Question

For the galvanic cell: H\(_2\)(g) | HCl(aq) | Cl\(_2\)(g), the standard electromotive force is \[ E^0 = 1.73 - (1.25\times10^{-3})T + (1.00\times10^{-6})T^2 \quad (\text{V, with }T\text{ in K}). \] Find the standard enthalpy change \(\Delta_r H^0\) at \(300\ \text{K}\) (kJ mol\(^{-1}\)). Assume the cell reaction is written per mole of HCl, i.e., \(\tfrac12\mathrm{H_2} + \tfrac12\mathrm{Cl_2} \rightarrow \mathrm{HCl(aq)\) so that \(n=1\). (Given \(F=96500\ \text{C mol}^{-1}\)).}

Hint:

For EMF that depends on \(T\): \(\Delta H^0=-nF\big[E^0 - T(dE^0/dT)\big]\). Be clear whether the cell reaction is written per mole of product (here, per mole of HCl so \(n=1\)) or for the full \(2\text{e}^-\) process.

Answer 1

Correct Answer: -164

Step 1: Evaluate \(E^0\) and its temperature derivative at \(T=300\ \text{K}\). \[ E^0 = 1.73-(1.25\times10^{-3})(300)+(1.00\times10^{-6})(300)^2 =1.445\ \text{V}, \] \[ \left(\frac{dE^0}{dT}\right)_P = -1.25\times10^{-3}+2(1.00\times10^{-6})(300) = -6.5\times10^{-4}\ \text{V K}^{-1}. \] Step 2: Use thermodynamic relations. \[ \Delta G^0=-nFE^0,\qquad \Delta S^0=nF\left(\frac{dE^0}{dT}\right), \] \[ \Rightarrow\ \Delta H^0=\Delta G^0+T\Delta S^0 = -nFE^0 + nFT\left(\frac{dE^0}{dT}\right). \] Step 3: Insert \(n=1\). \[ \Delta H^0 = -(96500)(1.445) + (96500)(300)(-6.5\times10^{-4}) = -1.394\times10^{5} - 1.882\times10^{4}\ \text{J mol}^{-1} \] \[ = -1.5826\times10^{5}\ \text{J mol}^{-1} \approx -1.58\times10^{5}\ \text{J mol}^{-1} = \boxed{-158\ \text{kJ mol}^{-1}}. \]