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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
For the given cells: a. The standard cell potential, E°(Cell), is -2.414 V, which indicates a non-spontaneous reaction under standard conditions. b. The standard cell potential, E°(Cell), is 0.699 V, which indicates a spontaneous reaction under standard conditions.

Step by step solution

01

a. Reduction and Oxidation half-reactions for H2

To find E° for the given cell, we need to identify the two half-reactions: Oxidation half-reaction: \(\mathrm{H}_{2}(g) \longrightarrow 2\mathrm{H}^{+}(a q)+2e^{-}\) Reduction half-reaction: \(2e^{-}+\mathrm{H}^{-}(a q) \longrightarrow \mathrm{H}_{2}(g)\)
02

a. Standard reduction potentials

From Table 18.1, the standard reduction potential for each half-reaction is: E°(Oxidation) = 0 V E°(Reduction) = -2.414 V
03

a. Calculation of E° of the cell

Now, let's calculate the E° of the overall cell reaction: E°(Cell) = E°(Reduction) – E°(Oxidation) = (-2.414) - (0) = -2.414 V Since E°(Cell) < 0, the reaction is non-spontaneous under standard conditions.
04

b. Reduction and Oxidation half-reactions for Au3+ and Ag

To find E° for the given cell, we need to identify the two half-reactions: Oxidation half-reaction: \(\mathrm{Ag}(s) \longrightarrow \mathrm{Ag}^{+}(a q)+e^{-}\) Reduction half-reaction: \(\mathrm{Au}^{3+}(a q)+3e^{-} \longrightarrow \mathrm{Au}(s)\)
05

b. Standard reduction potentials

From Table 18.1, the standard reduction potential for each half-reaction is: E°(Oxidation) = 0.799 V E°(Reduction) = 1.498 V
06

b. Calculation of E° of the cell

Now, let's calculate the E° of the overall cell reaction: E°(Cell) = E°(Reduction) – E°(Oxidation) = (1.498) - (0.799) = 0.699 V Since E°(Cell) > 0, the reaction is spontaneous under standard conditions.

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Most popular questions from this chapter

You have a concentration cell with Cu electrodes and \(\left[\mathrm{Cu}^{2+}\right]=1.00 M(\text { right side })\) and $1.0 \times 10^{-4} M(\text { left side })$ a. Calculate the potential for this cell at \(25^{\circ} \mathrm{C}\) b. The \(\mathrm{Cu}^{2+}\) ion reacts with \(\mathrm{NH}_{3}\) to form \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\) by the following equation: $$\begin{aligned} \mathrm{Cu}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q) & \\\ & K=1.0 \times 10^{13} \end{aligned}$$ Calculate the new cell potential after enough \(\mathrm{NH}_{3}\) is added to the left cell compartment such that at equilibrium \(\left[\mathrm{NH}_{3}\right]=2.0 \mathrm{M} .\)

Balance the following oxidation–reduction reactions that occur in basic solution using the half-reaction method. a. $\mathrm{PO}_{3}^{3-}(a q)+\mathrm{MnO}_{4}^{-}(a q) \rightarrow \mathrm{PO}_{4}^{3-}(a q)+\mathrm{MnO}_{2}(s)$ b. $\operatorname{Mg}(s)+\mathrm{OCl}^{-}(a q) \rightarrow \mathrm{Mg}(\mathrm{OH})_{2}(s)+\mathrm{Cl}^{-}(a q)$ c. $\mathrm{H}_{2} \mathrm{CO}(a q)+\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}+(a q) \rightarrow$ $$\mathrm{HCO}_{3}(a q)+\mathrm{Ag}(s)+\mathrm{NH}_{3}(a q)$$

Consider a galvanic cell based on the following theoretical half-reactions: $$\begin{array}{ll}{\text {}} & { \mathscr{E}^{\circ}(\mathbf{V}) } \\\ \hline{M^{4+}+4 e^{-} \longrightarrow M} & {0.66} \\ {N^{3+}+3 e^{-} \longrightarrow N} & {0.39}\end{array}$$ What is the value of \(\Delta G^{\circ}\) and \(K\) for this cell?

Consider the following galvanic cell: What happens to \(\mathscr{E}\) as the concentration of \(\mathrm{Zn}^{2+}\) is increased? As the concentration of \(\mathrm{Ag}^{+}\) is increased? What happens to \(\mathscr{E}^{\circ}\) in these cases?

An experimental fuel cell has been designed that uses carbon monoxide as fuel. The overall reaction is $$2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)$$ The two half-cell reactions are $$\mathrm{CO}+\mathrm{O}^{2-} \longrightarrow \mathrm{CO}_{2}+2 \mathrm{e}^{-}$$ $$\mathrm{O}_{2}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{O}^{2-}$$ The two half-reactions are carried out in separate compartments connected with a solid mixture of \(\mathrm{CeO}_{2}\) and \(\mathrm{Gd}_{2} \mathrm{O}_{3}\) . Oxide ions can move through this solid at high temperatures (about \(800^{\circ} \mathrm{C} ) . \Delta G\) for the overall reaction at $800^{\circ} \mathrm{C}\( under certain concentration conditions is \)-380 \mathrm{kJ}$ . Calculate the cell potential for this fuel cell at the same temperature and concentration conditions.
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