When magnesium metal is added to a beaker of \(\mathrm{HCl}(a q)\), a gas is produced. Knowing that magnesium is oxidized and that hydrogen is reduced, write the balanced equation for the reaction. How many electrons are transferred in the balanced equation? What quantity of useful work can be obtained when \(\mathrm{Mg}\) is added directly to the beaker of \(\mathrm{HCl}\) ? How can you harness this reaction to do useful work?

First, we need to write the unbalanced half-reactions for the reduction of hydrogen ions and the oxidation of magnesium metal. Reduction half-reaction: \(H^+ + e^- \rightarrow H\) Oxidation half-reaction: \(Mg \rightarrow Mg^{2+} + 2e^-\)

Now we need to balance these half-reactions by recognizing that magnesium loses 2 electrons and hydrogen gains 2 electrons in the reaction. Reduction half-reaction: \(2H^+ + 2e^- \rightarrow H_2\) Oxidation half-reaction: \(Mg \rightarrow Mg^{2+} + 2e^-\)

Let's combine the half-reactions to find the balanced overall reaction equation. Since the number of electrons transferred is the same in both half-reactions, they can be combined directly: Overall reaction: \(Mg + 2H^+ \rightarrow Mg^{2+} + H_2\) We can add the chloride ions back into the overall reaction to show the balanced equation involving aqueous hydrochloric acid: Balanced equation: \(Mg + 2HCl \rightarrow MgCl_2 + H_2\)

The number of electrons transferred can be found by looking at the balanced half-reactions. In the oxidation half-reaction, magnesium loses 2 electrons: \(Mg \rightarrow Mg^{2+} + 2e^-\) So, 2 electrons are transferred in the balanced equation.

To calculate the useful work obtained from the reaction, we need to know the Gibbs free energy change of the reaction, ΔG. The formula to calculate the work done under standard conditions is: \(W_{max} = -nFE\) Where, \(W_{max}\) = useful work obtained \(n\) = number of moles of electrons transferred (2 moles) \(F\) = Faraday's constant (96485 C/mol) \(E\) = standard cell potential We need to find the standard cell potential for the magnesium-hydrogen reaction. Looking up the standard reduction potentials, we find: \(E^0_{H^+}/_{H_2} = 0 V\) \(E^0_{Mg^{2+}}/_{Mg} = -2.37 V\) Since magnesium is being oxidized, the standard potential for magnesium should be reversed: \(-E^0_{Mg}/_{Mg^{2+}} = 2.37 V\) The total standard cell potential is the sum of the two standard potentials: \(E = E^0_{cathode} - E^0_{anode} = -E^0_{Mg}/_{Mg^{2+}} + E^0_{H^+}/_{H_2} = 2.37 V\) Now we can plug this value into the equation for useful work: \(W_{max} = -nFE = -(2)(96485)(2.37)\) \(W_{max} = -455.26 kJ/mol\) Note that the negative sign indicates that the reaction is spontaneous and will produce work.

One possible way to harness the reaction of magnesium with hydrochloric acid to do useful work is to create a galvanic (voltaic) cell, which converts the chemical energy of the redox reaction into electrical energy. The cell would have a magnesium anode and a hydrogen cathode. The reaction between magnesium and hydrochloric acid would produce a voltage across the cell, which can be used to power an electrical circuit or charge a battery.