For the electrochemical cell, M \ — Electrochemistry Chemistry Question

💡 Solution & Explanation

\subsection*{Part I: Detailed Solution for Question 51} \subsubsection*{Step 1 - Identifying the Half-Cell Reactions} The electrochemical cell is represented using standard IUPAC notation as: $$\ce{M | M+ || X- | X}$$ By convention, the left side represents the anode (where oxidation occurs) and the right side represents the cathode (where reduction occurs) [1]: \begin{itemize} \item \textbf{Anode (Oxidation):} $$\ce{M(s) -> M+(aq) + e-}$$ \item \textbf{Cathode (Reduction):} The electrode on the right is the $\ce{X-|X}$ half-cell. For reduction to occur, the oxidized form $\ce{X}$ must accept an electron to form its reduced form $\ce{X-}$: $$\ce{X + e- -> X-}$$ \end{itemize} \subsubsection*{Step 2 - Calculating the Standard Cell Potential ($E^\circ_{\text{cell}}$)} The standard cell potential is given by the difference between the standard reduction potentials of the cathode and the anode [1]: $$E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}$$ From the given values: \begin{itemize} \item Standard reduction potential of the anode, $E^\circ_{\text{anode}} = E^\circ_{\ce{M+|M}} = 0.44\text{ V}$ [1] \item Standard reduction potential of the cathode, $E^\circ_{\text{cathode}} = E^\circ_{\ce{X|X-}} = 0.33\text{ V}$ [1] \end{itemize} Substituting these values: $$E^\circ_{\text{cell}} = 0.33\text{ V} - 0.44\text{ V} = -0.11\text{ V}$$ \subsubsection*{Step 3 - Determining Spontaneity} The overall cell reaction is the sum of both half-reactions: $$\ce{M(s) + X -> M+(aq) + X-(aq)}$$ The standard Gibbs free energy change ($\Delta G^\circ$) is related to $E^\circ_{\text{cell}}$ by: $$\Delta G^\circ = -nFE^\circ_{\text{cell}}$$ Since $E^\circ_{\text{cell}} = -0.11\text{ V}$ is negative, the standard Gibbs free energy change is positive ($\Delta G^\circ > 0$), meaning the forward cell reaction: $$\ce{M + X -> M+ + X-}$$ is \textbf{non-spontaneous}. Conversely, the reverse reaction: $$\ce{M+ + X- -> M + X}$$ will have a positive potential of $E^\circ_{\text{reverse}} = +0.11\text{ V}$ and is therefore thermodynamically \textbf{spontaneous} ($\Delta G^\circ < 0$). Thus, the correct option for Question 51 is \textbf{(b)} [2]. \subsection*{Part II: Detailed Solution for Question 61 (Matching the Hint)} \subsubsection*{Step 4 - Analyzing the Iron-Cerium Cell (Question 61)} The cell is set up as follows [1]: $$\ce{Pt(s) | Fe^3+, Fe^2+ (a=1) || Ce^4+, Ce^3+ (a=1) | Pt(s)}$$ The given standard reduction potentials are [1]: \begin{itemize} \item $E^\circ_{\ce{Ce^4+|Ce^3+}} = 1.61\text{ V}$ \item $E^\circ_{\ce{Fe^3+|Fe^2+}} = 0.77\text{ V}$ \end{itemize} Since $E^\circ_{\ce{Ce^4+|Ce^3+}} > E^\circ_{\ce{Fe^3+|Fe^2+}}$, the cerium half-cell undergoes reduction (cathode) and the iron half-cell undergoes oxidation (anode) [1]: \begin{itemize} \item \textbf{Cathode:} $\ce{Ce^4+ + e- -> Ce^3+}$ \item \textbf{Anode:} $\ce{Fe^2+ -> Fe^3+ + e-}$ \end{itemize} \subsubsection*{Step 5 - Flow of Electrons and Current Direction} \begin{itemize} \item \textbf{Electron Flow:} Electrons are lost at the anode ($\ce{Fe}$ compartment) and flow through the external circuit to the cathode ($\ce{Ce}$ compartment). \item \textbf{Conventional Current:} By convention, electrical current flows in the direction opposite to electron flow, which is from the \textbf{Ce electrode to the Fe electrode}. \item \textbf{Time Dependence:} As the cell operates, the reactants ($\ce{Ce^4+}$ and $\ce{Fe^2+}$) are depleted, which causes the cell potential (EMF) to gradually decrease toward zero. Consequently, the electric current \textbf{decreases} with time. \end{itemize} Thus, the correct option for Question 61 is \textbf{(a)} [2].