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Chapter 17: Problem 53

Estimate \(\mathscr{E}^{\circ}\) for the half-reaction $$ 2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}+2 \mathrm{OH}^{-} $$ given the following values of \(\Delta G_{\mathrm{f}}^{\circ}\) $$ \begin{aligned} \mathrm{H}_{2} \mathrm{O}(l) &=-237 \mathrm{kJ} / \mathrm{mol} \\ \mathrm{H}_{2}(g) &=0.0 \\ \mathrm{OH}^{-}(a q) &=-157 \mathrm{kJ} / \mathrm{mol} \\ \mathrm{e}^{-} &=0.0 \end{aligned} $$ Compare this value of \(\mathscr{E}^{\circ}\) with the value of \(\mathscr{E}^{\circ}\) given in Table 17-1.

Short Answer

Expert verified
The standard electrode potential (\(\mathscr{E}^{\circ}\)) for the given half-reaction can be estimated using the provided Gibbs free energies of formation. For the balanced half-reaction, the number of transferred electrons is 2. The standard change of Gibbs free energy is calculated as -314 kJ/mol. Using the relationship between \(\mathscr{E}^{\circ}\) and \(\Delta G_{\mathrm{r}}^{\circ}\), we find that \(\mathscr{E}^{\circ} \approx +1.63\, \mathrm{V}\). This value is higher by 0.4 V than the value given in Table 17-1 (1.23 V), indicating that our estimate might not be as accurate but still provides a reasonable approximation of the standard electrode potential for the given half-reaction.

Step by step solution

01

Identify the number of electrons transferred

The balanced half-reaction is already given: \(2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}+2 \mathrm{OH}^{-}\) Here, two electrons are transferred, therefore: $$ n = 2 $$
02

Calculate the standard change of Gibbs free energy

To find the standard change of Gibbs free energy (\(\Delta G_{\mathrm{r}}^{\circ}\)), we will subtract the Gibbs free energy of the reactants from that of the products: $$ \Delta G_{\mathrm{r}}^{\circ} = [\Delta G^{\circ}_{\mathrm{H}_2} + 2\Delta G^{\circ}_{\mathrm{OH}^{-}}] - [2\Delta G^{\circ}_{\mathrm{H}_2 \mathrm{O}} + 0] $$ Substitute the provided values: $$ \Delta G_{\mathrm{r}}^{\circ} = [(0 \,\mathrm{kJ/mol}) + 2(-157 \,\mathrm{kJ/mol})] - [2(-237 \,\mathrm{kJ/mol}) + 0] = -314\,\mathrm{kJ/mol} $$
03

Calculate the standard electrode potential

Now, we will use the relationship between \(\mathscr{E}^{\circ}\) and \(\Delta G_{\mathrm{r}}^{\circ}\), $$ \Delta G_{\mathrm{r}}^{\circ} = -nFE_{\text{cell}}^{\circ} $$ Rearrange the equation to get the standard electrode potential: $$ \mathscr{E}^{\circ} = -\frac{\Delta G_{\mathrm{r}}^{\circ}}{nF} $$ Using the value of Faraday's constant F = 96,485 C/mol, $$ \mathscr{E}^{\circ} = -\frac{-314 \,\mathrm{kJ/mol}}{2 \times 96485 \,\mathrm{C/mol}} \times \frac{1000\,\mathrm{J}}{\mathrm{kJ}} \approx +1.63\, \mathrm{V} $$ Comparing this result with the value of \(\mathscr{E}^{\circ}\) given in Table 17-1 (1.23 V), we see that the calculated value is higher by 0.4 V, indicating that our estimate might not be as accurate but still gives a decent idea of the standard electrode potential for the given half-reaction.

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Most popular questions from this chapter

Consider the following electrochemical cell: a. If silver metal is a product of the reaction, is the cell a galvanic cell or electrolytic cell? Label the cathode and anode, and describe the direction of the electron flow. b. If copper metal is a product of the reaction, is the cell a galvanic cell or electrolytic cell? Label the cathode and anode, and describe the direction of the electron flow. c. If the above cell is a galvanic cell, determine the standard cell potential. d. If the above cell is an electrolytic cell, determine the minimum external potential that must be applied to cause the reaction to occur.

Consider the cell described below: $$ \mathrm{Zn}\left|\mathrm{Zn}^{2+}(1.00 M)\right|\left|\mathrm{Cu}^{2+}(1.00 M)\right| \mathrm{Cu} $$ Calculate the cell potential after the reaction has operated long enough for the \(\left[\mathrm{Zn}^{2+}\right]\) to have changed by \(0.20 \mathrm{mol} / \mathrm{L}\). (Assume \(T=25^{\circ} \mathrm{C} .\) )

a. Draw this cell under standard conditions, labeling the anode, the cathode, the direction of electron flow, and the concentrations, as appropriate. b. When enough \(\mathrm{NaCl}(s)\) is added to the compartment containing gold to make the \(\left[\mathrm{Cl}^{-}\right]=0.10 \mathrm{M},\) the cell potential is observed to be 0.31 V. Assume that \(\mathrm{Au}^{3+}\) is reduced and assume that the reaction in the compartment containing gold is $$ \mathrm{Au}^{3+}(a q)+4 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{AuCl}_{4}^{-}(a q) $$ Calculate the value of \(K\) for this reaction at \(25^{\circ} \mathrm{C}\).

The overall reaction in the lead storage battery is $$ \begin{array}{r} \mathrm{Pb}(s)+\mathrm{PbO}_{2}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{HSO}_{4}^{-}(a q) \longrightarrow \\ 2 \mathrm{PbSO}_{4}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \end{array} $$ Calculate \(\mathscr{E}\) at \(25^{\circ} \mathrm{C}\) for this battery when \(\left[\mathrm{H}_{2} \mathrm{SO}_{4}\right]=4.5 \mathrm{M}\) that is, \(\left[\mathrm{H}^{+}\right]=\left[\mathrm{HSO}_{4}^{-}\right]=4.5 \mathrm{M} .\) At \(25^{\circ} \mathrm{C}, 8^{\circ}=2.04 \mathrm{V}\) for the lead storage battery.

Explain the following relationships: \(\Delta G\) and \(w,\) cell potential and \(w,\) cell potential and \(\Delta G,\) cell potential and \(Q .\) Using these relationships, explain how you could make a cell in which both electrodes are the same metal and both solutions contain the same compound, but at different concentrations. Why does such a cell run spontaneously?
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