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Chapter 18: Problem 60

Consider the reaction $$ \mathrm{S}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{Ag}(s)+2 \mathrm{Br}^{-}(a q) \longrightarrow 2 \mathrm{AgBr}(s)+\mathrm{H}_{2} \mathrm{~S}(a q) $$ At what \(\mathrm{pH}\) is the voltage zero if all other species are at standard concentrations?

Short Answer

Expert verified
\( \mathrm{S}(s) + 2\,\mathrm{AgBr}(s) \longrightarrow \mathrm{H}_2\,\mathrm{S}(a q) + 2\,\mathrm{Ag}(s) + 2\,\mathrm{Br}^{-}(a q) \) (All species are at standard concentrations except for H鈦� ions) Answer: To find the pH at which the potential difference becomes zero, we will use the Nernst equation and two half-reactions. The oxidation half-reaction is \(\mathrm{S}(s) + 2\,\mathrm{H}^{+}(a q) \longrightarrow \mathrm{H}_2\,\mathrm{S}(a q) + 2\,e^鈭抃) and the reduction half-reaction is \(2\,\mathrm{Ag}(s) + 2\,\mathrm{Br}^{-}(a q) + 2\,e^鈭� \longrightarrow 2\,\mathrm{AgBr}(s)\). Applying the Nernst equation to both half-reactions and setting their difference to zero, we can find the H鈦� concentration from the equation: $$ [\mathrm{H}^{+}] = \sqrt{\exp{\left(-\dfrac{2 \cdot 96485\,\text{C/mol} \cdot E^{\circ}_{red}}{8.314\,\text{J/mol路K} \cdot 298\,\text{K}}\right)}} $$ Finally, we can calculate the pH using the equation: $$ \text{pH} = -\log_{10}{[\mathrm{H}^{+}]} $$

Step by step solution

01

Identify the half-reactions

By examining the given chemical reaction, we can identify the two half-reactions as follows: 1. Oxidation half-reaction: \(\mathrm{S}(s) + 2\,\mathrm{H}^{+}(a q) \longrightarrow \mathrm{H}_2\,\mathrm{S}(a q) + 2\,e^鈭抃) 2. Reduction half-reaction: \(2\,\mathrm{Ag}(s) + 2\,\mathrm{Br}^{-}(a q) + 2\,e^鈭� \longrightarrow 2\,\mathrm{AgBr}(s)\)
02

Write the Nernst equation

The Nernst equation gives the potential of a half-cell as a function of concentrations: $$ E = E^{\circ} - \dfrac{RT}{nF} \ln{Q} $$ where \(E\) is the potential, \(E^{\circ}\) is the standard potential, \(R\) is the gas constant (8.314 J/mol路K), \(T\) is the temperature in Kelvin, \(n\) is the number of electrons transferred, \(F\) is the Faraday constant (96485 C/mol), and \(Q\) is the reaction quotient. For this problem, we will assume a temperature of 25掳C (298 K).
03

Apply the Nernst equation to the oxidation half-reaction

For the oxidation half-reaction, we have \(n = 2\). Also, \(E^{\circ}_{ox} = 0\) since the oxidation of sulfur (S) in its standard state has zero potential. Thus, the Nernst equation for the oxidation half-reaction is: $$ E_{ox} = 0 - \dfrac{8.314\,\text{J/mol路K} \cdot 298\,\text{K}}{2 \cdot 96485\,\text{C/mol}} \ln{\dfrac{[\mathrm{H}_2\,\mathrm{S}]}{[\mathrm{S}] [\mathrm{H}^{+}]^2}} $$ Since all species are at standard concentrations (1 M), except H鈦�, we have: $$ E_{ox} = - \dfrac{8.314\,\text{J/mol路K} \cdot 298\,\text{K}}{2 \cdot 96485\,\text{C/mol}} \ln(([\mathrm{H}^{+}]^{-2})) $$
04

