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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, Co鲁鈦 will be the primary product. It is not possible to change this result by increasing the concentration of HNO鈧, as even with concentrated HNO鈧 (16 M), Co鲁鈦 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, Co虏鈦 or Co鲁鈦, will be the primary product.
02

Calculate E鈧cell for both Co虏鈦 and Co鲁鈦

Now we'll calculate the E鈧cell for both possible reactions: For the Co鲁鈦/Co reaction: \[E^\circ_{cell} = E^\circ_{Co^{3+}/Co} = 1.26\ \mathrm{V}\] For the Co虏鈦/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 HNO鈧

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 Co鲁鈦/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 Co虏鈦/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 Co鲁鈦/Co reaction, so Co鲁鈦 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, Co鲁鈦 will be the primary product.
05

Repeat the Nernst equation calculations for 16 M HNO鈧

Now let's see if the result would change using concentrated HNO鈧 (16 M). For Co鲁鈦/Co, keep its E unchanged: \[E_{Co^{3+}/Co} = 1.26\ \mathrm{V}\] For Co虏鈦/Co, use the Nernst equation with 16 M HNO鈧: \[E_{Co^{2+}/Co} = -0.28\ \mathrm{V} - \frac{8.314 \times 298\ \mathrm{K}}{2 \times 96485\ \mathrm{C/mol}} \ln\frac{[Co^{2+}]}{[Co]}\] The change in concentration of nitric acid does not affect the Nernst equation for Co虏鈦/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 HNO鈧. Even with concentrated HNO鈧 (16 M), Co鲁鈦 will still be the primary product when cobalt metal dissolves in nitric acid.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reduction Potentials
Reduction potentials are essential in electrochemistry, helping to predict the direction of redox reactions. In simple terms, a reduction potential, denoted as \(E^\circ\), measures the tendency of a chemical species to be reduced. It is often given in volts (V).

A positive standard reduction potential indicates a greater tendency to gain electrons, thus being reduced. Conversely, a negative value suggests a preference to lose electrons, or oxidize. Hence, in our cobalt example,
  • The reaction \(\text{Co}^{3+} + 3\text{e}^- \rightarrow \text{Co}\) has a reduction potential of \(1.26\,\text{V}\).
  • The reaction \(\text{Co}^{2+} + 2\text{e}^- \rightarrow \text{Co}\) has a reduction potential of \(-0.28\,\text{V}\).
This means that \(\text{Co}^{3+}\) has a stronger tendency to be reduced to cobalt metal than \(\text{Co}^{2+}\) does. As a result, when cobalt metal interacts with nitric acid, \(\text{Co}^{3+}\) forms more readily.

Knowing reduction potentials allows us to predict which species will behave as oxidizing agents (accepting electrons) and which will behave as reducing agents (donating electrons).
Nernst Equation
The Nernst equation is a powerful tool in electrochemistry used to calculate the cell potential under non-standard conditions. It accounts for the effect of concentration on cell potentials. The equation is expressed as:\[ E = E^\circ - \frac{RT}{nF} \ln Q \] where:
  • \(E\) is the cell potential under non-standard conditions.
  • \(E^\circ\) is the standard cell potential.
  • \(R\) is the universal gas constant (8.314 J/mol K).
  • \(T\) is the temperature in Kelvin.
  • \(n\) is the number of moles of electrons exchanged.
  • \(F\) is the Faraday constant (96485 C/mol).
  • \(Q\) is the reaction quotient, representing the ratio of product concentrations to reactant concentrations.

In the given problem, the Nernst equation demonstrates that changing the concentration of \(\text{HNO}_3\) doesn鈥檛 affect the potential for the \(\text{Co}^{2+}/\text{Co}\) couple because it doesn鈥檛 appear in the reaction quotient for this reaction. Consequently, even with the increased concentration of \(\text{HNO}_3\), the potential for \(\text{Co}^{2+}/\text{Co}\) remains constant, confirming \(\text{Co}^{3+}\) as the dominant product.
Redox Reactions
Redox reactions, short for reduction-oxidation reactions, are chemical processes where electrons are transferred between substances. They play a crucial role in energy production and resource cycling in nature and technology.Key aspects of redox reactions include:
  • Reduction: The gain of electrons by a molecule, atom, or ion.
  • Oxidation: The loss of electrons.
  • Oxidizing agent: A substance that gains electrons (is reduced itself).
  • Reducing agent: A substance that loses electrons (is oxidized itself).

In the context of the cobalt and nitric acid reaction, cobalt metal acts as a reducing agent. When it dissolves in nitric acid, cobalt ions (\(\text{Co}^{3+}\) and \(\text{Co}^{2+}\)) are generated. The net transfer of electrons determines which cobalt species predominates.Identifying and balancing redox reactions require understanding the electron flow, which is depicted through oxidation states and half-equations. This aids in predicting reaction spontaneity and product formation.

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Most popular questions from this chapter

A solution at \(25^{\circ} \mathrm{C}\) contains 1.0\(M \mathrm{Cu}^{2+}\) and \(1.0 \times 10^{-4} M\) \(\mathrm{Ag}^{+}\). Which metal will plate out first as the voltage is gradually increased when this solution is electrolyzed? (Hint: Use the Nernst equation to calculate \(\mathscr{E}\) for each half-reaction.)

What reaction will take place at the cathode and the anode when each of the following is electrolyzed? (Assume standard conditions.) a. 1.0 \(M \mathrm{KF}\) solution b. 1.0\(M \mathrm{CuCl}_{2}\) solution c. 1.0 \( M \mathrm{MgI}_{2}\) solution

An electrochemical cell consists of a standard hydrogen electrode and a copper metal electrode. If the copper electrode is placed in a solution of 0.10\(M \mathrm{NaOH}\) that is saturated with \(\mathrm{Cu}(\mathrm{OH})_{2},\) what is the cell potential at \(25^{\circ} \mathrm{C} ?\left[\text { For } \mathrm{Cu}(\mathrm{OH})_{2}\right.\) \(K_{\mathrm{sp}}=1.6 \times 10^{-19} . ]\)

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. \(\mathrm{Au}^{3+}(a q)+\mathrm{Ag}(s) \longrightarrow \mathrm{Ag}^{+}(a q)+\mathrm{Au}(s)\)

The equation \(\Delta G^{\circ}=-\mathrm{nF} \mathscr{E}^{\circ}\) also can be applied to half-reactions. Use standard reduction potentials to estimate \(\Delta G_{\mathrm{f}}^{\circ}\) for \(\mathrm{Fe}^{2+}(a q)\) and \(\mathrm{Fe}^{3+}(a q) .\left(\Delta G_{\mathrm{f}}^{\circ} \text { for } \mathrm{e}^{-}=0 .\right)\)

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