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A voltaic cell is constructed with two silver-silver chloride electrodes, each of which is based on the following half-reaction: $$ \operatorname{AgCl}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{Cl}^{-}(a q) $$ The two half-cells have \(\left[\mathrm{Cl}^{-}\right]=0.0150 \mathrm{M}\) and \(\left[\mathrm{Cl}^{-}\right]=\) \(2.55 M,\) respectively. (a) Which electrode is the cathode of the cell? (b) What is the standard emf of the cell? (c) What is the cell emf for the concentrations given? (d) For each electrode, predict whether \(\left[\mathrm{Cl}^{-}\right]\) will increase, decrease, or stay the same as the cell operates.

Short Answer

Expert verified
The cathode of the cell has a chloride concentration of \(0.0150\, \mathrm{M}\). The standard emf of the cell is \(0 \,\text{V}\), and the cell emf for the given concentrations is approximately \(+0.218\, \text{V}\). As the cell operates, the chloride concentration will increase at the anode and decrease at the cathode.

Step by step solution

01

(a) Identifying the cathode of the cell

Since the cell is composed of two silver-silver chloride half-cells with different chloride ion concentrations, the electrode with the higher concentration of chloride ions will serve as the anode, and the other electrode will serve as the cathode. In this case, the electrode with \(\left[\mathrm{Cl}^{-}\right]=2.55 M\) will be the anode, and the electrode with \(\left[\mathrm{Cl}^{-}\right]=0.0150 M\) will be the cathode.
02

(b) Calculating the standard emf of the cell

The standard emf of the cell, \(E_{cell}^0\), can be calculated using the half-reaction standard reduction potential, \(E^0 = +0.222 \,\text{V}\) (found in a standard reduction potential table): $$ E_{cell}^0 = E_{cathode}^0 - E_{anode}^0 $$ Since both electrodes involve the same half-reaction, we have: $$ E_{cell}^0 = E^0 - E^0 = 0 \,\text{V} $$
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(c) Calculating the cell emf for the given concentrations

To calculate the cell emf for the given concentrations, we will use the Nernst equation: $$ E_{cell} = E_{cell}^0 - \frac{RT}{nF} \ln{Q} $$ For this cell: - \(R = 8.314 \,\text{J/mol K}\) is the gas constant. - \(T = 298 \,\text{K}\) is the standard temperature. - \(n = 1\) is the number of moles of electrons transferred in the half-reaction. - \(F = 96,485 \,\text{C/mol}\) is the Faraday constant. - \(Q\) is the reaction quotient that can be calculated as \(\frac{[\mathrm{Cl}^{-}]_{anode}}{[\mathrm{Cl}^{-}]_{cathode}}\) in our case. Plugging in the values, we get: $$ E_{cell} = 0\, \text{V} - \frac{8.314 \,\text{J/mol K} \cdot 298 \,\text{K}}{(1)(96485 \, \text{C/mol})} \ln\left(\frac{2.55 \,\text{M}}{0.0150 \,\text{M}}\right) $$ Calculating the cell emf: $$ E_{cell} \approx +0.218 \,\text{V} $$
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(d) Predicting the change in chloride concentration for each electrode

During the operation of the cell, the anode's half-reaction will occur spontaneously (AgCl will lose an electron and produce Ag and Cl-), while the cathode's half-reaction will occur non-spontaneously (Ag and Cl- combine to form AgCl and release an electron). Thus, - For the anode, since Ag and Cl- are being produced, \(\left[\mathrm{Cl}^{-}\right]\) will increase. - For the cathode, since AgCl is being formed, \(\left[\mathrm{Cl}^{-}\right]\) will decrease. In summary, the cathode has a chloride concentration of \(0.0150\, \mathrm{M}\), the standard emf of the cell is \(0 \,\text{V}\), the cell emf for the given concentrations is approximately \(+0.218\, \text{V}\), and the chloride concentration will increase at the anode and decrease at the cathode as the cell operates.