Apply the Nernst equation to the reduction half-reaction

For the reduction half-reaction, we also have \(n = 2\). Assuming the standard potential for the reduction of Ag鈦� to Ag is \(E^{\circ}_{red}\). The Nernst equation for the reduction half-reaction is: $$ E_{red} = E^{\circ}_{red} - \dfrac{8.314\,\text{J/mol路K} \cdot 298\,\text{K}}{2 \cdot 96485\,\text{C/mol}} \ln{\dfrac{[\mathrm{Ag}]^2}{[\mathrm{Ag}^+]^2 [\mathrm{Br}^{-}]^2}} $$ Since all species are at standard concentrations (1 M), except H鈦�, we have: $$ E_{red} = E^{\circ}_{red} $$
05

Find the total potential and set it to zero

The total potential of the reaction is the difference between the reduction and oxidation potentials: $$ E_{total} = E_{red} - E_{ox} $$ Setting \(E_{total} = 0\), we get: $$ E^{\circ}_{red} = E_{ox} = - \dfrac{8.314\,\text{J/mol路K} \cdot 298\,\text{K}}{2 \cdot 96485\,\text{C/mol}} \ln(([\mathrm{H}^{+}]^{-2})) $$
06

Solve for the pH

To solve for the pH (which is \(-\log_{10}{[\mathrm{H}^{+}]}\)), we need to find the concentration of H鈦� from the above equation. Rearrange the equation and take the natural logarithm of both sides, then solve for \([\mathrm{H}^{+}]\): $$ [\mathrm{H}^{+}] = \sqrt{\exp{\left(-\dfrac{2 \cdot 96485\,\text{C/mol} \cdot E^{\circ}_{red}}{8.314\,\text{J/mol路K} \cdot 298\,\text{K}}\right)}} $$ Finally, take the negative logarithm of the H鈦� concentration to find the pH: $$ \text{pH} = -\log_{10}{[\mathrm{H}^{+}]} $$

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Key Concepts

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

Chemical Reactions
Chemical reactions are processes in which substances, known as reactants, are transformed into different substances called products. During these transformations, the chemical bonds between atoms are broken and new ones are formed, often releasing or absorbing energy in the process. The balance of atoms is maintained, as dictated by the law of conservation of mass, meaning the number of atoms of each element in the reactants must equal the number in the products.

Understanding a chemical reaction involves identifying the reactants and products and knowing how they interact and convert from one form to another. In the given exercise, sulfur (S), hydrogen ions (H+), silver (Ag), and bromide ions (Br-) come together to produce silver bromide (AgBr) and hydrogen sulfide (H2S) in aqueous solution.
Half-Reactions
Half-reactions are a way of breaking down complex chemical reactions, especially redox reactions, into simpler parts that help in analyzing the movement of electrons. A redox reaction can generally be separated into two half-reactions: oxidation, where there is a loss of electrons, and reduction, where there is a gain of electrons.

In the exercise, the oxidation half-reaction shows sulfur being oxidized in the presence of hydrogen ions, while the reduction half-reaction involves silver being reduced by gaining electrons in the presence of bromide ions. Writing the half-reactions separately allows for a clearer understanding of the electron transfer during the chemical process. An important staple in electrochemistry, half-reactions are also integral when utilizing the Nernst equation.
Electrochemistry
Electrochemistry is the branch of chemistry that deals with the interrelation of electrical currents and chemical reactions. It encompasses the study of both the movement of electrons in a chemical reaction and the generation of electricity through chemical changes in cells and batteries.

The Nernst equation, which appears in the step-by-step solution, is a fundamental equation in electrochemistry that relates the reducing and oxidizing potential of a half-cell to the concentration of ions in solution. It is especially useful for determining the cell potential under non-standard conditions, which are often the cases encountered in practical situations.

