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Chapter 18: Problem 89

Calculate \(\mathscr{E}^{\circ}\) for the following half-reaction: $$\mathrm{AgI}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{I}^{-}(a q)$$ (Hint: Reference the \(K_{\mathrm{sp}}\) value for AgI and the standard reduction potential for \(\mathrm{Ag}^{+}\).)

Short Answer

Expert verified
The standard EMF (\(\mathscr{E}^{\circ}\)) for the given half-reaction is approximately \(0.536\) V.

Step by step solution

01

Write down the given half-reaction and constants

The given half-reaction is: $$\mathrm{AgI}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{I}^{-}(a q)$$ The constants we need to reference are: 1. \(K_{\mathrm{sp}}\) for AgI 2. Standard reduction potential for \(\mathrm{Ag}^{+}/\mathrm{Ag}\)
02

Write the half-reaction for the reduction potential of \(\mathrm{Ag}^{+}\)

We need the \(\mathrm{Ag}^{+}/\mathrm{Ag}\) half-reaction to determine the standard reduction potential for the given reaction. The half-reaction is: $$\mathrm{Ag}^{+}(aq) + \mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)$$
03

Use the Nernst equation to find the relationship between \(\mathscr{E}^{\circ}\) and \(K_{\mathrm{sp}}\)

The Nernst equation is given as: $$\mathscr{E}=\mathscr{E}^{\circ}-\frac{RT}{nF} \ln{Q}$$ For this problem, the reaction quotient (Q) will be the solubility product constant \(K_{\mathrm{sp}}\). The number of electrons transferred (n) in the half-reaction is 1. At standard conditions, the equation can be simplified to: $$\mathscr{E}^{\circ}=\mathscr{E}_{\mathrm{Ag}^{+}/\mathrm{Ag}}-\frac{RT}{F} \ln{K_{\mathrm{sp}}}$$
04

Calculate \(\mathscr{E}^{\circ}\) using the given constants

At this point, we need the values for \(K_{\mathrm{sp}}\) of AgI and \(\mathscr{E}_{\mathrm{Ag}^{+}/\mathrm{Ag}}\). $$K_{\mathrm{sp}}(\mathrm{AgI})=8.5\times10^{-17}$$ $$\mathscr{E}_{\mathrm{Ag}^{+}/\mathrm{Ag}}=0.7996\,\mathrm{V}$$ Now, we just need to plug in the values in the equation we derived in Step 3: $$\mathscr{E}^{\circ}=0.7996-\frac{RT}{F} \ln{8.5\times10^{-17}}$$ At standard conditions (298 K), \(R=8.314\,\mathrm{J}\,\mathrm{mol}^{-1}\,\mathrm{K}^{-1}\), and \(F=96,485\,\mathrm{C}\,\mathrm{mol}^{-1}\). $$\mathscr{E}^{\circ}=0.7996-\frac{(8.314\,\mathrm{J}\,\mathrm{mol}^{-1}\,\mathrm{K}^{-1})(298\,\mathrm{K})}{(96,485\,\mathrm{C}\,\mathrm{mol}^{-1})} \ln{8.5\times10^{-17}}$$
05

Solve for \(\mathscr{E}^{\circ}\)

Calculate the result: $$\mathscr{E}^{\circ} \approx 0.536\,\mathrm{V}$$ Hence, the standard EMF (\(\mathscr{E}^{\circ}\)) for the given half-reaction is approximately \(0.536\) V.

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Most popular questions from this chapter

Specify which of the following equations represent oxidation–reduction reactions, and indicate the oxidizing agent, the reducing agent, the species being oxidized, and the species being reduced. a. $\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightarrow \mathrm{CO}(g)+3 \mathrm{H}_{2}(g)$ b. $2 \mathrm{AgNO}_{3}(a q)+\mathrm{Cu}(s) \rightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{Ag}(s)$ c. $\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \rightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)$ d. $2 \mathrm{H}^{+}(a q)+2 \mathrm{CrO}_{4}^{2-}(a q) \rightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$

Sketch the galvanic cells based on the following half- reactions. Show the direction of electron flow, show the direction of ion migration through the salt bridge, and identify the cathode and anode. Give the overall balanced equation, and determine \(\mathscr{E}^{\circ}\) for the galvanic cells. Assume that all concentrations are 1.0 \(M\) and that all partial pressures are 1.0 atm. a. $\mathrm{H}_{2} \mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O} \quad \mathscr{E}^{\circ}=1.78 \mathrm{V}$ $\mathrm{O}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{H}_{2} \mathrm{O}_{2} \quad \quad \mathscr{E}^{\circ}=0.68 \mathrm{V}$ b. $\mathrm{Mn}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn} \quad \quad \mathscr{E}^{\circ}=-1.18 \mathrm{V}$ $\mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} \quad \mathscr{E}^{\circ}=-0.036 \mathrm{V}$

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