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Consider a galvanic cell composed of the SHE and a half-cell using the reaction \(\mathrm{Ag}^{+}(a q)+e^{-} \longrightarrow \mathrm{Ag}(s)\). (a) Calculate the standard cell potential. (b) What is the spontaneous cell reaction under standard-state conditions? (c) Calculate the cell potential when \(\left[\mathrm{H}^{+}\right]\) in the hydrogen electrode is changed to (i) \(1.0 \times 10^{-2} M\) and (ii) \(1.0 \times 10^{-5} M\), all other reagents being held at standard- state conditions. (d) Based on this cell arrangement, suggest a design for a pH meter.

Short Answer

Expert verified
(a) 0.80 V. (b) \(2\mathrm{Ag}^{+} + H_2 \rightarrow 2\mathrm{Ag} + 2\mathrm{H}^{+}\). (c) (i) 0.92 V; (ii) 1.10 V. (d) Use cell potential changes to measure pH.

Step by step solution

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01

Identify the Standard Reduction Potentials

First, look up the standard reduction potentials of the half-reactions from the problem. The standard hydrogen electrode (SHE) has a potential of 0.00 V. The silver half-reaction is given in the problem:\[ \mathrm{Ag}^{+} + e^{-} \rightarrow \mathrm{Ag} \] Its standard reduction potential is 0.80 V.
02

Calculate the Standard Cell Potential (Part a)

To find the standard cell potential \( E^{\circ}_{\text{cell}} \), subtract the potential of the anode from the potential of the cathode. The silver electrode serves as the cathode, and the SHE as the anode:\[ E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = 0.80 \, \text{V} - 0.00 \, \text{V} = 0.80 \, \text{V} \]
03

Write the Spontaneous Cell Reaction (Part b)

The overall cell reaction combines the reactions at each electrode: - Anode (SHE): \( 2H^+ + 2e^- \rightarrow H_2 \)- Cathode (Ag): \( 2Ag^+ + 2e^- \rightarrow 2Ag \)Combine these to get the overall spontaneous cell reaction:\[ 2\mathrm{Ag}^{+}(aq) + H_2(g) \rightarrow 2\mathrm{Ag}(s) + 2\mathrm{H}^{+}(aq) \]
04

Calculate the Cell Potential for Different [H⁺] Concentrations (Part c)

We use the Nernst equation to adjust the cell potential for changes in the concentration of \([H^+]\):\[ E = E^{\circ}_{\text{cell}} - \frac{RT}{nF} \ln Q \]Here, \(Q\) is the reaction quotient, \(n=2\) for the number of electrons.*(i) For \([H^+] = 1.0 \times 10^{-2} M\):*\[ Q = \frac{[H^+]^2}{[Ag^+]^2} = \frac{(1.0 \times 10^{-2})^2}{1} = 1.0 \times 10^{-4} \]\[ E = 0.80 - \frac{0.05916}{2} \log{1.0 \times 10^{-4}} = 0.80 + 0.05916 \times 2 = 0.92 \, \text{V} \]*(ii) For \([H^+] = 1.0 \times 10^{-5} M\):*\[ Q = \frac{(1.0 \times 10^{-5})^2}{1} = 1.0 \times 10^{-10} \]\[ E = 0.80 - \frac{0.05916}{2} \log{1.0 \times 10^{-10}} = 0.80 + 0.05916 \times 5 = 1.10 \, \text{V} \]
05

Suggest a Design for a pH Meter (Part d)

A silver electrode coupled with a SHE can be used, where the potential difference is measured. Given that the potential change relates to \([H^+]\), and thus pH, use this setup for sensitive pH measurement: when \([H^+]\) changes, the voltage changes proportionally, allowing the derivation of pH from the measured voltage.

Key Concepts

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

standard cell potential
The standard cell potential, denoted as \(E^{\circ}_{\text{cell}}\), is a measure of the potential difference between two electrodes in a galvanic cell under standard conditions. Standard conditions typically mean all reactants and products are at a concentration of 1 M, a pressure of 1 atm, and a temperature of 25°C (298 K).

In a galvanic cell, the standard cell potential can be calculated by identifying the standard reduction potentials of the cathode and the anode. The standard hydrogen electrode (SHE) is often used as a reference electrode with a potential of 0.00 V. In the context of the silver electrode, it acts as the cathode with a standard reduction potential of 0.80 V.

The formula to calculate the standard cell potential is:
  • \(E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}\)
In this exercise, substituting the values gives \(E^{\circ}_{\text{cell}} = 0.80 \, \text{V}\). This positive potential indicates a spontaneous reaction under standard conditions.
Nernst equation
The Nernst equation allows us to calculate the cell potential under non-standard conditions by taking into account the concentration of the involved ions.

The equation is given by:
  • \(E = E^{\circ}_{\text{cell}} - \frac{RT}{nF} \ln Q\)
where \(E\) is the cell potential, \(R\) is the universal gas constant, \(T\) is the temperature in Kelvin, \(n\) is the number of moles of electrons exchanged, \(F\) is Faraday's constant, and \(Q\) is the reaction quotient.

