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\mathrm{~A}\( voltaic cell utilizes the following reaction: $$ \mathrm{Al}(\mathrm{s})+3 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Al}^{3}(a q)+3 \mathrm{Ag}(s) $$ What is the effect on the cell emf of each of the following changes? (a) Water is added to the anode half-cell, diluting the solution. (b) The size of the aluminum electrode is increased. (c) \)\mathrm{A}\( solution of \)\mathrm{AgNO}_{3}\( is added to the cathode half-cell, increasing the quantity of \)\mathrm{Ag}^{+}\(but not changing its concentration. (d) \)\mathrm{HCl}\( is added to the \)\mathrm{AgNO}_{3}\( solution, precipitating some of the \)\mathrm{Ag}^{+}\(as \)\mathrm{AgCl}$.

Short Answer

Expert verified
In summary, (a) diluting the solution at the anode half-cell will increase the cell emf as the concentration of Al³⁺ decreases; (b) increasing the size of the aluminum electrode will not affect the cell emf as it does not change the concentrations of reactants and products; (c) adding AgNO₃ to the cathode half-cell without changing Ag⁺ concentration will not affect the cell emf; and (d) adding HCl to the AgNO₃ solution and precipitating some Ag⁺ as AgCl will decrease the cell emf due to a decrease in Ag⁺ concentration.

Step by step solution

01

(a) Water is added to the anode half-cell, diluting the solution.

When the water is added to the anode half-cell, the concentration of the Al³⁺ ions in the solution would decrease. The cell emf is directly related to the concentration of the ions participating in the redox reaction, as described by the Nernst equation: \[ E_{cell} = E_{cell}^0 - \frac{RT}{nF} \ln \frac{[\text{Al}^{3+}]}{[\text{Ag}^+]} \] Where \(E_{cell}^0\) is the standard emf, R is the gas constant, T is the temperature, n is the number of electrons transferred per mole of reactants, F is the Faraday constant, and the concentrations are the equilibrium concentrations of the redox reactants and products. Since the concentration of Al³⁺ ions decreases, the cell emf will increase because the reaction will proceed more spontaneously to maintain the equilibrium.
02

(b) The size of the aluminum electrode is increased.

Increasing the size of the aluminum electrode does not affect the overall cell emf as it does not change the concentrations of the reacting species in the redox reaction. The cell emf will remain the same because it primarily depends on the concentrations of the reactants and products.
03

(c) A solution of \(\mathrm{AgNO}_{3}\) is added to the cathode half-cell.

Adding \(\mathrm{AgNO}_{3}\) solution to the cathode half-cell will increase the amount of \(\mathrm{Ag}^{+}\) ions in the solution but not their concentration. Since the overall concentration of the \(\mathrm{Ag}^{+}\) ions remains the same, the cell emf will not change, as it only depends on the concentration of the reactants and products.
04

(d) \(\mathrm{HCl}\) is added to the \(\mathrm{AgNO}_{3}\) solution, precipitating some of the \(\mathrm{Ag}^{+}\)as \(\mathrm{AgCl}\).

Adding \(\mathrm{HCl}\) to the \(\mathrm{AgNO}_{3}\) solution will precipitate some of the \(\mathrm{Ag}^{+}\) ions as \(\mathrm{AgCl}\). This would result in a decrease in the concentration of \(\mathrm{Ag}^{+}\) ions in the solution, which would affect the cell emf, as it depends on the concentration of the participating ions. According to the Nernst equation, the decrease in \(\mathrm{Ag}^{+}\) concentration will lead to a decrease in the cell emf, as the reaction will be less spontaneous.

