Problem 2

Molecular hydrogen (H2) is an alternative to carbon dioxide-emitting fuels, so lowering the cost and environmental impact of its production is a major challenge; water splitting is a promising candidate technology. Data at 298 K:

\Delta_f H^\circ

(kJ mol

^{-1}

): H2(g) 0, H2O(l) −285.8, H2O(g) −241.8, O2(g) 0;

S_m^\circ

(J mol

^{-1}

^{-1}

): H2(g) 130.6, H2O(l) 69.9, H2O(g) 188.7, O2(g) 205.2. (a) Write the balanced equation for the splitting of liquid water with a stoichiometric coefficient of 1 for water and show numerically whether the reaction is thermodynamically favourable at 298 K. (b) Water splitting can be performed electrochemically with two electrodes in an acidic water bath; write the half-reactions at each electrode and derive the condition on the applied voltage

\Delta E_{\text{applied}}

relative to the thermodynamic threshold

\Delta E_{\text{th}}

for the process to be favourable at 298 K. (c) For a Pt cathode the minimum voltage

\Delta E_{\min}

depends on the anode: IrOx 1.6 V, NiOx 1.7 V, CoOx 1.7 V, Fe2O3 1.9 V. Give the expression for the power efficiency

\eta_{\text{elec}}

of water electrolysis, calculate it for Pt/Fe2O3 and name the most efficient anode.

Step 2 of 4: Reaction enthalpy and entropy

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Intuition

Use products minus reactants for both ΔrH° and ΔrS°, then combine them through ΔrG° = ΔrH° − TΔrS°.

\Delta_r H^\circ = 285.8\ \text{kJ mol}^{-1},\quad \Delta_r S^\circ = 163.3\ \text{J mol}^{-1} \text{K}^{-1}

Analysis

$\Delta_r H^\circ = 0 + 0 - (-285.8) = +285.8$ kJ mol $^{-1}$ and $\Delta_r S^\circ = 130.6 + \tfrac{1}{2}(205.2) - 69.9 = +163.3$ J mol $^{-1}$ K $^{-1}$ . Then $\Delta_r G^\circ = 285.8 - 298\times 0.1633 = +237.1$ kJ mol $^{-1} > 0$ : the reaction is not spontaneous (equivalently $K^\circ \approx 10^{-41.6} \ll 1$ ).

Common pitfall. Do not forget the factor

\tfrac{1}{2}

in front of

S^\circ(\ce{O2})

; omitting it gives the wrong ΔrG°.