JEE Challenger
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ChemHardAdv 2018 • Paper 2

Standard Entropy Change Calculation for Given Electrochemical Cell

Consider an electrochemical cell: A(s)∣An+(aq,2 M)∥B2n+(aq,1 M)∣B(s)\text{A(s)} \mid \text{A}^{n+}(\text{aq}, 2\text{ M}) \parallel \text{B}^{2n+}(\text{aq}, 1\text{ M}) \mid \text{B(s)} The value of ΔH⊖\Delta H^\ominus for the overall cell reaction is twice that of ΔG⊖\Delta G^\ominus at 300 K300\text{ K}. If the electromotive force (emf) of the cell is zero, find the value of ΔS⊖\Delta S^\ominus (in J K−1mol−1\text{J K}^{-1}\text{mol}^{-1}) of the cell reaction per mole of B\text{B} formed at 300 K300\text{ K}.

(Given: ln⁡(2)=0.7\ln(2) = 0.7, R (universal gas constant)=8.3 J K−1mol−1R \text{ (universal gas constant)} = 8.3\text{ J K}^{-1}\text{mol}^{-1}. HH, SS and GG denote enthalpy, entropy and Gibbs energy, respectively.)

Official Numerical Answer-11.62

Topics & Concepts

ElectrochemistryElectrochemical CellNernst EquationGibbs Free EnergyEntropy and Second Law

Step-by-Step Solution

To find the standard entropy change (ΔS⊖\Delta S^\ominus) for the cell reaction per mole of B\text{B} formed, we proceed as follows:

  1. Cell Reaction and Reaction Quotient: The half-cell reactions are: Anode (Oxidation): 2A(s)⟶2An+(aq)+2ne−\text{Anode (Oxidation): } 2\text{A(s)} \longrightarrow 2\text{A}^{n+}(\text{aq}) + 2n\text{e}^- Cathode (Reduction): B2n+(aq)+2ne−⟶B(s)\text{Cathode (Reduction): } \text{B}^{2n+}(\text{aq}) + 2n\text{e}^- \longrightarrow \text{B(s)}

    The overall cell reaction per mole of B\text{B} formed is: 2A(s)+B2n+(aq)⟶2An+(aq)+B(s)2\text{A(s)} + \text{B}^{2n+}(\text{aq}) \longrightarrow 2\text{A}^{n+}(\text{aq}) + \text{B(s)}

    The number of moles of electrons transferred in this reaction is: z=2nz = 2n

    The reaction quotient QQ is given by: Q=[An+]2[B2n+]=(2)21=4Q = \frac{[\text{A}^{n+}]^2}{[\text{B}^{2n+}]} = \frac{(2)^2}{1} = 4

  2. Standard Cell Potential (Ecell⊖E^\ominus_{\text{cell}}): Using the Nernst equation for the cell at temperature T=300 KT = 300\text{ K}: Ecell=Ecell⊖−RTzFln⁡QE_{\text{cell}} = E^\ominus_{\text{cell}} - \frac{RT}{zF}\ln Q

    Given that the electromotive force of the cell is zero (Ecell=0E_{\text{cell}} = 0): Ecell⊖=RT2nFln⁡(4)=RT2nF(2ln⁡2)=RTnFln⁡2E^\ominus_{\text{cell}} = \frac{RT}{2nF}\ln(4) = \frac{RT}{2nF}(2\ln 2) = \frac{RT}{nF}\ln 2

  3. Standard Gibbs Free Energy Change (ΔG⊖\Delta G^\ominus): ΔG⊖=−zFEcell⊖=−(2n)F(RTln⁡2nF)=−2RTln⁡2\Delta G^\ominus = -zFE^\ominus_{\text{cell}} = -(2n)F\left(\frac{RT\ln 2}{nF}\right) = -2RT\ln 2

  4. Standard Entropy Change (ΔS⊖\Delta S^\ominus): Using the thermodynamic relation: ΔG⊖=ΔH⊖−TΔS⊖\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus

    Given that ΔH⊖=2ΔG⊖\Delta H^\ominus = 2\Delta G^\ominus: ΔG⊖=2ΔG⊖−TΔS⊖\Delta G^\ominus = 2\Delta G^\ominus - T\Delta S^\ominus TΔS⊖=ΔG⊖T\Delta S^\ominus = \Delta G^\ominus ΔS⊖=ΔG⊖T=−2RTln⁡2T=−2Rln⁡2\Delta S^\ominus = \frac{\Delta G^\ominus}{T} = \frac{-2RT\ln 2}{T} = -2R\ln 2

  5. Calculation: Substituting the given values (R=8.3 J K−1mol−1R = 8.3\text{ J K}^{-1}\text{mol}^{-1} and ln⁡2=0.7\ln 2 = 0.7): ΔS⊖=−2×8.3×0.7=−11.62 J K−1mol−1\Delta S^\ominus = -2 \times 8.3 \times 0.7 = -11.62\text{ J K}^{-1}\text{mol}^{-1}

Final Answer: The value of ΔS⊖\Delta S^\ominus is −11.62-11.62.

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