After some time, the voltage of an electrochemical cell becomes zero. This is because — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Working of an Electrochemical Cell In a galvanic or electrochemical cell, a spontaneous chemical reaction produces an electric current. The overall cell potential or electromotive force ($E_{\ce{cell}}$) is the difference between the reduction potential of the cathode ($E_{\ce{cathode}}$) and the reduction potential of the anode ($E_{\ce{anode}}$): $$E_{\ce{cell}} = E_{\ce{cathode}} - E_{\ce{anode}}$$ By convention, both $E_{\ce{cathode}}$ and $E_{\ce{anode}}$ are represented as standard reduction potentials. Step 2 - Analyze the Changes in Electrode Potentials Over Time As the cell operates and current flows through the external circuit: 1. **At the Cathode:** Cations are continuously reduced and deposited. Consequently, the concentration of electroactive cations in the cathode half-cell decreases. According to the Nernst equation, as the concentration of reactant cations decreases, the reduction potential of the cathode ($E_{\ce{cathode}}$) decreases: $$E_{\ce{cathode}} = E^\circ_{\ce{cathode}} - \frac{2.303 RT}{nF} \log \frac{1}{[\text{Cathode Cation}]}$$ 2. **At the Anode:** Metal atoms are oxidized to form cations, which enter the solution. Consequently, the concentration of metal cations in the anode half-cell increases. According to the Nernst equation, as the concentration of product cations increases, the reduction potential of the anode ($E_{\ce{anode}}$) increases (becomes more positive or less negative): $$E_{\ce{anode}} = E^\circ_{\ce{anode}} - \frac{2.303 RT}{nF} \log \frac{1}{[\text{Anode Cation}]}$$ Step 3 - Define the State of Equilibrium Since $E_{\ce{cathode}}$ decreases and $E_{\ce{anode}}$ increases over time, the difference between them ($E_{\ce{cell}}$) gradually diminishes. Eventually, the chemical system inside the cell reaches a state of **dynamic chemical equilibrium**, where the rates of oxidation and reduction become equal. At equilibrium, no net reaction occurs, and the cell can no longer generate any electrical work. Consequently, the measured voltage of the cell becomes exactly zero: $$E_{\ce{cell}} = 0$$ Substituting $E_{\ce{cell}} = 0$ into our potential difference formula: $$0 = E_{\ce{cathode}} - E_{\ce{anode}}$$ $$\implies E_{\ce{cathode}} = E_{\ce{anode}}$$ This mathematical relationship dictates that the reduction potential of the cathode is exactly equal to the reduction potential of the anode. For two physical values to be identical, they must have both the **same magnitude** and the **same algebraic sign**. Step 4 - Evaluate and Explain the Options * **Option (A) is incorrect:** The individual electrode potentials do not need to be zero. For example, if both electrodes have a reduction potential of $+0.34\text{ V}$, then $E_{\ce{cell}} = +0.34\text{ V} - (+0.34\text{ V}) = 0\text{ V}$. * **Option (B) is incorrect:** If the reduction potentials are equal but have opposite signs (e.g., $E_{\ce{cathode}} = +0.76\text{ V}$ and $E_{\ce{anode}} = -0.76\text{ V}$), the cell potential would be $E_{\ce{cell}} = +0.76\text{ V} - (-0.76\text{ V}) = +1.52\text{ V}$, which is not zero. * **Option (C) is correct:** As demonstrated, when the reduction potentials of both electrodes become equal in magnitude and carry the same sign ($E_{\ce{cathode}} = E_{\ce{anode}}$), their difference becomes zero, and the cell voltage drops to zero. * **Option (D) is incorrect:** The ions in the salt bridge continue to move to maintain electrical neutrality. The cell voltage drops to zero due to thermodynamic equilibrium, not because the ions in the salt bridge physically stop moving. $$\text{Correct Option: } \boxed{\text{C}}$$