}$$ Comparing the moles: $$\text{Moles of }\ce{Cu^{2+}}\ (1.0\text{ mol}) < \text{Moles of Zn}\ (1.53\text{ mol})$$ Since the stoichiometric ratio is $1:1$, the copper(II) ions are completely consumed first. Therefore, **$\ce{Cu^{2+}}$ is the limiting reactant**, and the maximum amount of reaction that can take place is limited to $1.0\text{ mole}$ of $\ce{Cu^{2+}}$ reduction. Step 3 - Calculate the Total Quantity of Electricity ($Q$) Transferred Since the cell reaction can proceed until $1.0\text{ mole}$ of $\ce{Cu^{2+}}$ is reduced, and each mole of $\ce{Cu^{2+}}$ reduced requires $2\text{ moles}$ of electrons, the total moles of electrons ($n_{e^-}$) transferred during the lifetime of the cell is: $$n_{e^-} = 2 \times 1.0\text{ mol} = 2.0\text{ mol of electrons}$$ Using Faraday's constant ($F \approx 96,500\text{ C/mol}$), we calculate the total electrical charge ($Q$) delivered: $$Q = n_{e^-} \times F$$ $$Q = 2.0\text{ mol} \times 96,500\text{ C/mol} = 193,000\text{ C}$$ Step 4 - Calculate the Running Time ($t$) of the Cell The relationship between total charge ($Q$), steady current ($I$), and time ($t$) in seconds is: $$Q = I \times t \implies t = \frac{Q}{I}$$ Substituting the values of charge ($Q = 193,000\text{ C}$) and steady current ($I = 1.0\text{ A}$): $$t = \frac{193,000\text{ C}}{1.0\text{ A}} = 193,000\text{ s}$$ Now, convert the time from seconds to hours: $$t = \frac{193,000\text{ s}}{3,600\text{ s/h}} \approx 53.61\text{ h}$$ Thus, the cell would run for approximately $53.6\text{ h}$. Step 5 - Evaluate and Explain the Options * **Option (A) is correct:** As calculated, the copper ions act as the limiting reactant, allowing the cell to deliver $1.0\text{ A}$ for approximately $53.6\text{ h}$. * **Option (B) is incorrect:** This value ($26.8\text{ h}$) is half of the correct value, which occurs if one incorrectly assumes that only $1\text{ mole}$ of electrons is transferred per mole of copper reduced ($n = 1$ instead of $n = 2$). * **Option (C) is incorrect:** This value ($81.97\text{ h}$) is obtained if one incorrectly assumes that zinc is the limiting reactant and is completely consumed: $$t = \frac{1.53\text{ mol} \times 2 \times 96,500\text{ C/mol}}{1.0\text{ A} \times 3,600\text{ s/h}} \approx 82\text{ h}$$ * **Option (D) is incorrect:** This value ($40.99\text{ h}$) represents half of the time calculated if zinc were the limiting reactant. $$\text{Correct Option: } \boxed{\text{A}}$$