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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

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鈦篯 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.

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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掳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掳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

Discuss the spontaneity of an electrochemical reaction in terms of its standard emf \(\left(E_{\mathrm{cell}}^{\circ}\right)\).

A galvanic cell consists of a silver electrode in contact with \(346 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{AgNO}_{3}\) solution and a magnesium electrode in contact with \(288 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) solution. (a) Calculate \(E\) for the cell at \(25^{\circ} \mathrm{C}\). (b) A current is drawn from the cell until \(1.20 \mathrm{~g}\) of silver has been deposited at the silver electrode. Calculate \(E\) for the cell at this stage of operation.

Calculate the amounts of \(\mathrm{Cu}\) and \(\mathrm{Br}_{2}\) produced in \(1.0 \mathrm{~h}\) at inert electrodes in a solution of \(\mathrm{CuBr}_{2}\) by a current of \(4.50 \mathrm{~A}\).

Oxalic acid \(\left(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) is present in many plants and vegetables. (a) Balance the following equation in acid solution: $$\mathrm{MnO}_{4}^{-}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{Mn}^{2+}+\mathrm{CO}_{2}$$ (b) If a \(1.00-\mathrm{g}\) sample of plant matter requires \(24.0 \mathrm{~mL}\) of \(0.0100 \mathrm{M} \mathrm{KMnO}_{4}\) solution to reach the equivalence point, what is the percent by mass of \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) in the sample?

The cathode reaction in the Leclanch茅 cell is given by: $$ 2 \mathrm{MnO}_{2}(s)+\mathrm{Zn}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{ZnMn}_{2} \mathrm{O}_{4}(s) $$ If a Leclanch茅 cell produces a current of \(0.0050 \mathrm{~A}\), calculate how many hours this current supply will last if there is initially \(4.0 \mathrm{~g}\) of \(\mathrm{MnO}_{2}\) present in the cell. Assume that there is an excess of \(\mathrm{Zn}^{2+}\) ions.The cathode reaction in the Leclanch茅 cell is given by: $$2 \mathrm{MnO}_{2}(s)+\mathrm{Zn}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{ZnMn}_{2} \mathrm{O}_{4}(s)$$ If a Leclanch茅 cell produces a current of \(0.0050 \mathrm{~A}\), calculate how many hours this current supply will last if there is initially \(4.0 \mathrm{~g}\) of \(\mathrm{MnO}_{2}\) present in the cell. Assume that there is an excess of \(\mathrm{Zn}^{2+}\) ions.

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