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Chapter 18: Problem 48

Calculate \(\mathscr{E}^{\circ}\) values for the following cells. Which reactions are spontaneous as written (under standard conditions)? Balance the equations that are not already balanced. Standard reduction potentials are found in Table \(18.1 .\) a. \(\mathrm{H}_{2}(g) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{H}^{-}(a q)\) b. \(\mathrm{Au}^{3+}(a q)+\mathrm{Ag}(s) \longrightarrow \mathrm{Ag}^{+}(a q)+\mathrm{Au}(s)\)

Short Answer

Expert verified
Under standard conditions, the first cell involving hydrogen gas forming hydride and proton is not spontaneous, with a standard cell potential of \(\mathscr{E}^\circ = 0.00 \ V\). On the other hand, the second cell involving gold(III) being reduced in the presence of silver is spontaneous, with a standard cell potential of \(\mathscr{E}^\circ = 0.701 \ V\).

Step by step solution

01

Identify the half-reactions

For each cell, we need to separate the given reaction into the oxidation half-reaction and the reduction half-reaction. a. In the first case, the reaction is \(H_{2}(g) \longrightarrow H^{+}(aq) + H^{-}(aq)\) Here, hydrogen gas is being oxidized to form a proton and a hydride ion \(H^-\). So, we have two half reactions: Oxidation: \(H_2(g) \longrightarrow 2H^+(aq) + 2e^-\) Reduction: \(2e^- + 2H^+(aq) \longrightarrow H_2(g) + 2H^-\) b. In the second case, the reaction is \(Au^{3+}(aq) + Ag(s) \longrightarrow Ag^+(aq) + Au(s)\) Here, silver is being oxidized, and gold(III) is being reduced: Oxidation: \(Ag(s) \longrightarrow Ag^+(aq) + e^-\) Reduction: \(Au^{3+}(aq) + 3e^- \longrightarrow Au(s)\)
02

Find the standard reduction potentials

Now, we need to look up the standard reduction potentials for each half-reaction in Table 18.1: a. \(H_2(g)\) is the standard state, so its standard reduction potential is 0.00 V: \(2e^- + 2H^+(aq) \longrightarrow H_2(g)\), \(E^\circ = 0.00 \ V\) b. From Table 18.1, we get the following standard reduction potentials: \(Ag^+ (aq) + e^- \longrightarrow Ag(s)\), \(E^\circ = 0.799 \ V\) \(Au^{3+}(aq) + 3e^- \longrightarrow Au(s)\), \(E^\circ = 1.50 \ V\)
03

Calculate the cell potentials

We calculate the standard cell potential by adding the standard reduction potentials for the oxidation and reduction half-reactions: a. For the first cell, since the standard reduction potential for hydrogen is 0.00 V, the cell potential will also be 0.00 V: \(\mathscr{E}^\circ = 0.00 \ V\) b. For the second cell, the standard cell potential is given by: \(\mathscr{E}^\circ = E_{reduction}^\circ - E_{oxidation}^\circ\) Here, the reduction half-reaction is that of gold, and the oxidation half-reaction is that of silver. So, we have: \(\mathscr{E}^\circ = (1.50 \ V) - (0.799 \ V) = 0.701 \ V\)
04

Determine the spontaneity of the reactions

A reaction is spontaneous under standard conditions if its standard cell potential is positive: a. For the first cell, we have \(\mathscr{E}^\circ = 0.00 \ V\), which is not positive. Thus, the reaction is not spontaneous. b. For the second cell, we have \(\mathscr{E}^\circ = 0.701 \ V\), which is positive. Thus, the reaction is spontaneous.
05

Summary

Under standard conditions, the first cell (hydrogen gas forming hydride and proton) is not spontaneous, while the second cell (gold(III) being reduced in the presence of silver) is spontaneous. The standard cell potentials for these reactions are 0.00 V and 0.701 V, respectively.

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Most popular questions from this chapter

A galvanic cell is based on the following half-reactions: $$ \begin{aligned} \mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe}(s) & \mathscr{E}^{\circ} &=-0.440 \mathrm{~V} \\ 2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}(g) & \mathscr{E}^{\circ} &=0.000 \mathrm{~V} \end{aligned} $$ where the iron compartment contains an iron electrode and \(\left[\mathrm{Fe}^{2+}\right]=1.00 \times 10^{-3} M\) and the hydrogen compartmentcontains a platinum electrode, \(P_{\mathrm{H}_{2}}=1.00 \mathrm{~atm}\), and a weak acid, HA, at an initial concentration of \(1.00 M .\) If the observed cell potential is \(0.333 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\), calculate the \(K_{\mathrm{a}}\) value for the weak acid HA.

The compound with the formula \(\mathrm{TII}_{3}\) is a black solid. Given the following standard reduction potentials, $$ \begin{aligned} \mathrm{Tl}^{3+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Tl}^{+} & \mathscr{E}^{\circ}=1.25 \mathrm{~V} \\ \mathrm{I}_{3}^{-}+2 \mathrm{e}^{-} \longrightarrow 3 \mathrm{I}^{-} & \mathscr{E}^{\circ}=0.55 \mathrm{~V} \end{aligned} $$ would you formulate this compound as thallium(III) iodide or thallium(I) triiodide?

How can one construct a galvanic cell from two substances, each having a negative standard reduction potential?

Given the following two standard reduction potentials, $$ \begin{array}{ll} \mathrm{M}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{M} & \mathscr{E}^{\circ}=-0.10 \mathrm{~V} \\ \mathrm{M}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{M} & \mathscr{E}^{\circ}=-0.50 \mathrm{~V} \end{array} $$ solve for the standard reduction potential of the half-reaction $$ \mathrm{M}^{3+}+\mathrm{e}^{-} \longrightarrow \mathrm{M}^{2+} $$ (Hint: You must use the extensive property \(\Delta G^{\circ}\) to determine the standard reduction potential.)

When copper reacts with nitric acid, a mixture of \(\mathrm{NO}(g)\) and \(\mathrm{NO}_{2}(g)\) is evolved. The volume ratio of the two product gases depends on the concentration of the nitric acid according to the equilibrium \(2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q)+\mathrm{NO}(g) \rightleftharpoons 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) Consider the following standard reduction potentials at \(25^{\circ} \mathrm{C}\) : \(3 \mathrm{e}^{-}+4 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) \(\mathscr{E}^{\circ}=0.957 \mathrm{~V}\) \(\mathrm{e}^{-}+2 \mathrm{H}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) \(\mathscr{E}^{\circ}=0.775 \mathrm{~V}\) a. Calculate the equilibrium constant for the above reaction. b. What concentration of nitric acid will produce a \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) mixture with only \(0.20 \% \mathrm{NO}_{2}\) (by moles) at \(25^{\circ} \mathrm{C}\) and \(1.00\) atm? Assume that no other gases are present and that the change in acid concentration can be neglected.
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