When an electric current is drawn from a galvanic cell — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Define the Electromotive Force (EMF) of a Galvanic Cell The electromotive force (EMF) or cell potential ($E_{\ce{cell}}$) of a galvanic cell is the potential difference between its two electrodes (the cathode and the anode). It is mathematically defined as: $$E_{\ce{cell}} = E_{\ce{cathode}} - E_{\ce{anode}}$$ where $E_{\ce{cathode}}$ and $E_{\ce{anode}}$ represent the reduction potentials of the cathode and anode half-cells under non-standard conditions. Step 2 - Analyze Concentration Changes Using the Nernst Equation Consider a general spontaneous redox reaction taking place inside an operating galvanic cell: $$\ce{aA(aq) + bB(s) -> cC(aq) + dD(s)}$$ According to the Nernst equation, the cell potential ($E_{\ce{cell}}$) at any given moment at a temperature $T$ is given by: $$E_{\ce{cell}} = E^\circ_{\ce{cell}} - \frac{2.303 RT}{nF} \log Q$$ Where: * $E^\circ_{\ce{cell}}$ is the standard cell potential. * $n$ is the number of moles of electrons transferred. * $F$ is Faraday's constant. * $Q$ is the reaction quotient, defined as: $$Q = \frac{[\ce{C}]^c}{[\ce{A}]^a}$$ When an electric current is drawn from the cell: 1. Reactants ($\ce{A}$) are continuously consumed, causing their concentration $[\ce{A}]$ to **decrease**. 2. Products ($\ce{C}$) are continuously generated, causing their concentration $[\ce{C}]$ to **increase**. Step 3 - Determine the Trend of EMF Over Time As the active chemical species are consumed and formed during cell discharge: * The reaction quotient $Q = \frac{[\ce{C}]^c}{[\ce{A}]^a}$ **increases continuously**. * This increase in $Q$ causes the logarithmic term $\frac{2.303 RT}{nF} \log Q$ to become **larger and more positive**. * In the Nernst equation, subtracting a progressively larger positive term from $E^\circ_{\ce{cell}}$ causes the cell potential ($E_{\ce{cell}}$) to **gradually decrease**. This drop in voltage continues until the electrochemical system reaches a state of **dynamic chemical equilibrium**, where the rate of the forward reaction equals the rate of the reverse reaction. At equilibrium: $$Q = K_{\ce{eq}}$$ $$E_{\ce{cell}} = \boxed{0\text{ V}}$$ At this point, the cell is fully discharged, can no longer perform electrical work, and the cell voltage drops completely to zero. Step 4 - Evaluate and Explain the Options * **Option (A) is incorrect:** The EMF does not suddenly increase because the driving force for the reaction decreases as reactants are depleted. * **Option (B) is incorrect:** The EMF does not increase to a maximum value; instead, it undergoes a continuous thermodynamic decline. * **Option (C) is correct:** As demonstrated mathematically by the Nernst equation, drawing current shifts the concentrations of reactants and products, causing the cell potential to decrease gradually and eventually fall to zero at chemical equilibrium. * **Option (D) is incorrect:** The EMF cannot remain constant during current discharge because the concentrations of the electroactive species are continuously changing. $$\text{Correct Option: } \boxed{\text{C}}$$