Applying the Nernst Equation

When considering how to apply the Nernst equation, it's critical to identify the number of electrons transferred, account for the reaction conditions such as all species being at standard concentrations except H鈦� ions, and then solve the equation to find the pH at which the net cell voltage is zero. The understanding and application of the Nernst equation provide invaluable insights into the behavior of electrochemical systems, including those that involve pH-dependent reactions.

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Most popular questions from this chapter

Use the following half-equations to write three spontaneous reactions. Justify your answers by calculating \(E^{\circ}\) for the cells. (1) \(\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q)+5 e^{-} \longrightarrow \mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}\) \(E^{\circ}=+1.512 \mathrm{~V}\) (2) \(\mathrm{O}_{2}(\mathrm{~g})+4 \mathrm{H}^{+}(\mathrm{aq})+4 e^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O} \quad E^{\circ}=+1.229 \mathrm{~V}\) (3) \(\mathrm{Co}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Co}(s) \quad E^{\circ}=-0.282 \mathrm{~V}\)

Consider a voltaic cell at \(25^{\circ} \mathrm{C}\) in which the following reaction takes place. $$ 3 \mathrm{O}_{2}(g)+4 \mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 4 \mathrm{NO}_{3}^{-}(a q)+4 \mathrm{H}^{+}(a q) $$ (a) Calculate \(E^{\circ}\) (b) Write the Nernst equation for the cell. (c) Calculate \(E\) under the following conditions: \(\left[\mathrm{NO}_{3}{ }^{-}\right]=0.750 M\), \(P_{\mathrm{NO}}=0.993 \mathrm{~atm}, P_{\mathrm{O}_{2}}=0.515 \mathrm{~atm}, \mathrm{pH}=2.85\)

For the following half-reactions, answer the questions below. $$ \begin{array}{cc} \mathrm{Co}^{3+}(a q)+e^{-} \longrightarrow \mathrm{Co}^{2+}(a q) & E^{\circ}=+1.953 \mathrm{~V} \\ \mathrm{Fe}^{3+}(a q)+e^{-} \longrightarrow \mathrm{Fe}^{2+}(a q) & E^{\circ}=+0.769 \mathrm{~V} \\ \mathrm{I}_{2}(a q)+2 e^{-} \longrightarrow 2 \mathrm{I}^{-}(a q) & E^{o}=+0.534 \mathrm{~V} \\ \mathrm{~Pb}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Pb}(s) & E^{\circ}=-0.127 \mathrm{~V} \\ \mathrm{Cd}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Cd}(s) & E^{\circ}=-0.402 \mathrm{~V} \\ \mathrm{Mn}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Mn}(s) & E^{\circ}=-1.182 \mathrm{~V} \end{array} $$ (a) Which is the weakest reducing agent? (b) Which is the strongest reducing agent? (c) Which is the strongest oxidizing agent? (d) Which is the weakest oxidizing agent? (e) Will \(\mathrm{Pb}(s)\) reduce \(\mathrm{Fe}^{3+}(a q)\) to \(\mathrm{Fe}^{2+}(a q) ?\) (f) Will \(\mathrm{I}^{-}(a q)\) reduce \(\mathrm{Pb}^{2+}(a q)\) to \(\mathrm{Pb}(s) ?\) (g) Which ion(s) can be reduced by \(\mathrm{Pb}(s)\) ? (h) Which if any metal(s) can be oxidized by \(\mathrm{Fe}^{3+}(a q)\) ?

Consider the following reaction carried out at \(1000^{\circ} \mathrm{C}\). $$ \mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) $$ Assuming that all gases are at \(1.00 \mathrm{~atm}\), calculate the voltage produced at the given conditions. (Use Appendix 1 and assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not change with an increase in temperature.)

Which of the following species will react with \(1 \mathrm{M} \mathrm{HNO}_{3}\) ? (a) \(\mathrm{I}^{-}\) (b) Fe (c) \(\mathrm{Ag}\) (d) \(\mathrm{Pb}\)
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