Pick up the false statement(s): — Electrochemistry Chemistry Question

💡 Solution & Explanation

Step 1 - Analyze Statement (A): Net Chemical Change in a Galvanic Cell A galvanic (or voltaic) cell is an electrochemical system that converts the chemical energy of a spontaneous chemical reaction into electrical energy. For electrical current to flow, electrons must be transferred from the reducing agent to the oxidizing agent through an external circuit. This transfer is composed of two synchronized half-reactions: * An oxidation half-reaction at the anode (loss of electrons): $$\ce{Red1 -> Ox1 + n e^-}$$ * A reduction half-reaction at the cathode (gain of electrons): $$\ce{Ox2 + n e^- -> Red2}$$ The sum of these two processes yields the overall cell reaction: $$\ce{Red1 + Ox2 -> Ox1 + Red2}$$ Because any galvanic cell reaction must involve both oxidation and reduction to establish a potential difference and drive an electric current, the net chemical change is always a redox reaction. Therefore, Statement (A) is **true**. Step 2 - Analyze Statement (B): Cobalt and Cadmium Galvanic Cell To determine which electrode acts as the anode (where oxidation occurs) and which acts as the cathode (where reduction occurs) in a galvanic cell, we compare their standard reduction potentials ($E^\circ$): * For the cobalt electrode: $$\ce{Co^2+(aq) + 2e^- -> Co(s)} \quad E^\circ_{\ce{Co^2+/Co}} = -0.28\text{ V}$$ * For the cadmium electrode: $$\ce{Cd^2+(aq) + 2e^- -> Cd(s)} \quad E^\circ_{\ce{Cd^2+/Cd}} = -0.40\text{ V}$$ Since $E^\circ_{\ce{Co^2+/Co}} > E^\circ_{\ce{Cd^2+/Cd}}$, the cobalt ions ($\ce{Co^2+}$) have a greater thermodynamic tendency to accept electrons and be reduced than the cadmium ions ($\ce{Cd^2+}$). Thus, in a spontaneous cell: * The cobalt electrode acts as the **cathode** (reduction): $$\ce{Co^2+(aq) + 2e^- -> Co(s)}$$ * The cadmium electrode acts as the **anode** (oxidation): $$\ce{Cd(s) -> Cd^2+(aq) + 2e^-}$$ Therefore, the assertion that the cobalt electrode acts as the anode is **false**. Step 3 - Analyze Statement (C): Behavior of Standard Potential with Concentration The **standard potential** ($E^\circ$) of an electrode or an electrochemical cell is a thermodynamic constant defined strictly under standard-state conditions: * The concentration of all dissolved solute species is exactly $1.0\text{ M}$. * The partial pressure of any reacting gas is exactly $1.0\text{ bar}$ (or $1.0\text{ atm}$). * The temperature is maintained at a specified value, typically $298.15\text{ K}$ ($25^\circ\text{C}$). Because standard potential is defined at a fixed reference concentration of $1\text{ M}$, it is an intensive property and is completely independent of the actual concentration of the electrolyte in the cell. Only the **non-standard potential** ($E$) varies with concentration, as quantitatively described by the Nernst equation: $$E = E^\circ - \frac{2.303 RT}{nF} \log_{10} Q$$ Therefore, the assertion that standard potential increases with concentration is **false**. Step 4 - Analyze Statement (D): Calomel Electrode vs. Standard Hydrogen Electrode The calomel electrode ($\ce{Hg(l) \mid Hg2Cl2(s) \mid Cl^-(aq)}$) is a widely used secondary reference electrode. Its electrode potential is determined by the concentration of chloride ions in the cell and is not $0.00\text{ V}$. For instance, the standard potential of a saturated calomel electrode (SCE) at $298\text{ K}$ is: $$E^\circ_{\text{SCE}} \approx +0.244\text{ V}$$ The electrode that is universally and arbitrarily assigned a standard reduction potential of exactly $0.00\text{ V}$ at all temperatures is the Standard Hydrogen Electrode (SHE): $$\ce{2H+(aq) + 2e^- -> H2(g)} \quad E^\circ_{\ce{H+/H2}} = 0.00\text{ V}$$ Therefore, the assertion that the calomel electrode has a potential of $0.00\text{ V}$ is **false**. Step 5 - Conclusion Evaluating the options: * Statement (A) is **true**. * Statement (B) is **false**. * Statement (C) is **false**. * Statement (D) is **false**. Since the question asks to identify the false statement(s), the correct choices are (B), (C), and (D). $$\text{Correct Options: } \boxed{B,C,D}$$