If a reaction can be carried out only by electrolysis, which of the following changes in a thermodynamic property must apply: (a) \(\Delta H>0 ;\) (b) \(\Delta S>0\) (c) \(\Delta G=\Delta H ;\) (d) \(\Delta G>0 ?\) Explain.

Before addressing the specific options in the exercise, it's essential to understand what each sign of these properties means: \n- \(\Delta H\) represents the change in enthalpy (the heat content) of a reaction. If \(\Delta H > 0\), heat is absorbed (endothermic process) and if \(\Delta H < 0\), heat is released (exothermic process). - \(\Delta S\) stands for the change in entropy (the disorder of the system). If \(\Delta S > 0\), the system is becoming more disordered. If \(\Delta S < 0\), the system is becoming more ordered.- \(\Delta G\) denotes the change in Gibbs free energy, which is the maximum reversible work done at constant temperature and pressure. It's calculated as \(\Delta G = \Delta H - T\Delta S\). If \(\Delta G < 0\), the reaction is spontaneous. If \(\Delta G > 0\), the reaction is non-spontaneous and requires external energy.

Electrolysis involves the use of an electric current to drive a non-spontaneous chemical reaction - a reaction that otherwise wouldn’t occur without the addition of energy. So in terms of thermodynamics, for a reaction to only be feasible via electrolysis, it must be non-spontaneous, i.e., \(\Delta G > 0\).

With the above insights, it can be seen that the relevant answer is (d) \(\Delta G > 0\). Neither a positive \(\Delta H\), nor a positive \(\Delta S\), nor \(\Delta G = \Delta H\) would necessarily imply that the reaction could only occur via electrolysis.