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 3 of 4: Thermodynamic electrolysis voltage

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\Delta E_{\text{applied}} > \Delta E_{\text{th}} = \dfrac{\Delta_r G^\circ}{2F} = \dfrac{237.1\times 10^{3}}{2\times 96485} = 1.229\ \text{V}

Analysis

Half-reactions: cathode $\ce{2H+ + 2e- -> H2}$ , anode $\ce{H2O -> 2H+ + 1/2 O2 + 2e-}$ . Two electrons circulate per water molecule split, so the minimum applied voltage satisfies $2F\,\Delta E \ge \Delta_r G^\circ$ , i.e. $\Delta E_{\text{applied}} > \Delta E_{\text{th}} = 1.229$ V.