Which of the following pair of metals, when coupled, will give maximum EMF for a voltaic cell? — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Principle of EMF of a Voltaic Cell The electromotive force (EMF) of a voltaic (or galvanic) cell under standard state conditions ($E^\circ_{\text{cell}}$) is determined by the difference between the standard reduction potentials of the two coupled electrodes. The metal with the higher standard reduction potential acts as the cathode (where reduction occurs), and the metal with the lower standard reduction potential acts as the anode (where oxidation occurs): $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$ To obtain the maximum cell EMF from a given set of options, we must couple the metal with the most negative (lowest) standard reduction potential (anode) with the metal with the most positive (highest) standard reduction potential (cathode). Step 2 - List the Standard Reduction Potentials of the Metals Involved Let us identify the standard reduction potentials ($E^\circ$) at $25^\circ\text{C}$ for the metal systems presented in the options: 1. **Calcium ($\ce{Ca}$):** $$\ce{Ca^{2+}(aq) + 2e^- -> Ca(s)} \quad E^\circ_{\ce{Ca^{2+}/Ca}} \approx -2.87\text{ V}$$ 2. **Iron ($\ce{Fe}$):** $$\ce{Fe^{2+}(aq) + 2e^- -> Fe(s)} \quad E^\circ_{\ce{Fe^{2+}/Fe}} \approx -0.44\text{ V}$$ 3. **Lead ($\ce{Pb}$):** $$\ce{Pb^{2+}(aq) + 2e^- -> Pb(s)} \quad E^\circ_{\ce{Pb^{2+}/Pb}} \approx -0.13\text{ V}$$ 4. **Copper ($\ce{Cu}$):** $$\ce{Cu^{2+}(aq) + 2e^- -> Cu(s)} \quad E^\circ_{\ce{Cu^{2+}/Cu}} \approx +0.34\text{ V}$$ 5. **Gold ($\ce{Au}$):** $$\ce{Au^{3+}(aq) + 3e^- -> Au(s)} \quad E^\circ_{\ce{Au^{3+}/Au}} \approx +1.50\text{ V}$$ Step 3 - Calculate the Standard EMF ($E^\circ_{\text{cell}}$) for Each Option By pairing the two metals in each option such that the spontaneous direction yields a positive potential, we calculate the standard cell potential: * **Option (A) - \ce{Fe} and \ce{Cu} couple:** Copper ($\ce{Cu}$, $E^\circ = +0.34\text{ V}$) acts as the cathode and iron ($\ce{Fe}$, $E^\circ = -0.44\text{ V}$) acts as the anode: $$E^\circ_{\text{cell}} = E^\circ_{\ce{Cu^{2+}/Cu}} - E^\circ_{\ce{Fe^{2+}/Fe}}$$ $$E^\circ_{\text{cell}} = +0.34\text{ V} - (-0.44\text{ V}) = +0.78\text{ V}$$ * **Option (B) - \ce{Pb} and \ce{Au} couple:** Gold ($\ce{Au}$, $E^\circ = +1.50\text{ V}$) acts as the cathode and lead ($\ce{Pb}$, $E^\circ = -0.13\text{ V}$) acts as the anode: $$E^\circ_{\text{cell}} = E^\circ_{\ce{Au^{3+}/Au}} - E^\circ_{\ce{Pb^{2+}/Pb}}$$ $$E^\circ_{\text{cell}} = +1.50\text{ V} - (-0.13\text{ V}) = +1.63\text{ V}$$ * **Option (C) - \ce{Cu} and \ce{Au} couple:** Gold ($\ce{Au}$, $E^\circ = +1.50\text{ V}$) acts as the cathode and copper ($\ce{Cu}$, $E^\circ = +0.34\text{ V}$) acts as the anode: $$E^\circ_{\text{cell}} = E^\circ_{\ce{Au^{3+}/Au}} - E^\circ_{\ce{Cu^{2+}/Cu}}$$ $$E^\circ_{\text{cell}} = +1.50\text{ V} - 0.34\text{ V} = +1.16\text{ V}$$ * **Option (D) - \ce{Ca} and \ce{Cu} couple:** Copper ($\ce{Cu}$, $E^\circ = +0.34\text{ V}$) acts as the cathode and calcium ($\ce{Ca}$, $E^\circ = -2.87\text{ V}$) acts as the anode: $$E^\circ_{\text{cell}} = E^\circ_{\ce{Cu^{2+}/Cu}} - E^\circ_{\ce{Ca^{2+}/Ca}}$$ $$E^\circ_{\text{cell}} = +0.34\text{ V} - (-2.87\text{ V}) = +3.21\text{ V}$$ Step 4 - Evaluate and Explain the Options Comparing the standard EMF values calculated above: * **Option (A) is incorrect:** The combination of iron and copper yields a standard potential difference of only $0.78\text{ V}$. * **Option (B) is incorrect:** Although gold has a highly positive potential and lead has a negative potential, this couple yields $1.63\text{ V}$, which is not the maximum among the options. * **Option (C) is incorrect:** Since both gold and copper have positive standard reduction potentials, their difference is relatively small ($1.16\text{ V}$). * **Option (D) is correct:** The calcium and copper couple provides a very large potential difference of $3.21\text{ V}$. This occurs because calcium is a highly electropositive alkaline earth metal with an extremely low reduction potential, while copper is a stable coinage transition metal with a positive reduction potential. $$\text{Correct Option: } \boxed{\text{D}}$$