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Key Concepts

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

Cell EMF
The electromotive force (emf) of a cell, also known as the cell voltage, is the energy that drives the electric current through the circuit. It arises from the potential difference between the two electrodes in an electrochemical cell. This potential difference results from the chemical reactions occurring in the cell's two half-cells.
The cell emf can be calculated under standard conditions (25°C, 1M concentrations, and 1 atm pressure) using the standard electrode potentials of the half-reactions. However, under non-standard conditions, the cell emf is affected by several factors such as concentration changes, temperature, and pressure alterations.
Understanding the cell emf is essential in determining the efficiency and spontaneity of the redox reactions in electrochemical cells.
Nernst Equation
The Nernst equation allows us to calculate the cell emf under non-standard conditions by accounting for the concentrations of the ions involved in the redox reaction. The equation is expressed as:
\[ E_{cell} = E_{cell}^0 - \frac{RT}{nF} \ln Q \]
where:
  • \( E_{cell} \): cell potential under non-standard conditions
  • \( E_{cell}^0 \): standard cell potential
  • \( R \): universal gas constant (8.314 J/mol K)
  • \( T \): temperature in Kelvin
  • \( n \): number of moles of electrons transferred
  • \( F \): Faraday's constant (96,485 C/mol)
  • \( Q \): reaction quotient, which is the ratio of concentrations of products over reactants
This equation helps us understand how the emf is influenced by ion concentrations. For instance, dilution of solutions (as seen when water is added) affects \( Q \), thereby altering the cell emf. It indicates the spontaneous nature of the reaction, directly relating the cell's emf to its reaction conditions.
Redox Reaction
In electrochemical cells, the redox reaction is the heart of the process. It involves two key components: oxidation and reduction.
Oxidation occurs at the anode, where electrons are lost, and reduction happens at the cathode, where electrons are gained. The typical reaction, such as the one given in the exercise, illustrates aluminum getting oxidized (\( \mathrm{Al}(s) \) to \( \mathrm{Al}^{3+}(aq) \)) while silver ions are reduced (\( \mathrm{Ag}^{+}(aq) \) to \( \mathrm{Ag}(s) \)).
The overall redox reaction powers the flow of electrons through the external circuit, creating the electricity generated by the cell. By understanding which species are oxidized and which are reduced, we can predict how changes, such as the concentration of ions, affect the overall cell function and cell emf.
Concentration Effect in Electrochemistry
Concentration plays a pivotal role in the behavior of electrochemical cells, particularly influencing their emf and overall function. According to Le Chatelier's Principle, changes in concentration can shift the equilibrium position of a reaction, affecting the cell's operation.
In cell operations, adding water to dilute the solution at the anode or adding reagents to one half-cell to increase the quantity of ions (without changing concentration) might seem negligible under some perspectives but are crucial in others. They can vastly influence the cell's emf, as indicated by the Nernst equation. For example, decreasing the concentration of ions at the anode will increase the cell emf, making the reaction more spontaneous.
On the contrary, precipitating ions out of the solution, as in the addition of HCl triggering the formation of AgCl, directly reduces the concentration of active ions, diminishing cell emf. Therefore, attention to concentration details is imperative to understanding how real-life electrochemical cells operate effectively.

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Most popular questions from this chapter

A plumber's handbook states that you should not connect a copper pipe directly to a steel pipe because electrochemical reactions between the two metals will cause corrosion. The handbook recommends you use instead an insulating fitting to connect them. What spontaneous redox reaction(s) might cause the corrosion? Justify your answer with standard emf calculations.

Using the standard reduction potentials listed in Appendix E, calculate the equilibrium constant for each of the following reactions at \(298 \mathrm{~K}\) - (a) \(\mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Cu}^{2}+(a q)+2 \mathrm{Ag}(s)\) (b) \(3 \mathrm{Ce}^{4+}(a q)+\mathrm{Bi}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 3 \mathrm{Ce}^{3+}(a q)+\) \(\mathrm{BiO}^{+}(a q)+2 \mathrm{H}^{+}(a q)\) (c) \(\mathrm{N}_{2} \mathrm{H}_{5}^{+}(a q)+4 \mathrm{Fe}(\mathrm{CN})_{6}^{3-}(a q) \longrightarrow \mathrm{N}_{2}(\mathrm{~g})+\) \(5 \mathrm{H}^{+}(a q)+4 \mathrm{Fe}(\mathrm{CN})_{6}^{4}(a q)\)

