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A voltaic cell is constructed with all reactants and products in their standard states. Will the concentration of the reactants increase, decrease, or remain the same as the cell operates?

Short Answer

Expert verified
In a voltaic cell operating under standard conditions, the concentration of reactants will decrease, and the concentration of products will increase as the redox reaction proceeds. This occurs because reactants are transformed into products during the reaction. However, since the cell potential remains constant under standard conditions, these concentration changes will not affect the overall reaction.

Step by step solution

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1. Understand voltaic cells and redox reactions

A voltaic cell is an electrochemical cell that uses a spontaneous redox reaction to generate electricity. In a redox reaction, there are two half-reactions: the oxidation reaction, where electrons are lost, and the reduction reaction, where electrons are gained. These half-reactions occur at the two electrodes of the voltaic cell - the anode (where oxidation occurs) and the cathode (where reduction occurs).
02

2. Analyze the behavior of reactants and products under standard conditions

The standard state is a reference condition at which all reactants and products are considered to have a concentration of 1 M and a pressure of 1 atm. Under standard conditions, the concentration of the reactants will either increase or decrease as the cell operates, depending on the stoichiometry of the specific redox reaction. In a redox reaction, the reactants are transformed into products as the reaction progresses. Thus, the concentration of the reactants will decrease, and the concentration of the products will increase.
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3. Relate the concentration changes to the cell potential

As the concentration of the reactants decreases and the concentration of the products increases, the cell potential will decrease due to the Nernst equation. However, because we are considering the cell under standard conditions, the cell potential remains constant, and the concentration changes should not affect the overall reaction.
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4. Draw conclusions

Under standard conditions, the concentrations of the reactants will decrease, and the concentrations of the products will increase as the voltaic cell operates. Since the cell potential remains constant under standard conditions, these concentration changes will not impact the overall reaction.

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Key Concepts

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

Understanding Redox Reactions in Voltaic Cells
Redox reactions are the heart of voltaic cell operation. In a redox process, one species gives away electrons, known as oxidation, while another species receives electrons, known as reduction. A voltaic cell harnesses this exchange of electrons to generate electric power.

Inside a voltaic cell, two half-reactions occur at separate electrodes. At the anode, oxidation takes place, meaning electrons are released. These electrons travel through a wire to the cathode, where reduction happens—electrons are accepted by a different substance. Together, these half-reactions make up the overall cell reaction, turning chemical energy into electrical energy.

As the cell operates, the reactants in the anode compartment (oxidized substances) will be consumed while reduction continues at the cathode. Consequently, the concentration of the reactants naturally decreases. The exercise improvement advice would be to highlight that although reactants are being used up, the spontaneous nature of the redox reactions ensures the continued production of electricity until the reactants are depleted.
Impact of Standard State Conditions on Cell Operation
The use of standard state conditions is crucial when discussing the theoretical aspects of voltaic cells. This concept implies that all reactants and products are at a concentration of 1 molar (1 M) and gases are at a pressure of 1 atmosphere (1 atm).

These conditions provide a baseline to measure and compare the potentials of various electrochemical cells. It's important to clarify that in a real-world setting, these conditions are not static; concentrations and pressures change as the cell operates. However, for the purpose of the exercise, assuming that a voltaic cell is operating under standard state conditions allows us to deduce that the concentration of reactants will decrease due to their conversion into products.

Additionally, standard conditions assume a constant temperature, typically 25°C (298 K). It's worth teaching students that while these idealized conditions are a staple in textbooks, in practice, variations must be accounted for to understand the actual behavior of a voltaic cell.
The Role of the Nernst Equation in Voltaic Cell Chemistry
The Nernst equation provides a deeper understanding of how concentrations affect the cell potential. It's an equation that correlates the electrochemical cell's potential to the reaction quotient—a ratio that reflects the concentrations of the reactants and products at any given moment. The equation is as follows: \[ E = E^0 - \frac{RT}{nF} \ln Q \]where \( E \) is the cell potential, \( E^0 \) is the standard cell potential, \( R \) is the ideal gas constant, \( T \) is the temperature in kelvin, \( n \) is the number of moles of electrons transferred, \( F \) is the Faraday constant, and \( Q \) is the reaction quotient. In the context of our exercise, under standard state conditions, the \( E^0 \) is constant. However, as the cell operates and reactants are consumed, \( Q \) changes which, in a real scenario, would affect the voltage of the cell according to the Nernst equation. But for the sake of the exercise, where standard conditions imply a constant cell potential, this change in concentration does not alter the cell potential. Thus, we selectively apply the Nernst equation to understand theoretical behavior while acknowledging its limitations under the standard state assumption.