At 25°C, the Nernst equation often simplifies using base-10 logarithms:
  • \(E = E^{\circ}_{\text{cell}} - \frac{0.05916}{n} \log Q\)
By adjusting the concentration of ions like \([\text{H}^+]\), we can find the resulting cell potential for new conditions, thus illustrating the dynamic nature of electrochemical cells.
spontaneous cell reaction
A spontaneous cell reaction occurs when the cell potential is positive, resulting in a flow of electrons from the anode to the cathode within the galvanic cell.

For the galvanic cell in this problem, which involves the SHE and a silver electrochemical couple, the reactions at each electrode are critical:
  • Anode reaction (SHE): \(2\text{H}^+ + 2e^- \rightarrow \text{H}_2\)
  • Cathode reaction (Ag): \(2\text{Ag}^+ + 2e^- \rightarrow 2\text{Ag}\)
The electrons flow from hydrogen gas is oxidized to protons, and silver ions gain electrons to form metallic silver.

These equations combine to give the overall spontaneous reaction in the cell:
  • \(2\text{Ag}^+(aq) + \text{H}_2(g) \rightarrow 2\text{Ag}(s) + 2\text{H}^+(aq)\)
This overall reaction is energetically favorable when the standard cell potential is positive.
pH meter design
A pH meter is a practical application of galvanic cells and the principles behind cell potential changes with concentration. In this problem's context, a silver electrode alongside a SHE can measure potential differences influenced by changes in \([\text{H}^+]\) concentration, directly linked to pH.

The primary design involves:
  • Using a silver electrode to detect potential changes.
  • The SHE acts as a stable reference.
  • The potential difference (voltage) correlates with pH level, as the Nernst equation shows how cell potential varies with ion concentration.
When \([\text{H}^+]\) changes, the resulting voltage shifts can accurately reflect the sample's pH. Such designs leverage the predictable nature of the Nernst equation to translate minute potential differences into precise pH readings.

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Most popular questions from this chapter

Calculate the standard emf of a cell that uses the \(\mathrm{Mg} / \mathrm{Mg}^{2+}\) and \(\mathrm{Cu} / \mathrm{Cu}^{2+}\) half-cell reactions at \(25^{\circ} \mathrm{C}\). Write the equation for the cell reaction that occurs under standard-state conditions.

Calculate the emf of the following concentration cell at $$ \begin{array}{l} 25^{\circ} \mathrm{C}: \\ \quad \mathrm{Cu}(s)\left|\mathrm{C} \mathrm{u}^{2+}(0.080 \mathrm{M}) \| \mathrm{Cu}^{2+}(1.2 M)\right| \mathrm{Cu}(s) \end{array} $$

Given that: $$ \begin{array}{ll} 2 \mathrm{Hg}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Hg}_{2}^{2+}(a q) & E^{\circ}=0.92 \mathrm{~V} \\\ \mathrm{Hg}_{2}^{2+}(a q)+2 e^{-} \longrightarrow 2 \mathrm{Hg}(l) & E^{\circ}=0.85 \mathrm{~V} \end{array} $$ calculate \(\Delta G^{\circ}\) and \(K\) for the following process at \(25^{\circ} \mathrm{C}:\) $$\mathrm{Hg}_{2}^{2+}(a q) \longrightarrow \mathrm{Hg}^{2+}(a q)+\mathrm{Hg}(l)$$ (The preceding reaction is an example of a disproportionation reaction in which an element in one oxidation state is both oxidized and reduced.)

A construction company is installing an iron culvert (a long cylindrical tube) that is \(40.0 \mathrm{~m}\) long with a radius of \(0.900 \mathrm{~m}\). To prevent corrosion, the culvert must be galvanized. This process is carried out by first passing an iron sheet of appropriate dimensions through an electrolytic cell containing \(\mathrm{Zn}^{2+}\) ions, using graphite as the anode and the iron sheet as the cathode. If the voltage is \(3.26 \mathrm{~V}\), what is the cost of electricity for depositing a layer \(0.200 \mathrm{~mm}\) thick if the efficiency of the process is 95 percent? The electricity rate is \(\$ 0.12\) per kilowatt hour \((\mathrm{kWh})\), where \(1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s}\) and the density of \(\mathrm{Zn}\) is \(7.14 \mathrm{~g} / \mathrm{cm}^{3}\).

For a number of years, it was not clear whether mercury(I) ions existed in solution as \(\mathrm{Hg}^{+}\) or as \(\mathrm{Hg}_{2}^{2+}\). To distinguish between these two possibilities, we could set up the following system: $$ \operatorname{Hg}(l) \mid \text { soln } \mathrm{A} \| \operatorname{soln} \mathrm{B} \mid \operatorname{Hg}(l)$$ where soln A contained 0.263 g mercury(I) nitrate per liter and soln B contained \(2.63 \mathrm{~g}\) mercury(I) nitrate per liter. If the measured emf of such a cell is \(0.0289 \mathrm{~V}\) at \(18^{\circ} \mathrm{C},\) what can you deduce about the nature of the mercury(I) ions?

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