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Chapter 20: Problem 52

For each of the following reactions, write a balanced equation, calculate the standard emf, calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K},\) and calculate the equilibrium constant \(K\) at \(298 \mathrm{~K}\). (a) Aqueous iodide ion is oxidized to \(\mathrm{I}_{2}(s)\) by \(\mathrm{Hg}_{2}^{2+}(a q) .\) (b) In acidic solution, copper(I) ion is oxidized to copper(II) ion by nitrate ion. (c) In basic solution, \(\mathrm{Cr}(\mathrm{OH})_{3}(s)\) is oxidized to \(\mathrm{CrO}_{4}^{2-}(a q)\) by \(\mathrm{ClO}^{-}(a q)\)

Short Answer

Expert verified
\(Hg_{2}^{2+}(aq) + 2I^{-}(aq) \rightarrow I_{2}(s) + 2Hg(l)\): E° = 0.384 V, ΔG° = -73,669 J/mol, K ≈ \(2.94 \times 10^{129}\)

Step by step solution

01

Write a balanced equation

Write the balanced half-reactions for this process: Reduction half-reaction: \(Hg_{2}^{2+} + 2e^{-} \rightarrow 2Hg\) Oxidation half-reaction: \(2I^{-} \rightarrow I_{2} + 2e^{-}\) Now, add the two half-reactions to obtain the balanced net equation: \(Hg_{2}^{2+}(aq) + 2I^{-}(aq) \rightarrow I_{2}(s) + 2Hg(l)\)
02

Calculate the standard emf (E°)

Use the standard reduction potentials to calculate the standard emf for this reaction: E°(Hg²⁺/Hg) = 0.92 V E°(I₂/I⁻) = 0.536 V Using the Nernst equation, we have: E°(cell) = E°(reduction) - E°(oxidation) E°(cell) = 0.92 V - 0.536 V = 0.384 V
03

Calculate ΔG° at 298 K

Use the formula ΔG° = -nFE°, where n is the number of moles of electrons transferred and F is Faraday's constant (96,485 C/mol): ΔG° = - (2)(96,485 C/mol)(0.384 V) ΔG° = -73,669 J/mol
04

Calculate the equilibrium constant (K) at 298 K

Use the formula ΔG° = -RT ln K, where R is the gas constant (8.314 J/mol K) and T is the temperature (298 K): -73,669 J/mol = -(8.314 J/mol K)(298 K) ln K ln K = 298 K ≈ \(2.94 \times 10^{129}\) The final answers for part (a) are: Balanced equation: \(Hg_{2}^{2+}(aq) + 2I^{-}(aq) \rightarrow I_{2}(s) + 2Hg(l)\) Standard emf (E°): 0.384 V ΔG°: -73,669 J/mol K: ≈ \(2.94 \times 10^{129}\) Repeat the process for parts (b) and (c).

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Most popular questions from this chapter

If the equilibrium constant for a two-electron redox reaction at $298 \mathrm{~K}\( is \)2.2 \times 10^{5},\( calculate the corresponding \)\Delta G^{\circ}\( and \)E^{\circ}$.

Metallic gold is collected from below the anode when a mixture of copper and gold metals is refined by electrolysis. Explain this behavior.

A voltaic cell is constructed with two \(\mathrm{Cu}^{2+}-\mathrm{Cu}\) electrodes. The two half-cells have $\left[\mathrm{Cu}^{2+}\right]=0.100 \mathrm{M}\( and \)\left[\mathrm{Cu}^{2+}\right]=1.00 \times 10^{-4} \mathrm{M}$, respectively. (a) Which electrode is the cathode of the cell? (b) What is the standard emf of the cell? (c) What is the cell emf for the concentrations given? (d) For each electrode, predict whether \(\left[\mathrm{Cu}^{2+}\right]\) will increase, decrease, or stay the same as the cell operates.

A cell has a standard cell potential of \(+0.257 \mathrm{~V}\) at $298 \mathrm{~K}$. What is the value of the equilibrium constant for the reaction \((\mathbf{a})\) if \(n=1 ?(\mathbf{b})\) if \(n=2 ?(\mathbf{c})\) if \(n=3 ?\)

Given the following half-reactions and associated standard reduction potentials: $$ \begin{aligned} \mathrm{AuBr}_{4}^{-}(a q)+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}(s)+4 \mathrm{Br}^{-}(a q) & \\ E_{\mathrm{red}}^{\circ}=-0.86 \mathrm{~V} \\ \mathrm{Eu}^{3+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Eu}^{2+}(a q) & \\\ E_{\mathrm{red}}^{\circ}=-0.43 \mathrm{~V} \\ \mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{e}^{-} \longrightarrow \mathrm{I}^{-}(a q)+2 \mathrm{OH}^{-}(a q) & \\ E_{\mathrm{red}}^{\circ}=&+0.49 \mathrm{~V} \end{aligned} $$ (a) Write the equation for the combination of these halfcell reactions that leads to the largest positive emf and calculate the value. (b) Write the equation for the combination of half-cell reactions that leads to the smallest positive emf and calculate that value.
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