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Chapter 18: Problem 139

Consider the following reduction potentials: $$\begin{array}{ll}{\mathrm{Co}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Co}} & {\mathscr{E}^{\circ}=1.26 \mathrm{V}} \\ {\mathrm{Co}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Co}} & {\mathscr{E}^{\circ}=-0.28 \mathrm{V}}\end{array}$$ a. When cobalt metal dissolves in 1.0\(M\) nitric acid, will \(\mathrm{Co}^{3+}\) or \(\mathrm{Co}^{2+}\) be the primary product (assuming standard conditions)? b. Is it possible to change the concentration of \(\mathrm{HNO}_{3}\) to get a different result in part a? Concentrated \(\mathrm{HNO}_{3}\) is about 16 \(M\) .

Short Answer

Expert verified
When cobalt metal dissolves in 1.0 M nitric acid under standard conditions, Co³⁺ will be the primary product. It is not possible to change this result by increasing the concentration of HNO₃, as even with concentrated HNO₃ (16 M), Co³⁺ will still be the primary product.

Step by step solution

01

Write the redox reaction with nitric acid

First, let's write the redox reaction that occurs when cobalt metal dissolves in nitric acid: \[2\mathrm{Co} + 6\mathrm{H}^+ + 6\mathrm{NO}_3^- \longrightarrow 2\mathrm{Co}^n+ + 6\mathrm{NO}_2 + 3\mathrm{H}_2\mathrm{O}\] Here, n is either 2 or 3, so we'll have to find out which cobalt ion, Co²⁺ or Co³⁺, will be the primary product.
02

Calculate E₀cell for both Co²⁺ and Co³⁺

Now we'll calculate the E₀cell for both possible reactions: For the Co³⁺/Co reaction: \[E^\circ_{cell} = E^\circ_{Co^{3+}/Co} = 1.26\ \mathrm{V}\] For the Co²⁺/Co reaction: \[E^\circ_{cell} = E^\circ_{Co^{2+}/Co} = -0.28\ \mathrm{V}\]
03

Use the Nernst equation to calculate the reduction potential for both reactions at 1.0 M HNO₃

The Nernst equation is given by: \[E = E^\circ - \frac{RT}{nF} \ln Q\] Let's plug in the reduction potentials, concentrations, and constants for both reactions: For Co³⁺/Co: \[E_{Co^{3+}/Co} = 1.26\ \mathrm{V} - \frac{8.314 \times 298\ \mathrm{K}}{3 \times 96485\ \mathrm{C/mol}} \ln\frac{[Co^{3+}]}{[Co]} = 1.26\ \mathrm{V}\] For Co²⁺/Co: \[E_{Co^{2+}/Co} = -0.28\ \mathrm{V} - \frac{8.314 \times 298\ \mathrm{K}}{2 \times 96485\ \mathrm{C/mol}} \ln\frac{[Co^{2+}]}{[Co]} = -0.28\ \mathrm{V}\] Since the concentration of Co remains constant in both reactions, we can compare them directly. The larger potential is for the Co³⁺/Co reaction, so Co³⁺ is the primary product under standard conditions.
04

Answer to part (a)

When cobalt metal dissolves in 1.0 M nitric acid under standard conditions, Co³⁺ will be the primary product.
05

Repeat the Nernst equation calculations for 16 M HNO₃

Now let's see if the result would change using concentrated HNO₃ (16 M). For Co³⁺/Co, keep its E unchanged: \[E_{Co^{3+}/Co} = 1.26\ \mathrm{V}\] For Co²⁺/Co, use the Nernst equation with 16 M HNO₃: \[E_{Co^{2+}/Co} = -0.28\ \mathrm{V} - \frac{8.314 \times 298\ \mathrm{K}}{2 \times 96485\ \mathrm{C/mol}} \ln\frac{[Co^{2+}]}{[Co]}\] The change in concentration of nitric acid does not affect the Nernst equation for Co²⁺/Co because it doesn't appear in the equation. Therefore, the calculated potential remains -0.28 V.
06

Answer to part (b)

It is not possible to change the result in part (a) by changing the concentration of HNO₃. Even with concentrated HNO₃ (16 M), Co³⁺ will still be the primary product when cobalt metal dissolves in nitric acid.

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Most popular questions from this chapter

Consider the following galvanic cell at \(25^{\circ} \mathrm{C} :\) $$\text { Pt }\left|\mathrm{Cr}^{2+}(0.30 M), \mathrm{Cr}^{3+}(2.0 M)\right|\left|\mathrm{Co}^{2+}(0.20 M)\right| \mathrm{Co}$$ The overall reaction and equilibrium constant value are $$2 \mathrm{Cr}^{2+}(a q)+\mathrm{Co}^{2+}(a q) \rightleftharpoons_{2 \mathrm{Cr}^{3+}}(a q)+\mathrm{Co}(s) \quad K=2.79 \times 10^{7}$$ Calculate the cell potential, \(\mathscr{E},\) for this galvanic cell and \(\Delta G\) for the cell reaction at these conditions.

What is the maximum work that can be obtained from a hydrogen-oxygen fuel cell at standard conditions that produces 1.00 \(\mathrm{kg}\) water at $25^{\circ} \mathrm{C}$ ? Why do we say that this is the maximum work that can be obtained? What are the advantages and disadvantages in using fuel cells rather than the corresponding combustion reactions to produce electricity?

Look up the reduction potential for \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+} .\) Look up the reduction potential for \(\mathrm{Fe}^{2+}\) to Fe. Finally, look up the reduction potential for \(\mathrm{Fe}^{3+}\) to Fe. You should notice that adding the reduction potentials for the first two does not give the potential for the third. Why not? Show how you can use the first two potentials to calculate the third potential.

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 \(\Delta G_{\mathrm{f}}^{\circ} :\) $$\quad\quad\quad \mathrm{H}_{2} \mathrm{O}(l)=-237 \mathrm{kJ} / \mathrm{mol}$$ $$\mathrm{H}_{2}(g)=0.0$$ $$\quad\quad\quad \mathrm{OH}^{-}(a q)=-157 \mathrm{kJ} / \mathrm{mol}$$ $$\quad \mathrm{e}^{-}=0.0$$ Compare this value of \(\mathscr{E}^{\circ}\) with the value of \(\mathscr{E}^{\circ}\) given in Table 18.1

A standard galvanic cell is constructed so that the overall cell reaction is $$2 \mathrm{Al}^{3+}(a q)+3 \mathrm{M}(s) \longrightarrow 3 \mathrm{M}^{2+}(a q)+2 \mathrm{Al}(s)$$ where \(\mathrm{M}\) is an unknown metal. If \(\Delta G^{\circ}=-411 \mathrm{kJ}\) for the overall cell reaction, identify the metal used to construct the standard cell.
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