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Most popular questions from this chapter

From each of the following pairs of substances, use data in Appendix E to choose the one that is the stronger reducing agent: $$ \begin{array}{l}{\text { (a) } \mathrm{Fe}(s) \text { or } \mathrm{Mg}(s)} \\\ {\text { (b) } \mathrm{Ca}(s) \text { or } \mathrm{Al}(s)} \\ {\text { (c) } \mathrm{H}_{2}\left(g, \text { acidic solution ) or } \mathrm{H}_{2} \mathrm{S}(g)\right.} \\ {\text { (d) } \mathrm{BrO}_{3}^{-}(a q) \text { or } \mathrm{IO}_{3}^{-}(a q)}\end{array} $$

For each of the following balanced oxidation-reduction reactions, (i) identify the oxidation numbers for all the elements in the reactants and products and (ii) state the total number of electrons transferred in each reaction. $$ \begin{array}{l}{\text { (a) } \mathrm{I}_{2} \mathrm{O}_{5}(s)+5 \mathrm{CO}(g) \longrightarrow \mathrm{I}_{2}(s)+5 \mathrm{CO}_{2}(g)} \\\ {\text { (b) } 2 \mathrm{Hg}^{2+}(a q)+\mathrm{N}_{2} \mathrm{H}_{4}(a q) \longrightarrow 2 \mathrm{Hg}(l)+\mathrm{N}_{2}(g)+4 \mathrm{H}^{+}(a q)} \\\ {\text { (c) } 3 \mathrm{H}_{2} \mathrm{S}(a q)+2 \mathrm{H}^{+}(a q)+2 \mathrm{NO}_{3}^{-}(a q) \longrightarrow 3 \mathrm{S}(s)+} \\\\{\quad\quad 2 \mathrm{NO}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)}\end{array} $$

A voltaic cell that uses the reaction $$ \mathrm{T}^{3+}(a q)+2 \mathrm{Cr}^{2+}(a q) \longrightarrow \mathrm{Tl}^{+}(a q)+2 \mathrm{Cr}^{3+}(a q) $$ has a measured standard cell potential of \(+1.19 \mathrm{V}\) . (a) Write the two half-cell reactions. (b) By using data from Appendix E, determine \(E_{\text { red }}^{\circ}\) for the reaction involving Pd. (c) Sketch the voltaic cell, label the anode and cathode, and indicate the direction of electron flow.

Indicate whether the following balanced equations involve oxidation-reduction. If they do, identify the elements that undergo changes in oxidation number. $$ \begin{array}{l}{\text { (a) } 2 \mathrm{AgNO}_{3}(a q)+\mathrm{CoCl}_{2}(a q) \longrightarrow 2 \mathrm{AgCl}(s)+} \\\ {\quad\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}(a q)} \\ {\text { (b) } 2 \mathrm{PbO}_{2}(s) \longrightarrow 2 \mathrm{PbO}(s)+\mathrm{O}_{2}(g)} \\\ {\text { (c) } 2 \mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NaBr}(s) \rightarrow \mathrm{Br}_{2}(l)+\mathrm{SO}_{2}(g)+} \\ {\quad \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)}\end{array} $$

Complete and balance the following half-reactions. In each case indicate whether the half-reaction is an oxidation or a reduction. $$ \begin{array}{l}{\text { (a) } \mathrm{Mo}^{3+}(a q) \longrightarrow \mathrm{Mo}(s) \text { (acidic solution) }} \\ {\text { (b) } \mathrm{H}_{2} \mathrm{SO}_{3}(a q) \longrightarrow \mathrm{SO}_{4}^{2-}(a q) \text { (acidic solution) }} \\ {\text { (c) } \mathrm{NO}_{3}^{-}(a q) \longrightarrow \mathrm{NO}(g)(\text { acidic solution })} \\ {\text { (d) } \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \text { (acidic solution) }} \\ {\text { (e) } \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \text { (basic solution) }} \end{array} \\\ {\text { (f) } \mathrm{Mn}^{2+}(a q) \longrightarrow \mathrm{MnO}_{2}(s) \text { (basic solution) }} \\ {\text { (g) } \mathrm{Cr}(\mathrm{OH})_{3}(s) \longrightarrow \mathrm{CrO}_{4}^{2-}(a q) \text { (basic solution) }} $$

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