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Key Concepts

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

Nernst Equation
The Nernst Equation is a fundamental concept in electrochemistry used to calculate the electromotive force (emf) of a voltaic cell when it is not under standard conditions. It relates the cell potential to the concentration of ions participating in the electrochemical reaction. This equation is crucial when dealing with voltaic cells with non-standard ion concentrations. The general form of the Nernst Equation is:
  • \( E_{cell} = E_{cell}^0 - \frac{RT}{nF} \ln{Q} \)
Here:
  • \( E_{cell} \) is the cell potential under non-standard conditions.
  • \( E_{cell}^0 \) is the standard cell potential, which is often \( 0 \,\text{V} \) for similar electrodes as in our case.
  • \( R \) is the universal gas constant \( (8.314 \,\text{J/mol K}) \).
  • \( T \) is the temperature in Kelvin \( (298 \,\text{K} \) for standard conditions).
  • \( n \) is the number of electrons exchanged in the half-reaction \( (n = 1) \).
  • \( F \) is the Faraday constant \( (96,485 \,\text{C/mol}) \).
  • \( Q \) is the reaction quotient.
By using the Nernst Equation, we can determine that the cell emf changes due to fluctuating concentrations of the participating ions, which directly affect the overall voltage of the cell.
Electrode Potential
Electrode potential is essential in understanding how voltaic cells work. It represents the tendency of a chemical species to gain or lose electrons. When constructing a voltaic cell, each electrode is assigned an "electrode potential," which is specific for the half-reaction occurring at that electrode.
  • Standard electrode potential \( (E^0) \) is measured under standard conditions, which include a concentration of 1 M, a pressure of 1 atm, and a temperature of 298 K.
  • In our silver-silver chloride voltaic cell, both electrodes have the same standard electrode potential of \(+0.222 \,\text{V} \).
In practical scenarios, the actual electrode potential can differ from the standard potential due to factors like non-standard ion concentrations or temperature changes. The Nernst Equation helps calculate these deviations, essentially allowing the electrodes' potential to vary and influence the overall cell potential. In the exercise, the concentration differences at the electrodes cause the actual potential to differ from the standard potential.
Reaction Quotient
The Reaction Quotient \( (Q) \) is a key concept in electrochemistry that helps determine the direction in which a chemical reaction will proceed. It is especially important in calculating the cell potential of a voltaic cell using the Nernst Equation.
  • The reaction quotient \( Q \) is defined as the concentration of products divided by the concentration of reactants. In the context of a voltaic cell, it can be represented as \( Q = \frac{\text{[Cl}^{-}]_{anode}}{\text{[Cl}^{-}]_{cathode}} \).
  • \( Q \) provides insight into whether the reaction is at equilibrium or if it will proceed forward or backward to reach equilibrium.
When \( Q = 1 \), the reaction is at equilibrium, and no net change occurs. If \( Q < K \) (where \( K \) is the equilibrium constant), the forward reaction is favored, whereas if \( Q > K \), the reverse reaction is favored.
  • In our exercise, differing chloride ion concentrations yield a \( Q \) value that adjusts the cell potential from about \( 0 \, \text{V} \) (standard conditions) to \(+0.218 \, \text{V} \) based on the actual ratio of chloride concentrations.
Understanding \( Q \) is vital for predicting how changes within a voltaic cell, caused by variation in ion concentration, affect its ability to do electrical work.

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Most popular questions from this chapter

Copper corrodes to cuprous oxide, \(\mathrm{Cu}_{2} \mathrm{O},\) or cupric oxide, \(\mathrm{CuO},\) depending on environmental conditions. (a) What is the oxidation state of copper in cuprous oxide? (b) What is the oxidation state of copper in cupric oxide? (c) Copper peroxide is another oxidation product of elemental copper. Suggest a formula for copper peroxide based on its name. (d) Copper(III) oxide is another unusual oxidation product of elemental copper. Suggest a chemical formula for copper(III) oxide.

(a) Write the half-reaction that occurs at an oxygen electrode in acidic aqueous solution when it serves as the cathode of a voltaic cell. (b) Write the half-reaction that occurs at an oxygen electrode in acidic aqueous solution when it serves as the anode of a voltaic cell. (c) What is standard about the standard oxygen electrode?

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In a galvanic cell the cathode is an \(\mathrm{Ag}^{+}(1.00 \mathrm{M}) / \mathrm{Ag}(s)\) half-cell. The anode is a standard hydrogen electrode immersed in a buffer solution containing \(0.10 \mathrm{M}\) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\right)\) and \(0.050 \mathrm{M}\) sodium benzoate \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-} \mathrm{Na}^{+}\right)\). The measured cell voltage is \(1.030 \mathrm{~V}\). What is the \(\mathrm{p} K_{\mathrm{a}}\) of benzoic acid?

A voltaic cell utilizes the following reaction: $$ 2 \mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}^{2+}(a q)+2 \mathrm{H}^{+}(a q) $$ (a) What is the emf of this cell under standard conditions? (b) What is the emf for this cell when \(\left[\mathrm{Fe}^{3+}\right]=3.50 \mathrm{M}, P_{\mathrm{H}_{2}}=\) \(96.3 \mathrm{kPa},\left[\mathrm{Fe}^{2+}\right]=0.0010 \mathrm{M},\) and the \(\mathrm{pH}\) in both half-cells is \(4.00 ?\)

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