Solveeit Logo
Login

Question

Question: Consider the electrochemical cell M(s)|MI2(s)|MI2(aq) | M(s) where 'M' is a metal. At 298 K, the sta...

Consider the electrochemical cell M(s)|MI2(s)|MI2(aq) | M(s) where 'M' is a metal. At 298 K, the standard reduction potentials are EM2+(aq)/M(s)o=−0.12 V,EMI2(s)/M(s)o=−0.36 VE_{M^{2+}(aq)/M(s)}^o = -0.12 \ V, E_{MI_2(s)/M(s)}^o = -0.36 \ V and the temperature coefficient is (∂Ecello∂T)P=1.5×10−4 V K−1(\frac{\partial E_{cell}^o}{\partial T})_P = 1.5 \times 10^{-4} \ V \ K^{-1}. At this temperature the standard enthalpy change for the overall cell reaction, ΔrHP\Delta_r H^P, is __________ kJmol−1^{-1}. (Faraday constant F=96500Cmol−1F = 96500 Cmol^{-1})

A

-40

B

< -35

C

< -25

D

-25

Answer

< -35

Explanation

Solution

Here's how to solve this electrochemistry problem:

  1. Calculate the standard cell potential (EcelloE^o_{\text{cell}}):

    Ecello=Ecathodeo−EanodeoE^o_{\text{cell}} = E^o_{\text{cathode}} - E^o_{\text{anode}}
    Ecello=−0.12 V−(−0.36 V)=0.24 VE^o_{\text{cell}} = -0.12 \text{ V} - (-0.36 \text{ V}) = 0.24 \text{ V}

  2. Calculate the standard entropy change (ΔrSo\Delta_r S^o):

    ΔrSo=nF(∂Ecello∂T)P\Delta_r S^o = nF \left(\frac{\partial E^o_{\text{cell}}}{\partial T}\right)_P
    Where:

    • n=2n = 2 (number of moles of electrons transferred)
    • F=96500 C mol−1F = 96500 \text{ C mol}^{-1} (Faraday constant)
    • (∂Ecello∂T)P=1.5×10−4 V K−1\left(\frac{\partial E^o_{\text{cell}}}{\partial T}\right)_P = 1.5 \times 10^{-4} \text{ V K}^{-1} (temperature coefficient)

    ΔrSo=2×96500 C mol−1×1.5×10−4 V K−1=28.95 J K−1mol−1\Delta_r S^o = 2 \times 96500 \text{ C mol}^{-1} \times 1.5 \times 10^{-4} \text{ V K}^{-1} = 28.95 \text{ J K}^{-1} \text{mol}^{-1}

  3. Calculate the standard Gibbs free energy change (ΔrGo\Delta_r G^o):

    ΔrGo=−nFEcello\Delta_r G^o = -nFE^o_{\text{cell}}
    ΔrGo=−2×96500 C mol−1×0.24 V=−46320 J mol−1\Delta_r G^o = -2 \times 96500 \text{ C mol}^{-1} \times 0.24 \text{ V} = -46320 \text{ J mol}^{-1}

  4. Calculate the standard enthalpy change (ΔrHo\Delta_r H^o):

    ΔrHo=ΔrGo+TΔrSo\Delta_r H^o = \Delta_r G^o + T\Delta_r S^o
    ΔrHo=−46320 J mol−1+(298 K×28.95 J K−1mol−1)\Delta_r H^o = -46320 \text{ J mol}^{-1} + (298 \text{ K} \times 28.95 \text{ J K}^{-1} \text{mol}^{-1})
    ΔrHo=−46320 J mol−1+8627.1 J mol−1=−37692.9 J mol−1\Delta_r H^o = -46320 \text{ J mol}^{-1} + 8627.1 \text{ J mol}^{-1} = -37692.9 \text{ J mol}^{-1}
    ΔrHo≈−37.7 kJ mol−1\Delta_r H^o \approx -37.7 \text{ kJ mol}^{-1}

Since −37.7 kJ mol−1-37.7 \text{ kJ mol}^{-1} is less than -35 kJ/mol, the correct answer is:

< -35