From each of the following pairs of substances, use data in Appendix E to choose the one that is the stronger reducing agent: (a) Fe(s) or \(\mathrm{Mg}(s)\) (b) \(\mathrm{Ca}(s)\) or \(\mathrm{Al}(s)\) (c) \(\mathrm{H}_{2}\) (g, acidic solution) or \(\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})\) (d) \(\mathrm{BrO}_{3}^{-}(a q)\) or \(\mathrm{lO}_{3}^{-}(a q)\) 20.44 From each of the following pairs of substances, use data in Appendix E to choose the one that is the stronger oxidizing agent: (a) \(\mathrm{Cl}_{2}(g)\) or \(\mathrm{Br}_{2}(l)\) (b) \(\mathrm{Zn}^{2+}(a q)\) or \(\operatorname{Cd}^{2+}(a q)\) (c) \(\mathrm{Cl}^{-}(a q)\) or \(\mathrm{ClO}_{3}^{-}(a q)\) (d) \(\mathrm{H}_{2} \mathrm{O}_{2}(a q)\) or \(\mathrm{O}_{2}(\mathrm{~g})\) 20.45 By using the data in Appendix E, determine whether each of the following substances is likely to serve as an exidant or a reductant: (a) \(\mathrm{Cl}_{2}\) (g), (b) \(\mathrm{MnO}_{4}^{-}\)(aq, acidic solution), (c) \(\mathrm{Ba}\) (s), (d) \(\mathrm{Zn}(\) s). 20.46 Is each of the following substances likely to serve as an oxidant or a reductant: (a) \(\mathrm{Ce}^{3+}(\mathrm{aq})\), (b) \(\mathrm{Ca}(\mathrm{s})\), (c) \(\mathrm{CO}_{3}^{-}(\mathrm{aq})\), (d) \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) ? \(20.47\) (a) Assuming standard conditions, arrange the following in order of increasing strength as oxidizing agents in acidic solution: \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}, \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{Cu}^{2+}, \mathrm{C}_{2}, \mathrm{O}_{2}\) (b) Arrange the following in order of increasing strength as reducing agents in acidic solution: \(\mathrm{Zn}, \mathrm{I}^{-}, \mathrm{Sn}^{2+}, \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{AL}\). 20.48 Based on the data in Appendix E, (a) which of the following is the strongest oxidizing agent and which is the weakest in acidic solution: \(\mathrm{Br}_{2}, \mathrm{H}_{2} \mathrm{O}_{2}, \mathrm{Zn}, \mathrm{Cr}_{2} \mathrm{O}_{7}{ }^{2-} ?\) (b) Which of the following is the strongest reducing agent, and which is the weaket in acidic solution: \(\mathrm{F}^{-}, \mathrm{Zn}, \mathrm{N}_{1}{ }^{+}\), \(\mathrm{I}_{\mathrm{n}} \mathrm{NO}\) ?

(a) What is meant by the term reduction? (b) On which side of a reduction half-reaction do the electrons appear? (c) What is meant by the term reductant? (d) What is meant by the term reducing agent?

In a galvanic cell the cathode is an \(\mathrm{Ag}^{+}(1.00 \mathrm{M}) / \mathrm{Ag}^{(s)}\) halfcell. The anode is a standard hydrogen electrode immersed in a buffer solution containing \(0.10 \mathrm{M}\) benzoic acid ( \(\left.\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{COOH}\right)\) and \(0.050 \mathrm{M}\) sodium benzoate \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COON} \mathrm{Na}^{+}\right)\). The measured cell voltage is \(1.030 \mathrm{~V}\). What is the \(\mathrm{p} K_{4}\) of benzoic acid?

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