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Consider the half-cell reaction \(\mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+\) \(4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(I) .\) By what factor are \(n, Q, E,\) and \(E^{\circ}\) changed if all the stoichiometric coefficients are multiplied by the factor two? Justify your answers.

Short Answer

Expert verified
When all stoichiometric coefficients are multiplied by two, the number of electrons (n) is doubled. The reaction quotient (Q) is changed but not in a direct proportional manner. The standard cell potential (E°) remains unchanged, and the cell potential (E) is changed but not in a direct proportional manner.

Step by step solution

01

Double the stoichiometric coefficients

The new half-cell reaction will be: \(2O_2(g) + 8H^+(aq) + 8e^- \rightarrow 4H_2O(l)\) Now, let's analyze the effects on n, Q, E, and E°.
02

Number of electrons (n)

Original reaction: \(n_1 = 4\) New reaction: \(n_2 = 8\) The number of electrons in the new reaction is doubled.
03

Reaction Quotient (Q)

Original reaction: \(Q_1 = \frac{[H_2O]^2}{[O_2][H^+]^4}\) New reaction: \(Q_2 = \frac{[H_2O]^4}{[O_2]^2[H^+]^8}\) The reaction quotient is changed since the overall stoichiometry of the reaction has changed. It is not a simple scaling factor.
04

Cell Potential (E)

The cell potential E can be calculated using the Nernst equation: \(E = E° - \frac{RT}{nF} \ln Q\) For the original reaction, we have: \(E_1 = E°_1 - \frac{RT}{4F} \ln Q_1\) For the new reaction, we have: \(E_2 = E°_2 - \frac{RT}{8F} \ln Q_2\) Since the stoichiometry of the reaction has changed, we cannot directly say that E is doubled.
05

Standard Cell Potential (E°)

The standard cell potential depends only on the chemical species involved in the reaction and the standard conditions. Therefore, changing the stoichiometric coefficients will not change the standard cell potential: \(E°_1 = E°_2 = E°\) In conclusion, doubling the stoichiometric coefficients will double the number of electrons involved (n), change the reaction quotient (Q) but not in a direct proportional manner, and not change the standard cell potential (E°). The cell potential (E) will also change, but its relation to the original reaction cannot be directly determined through a scaling factor.

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Key Concepts

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

Electrode Reaction
Electrode reactions are the chemical processes that occur at the interface of an electrode and the electrolyte in an electrochemical cell. They involve the transfer of electrons between the electrode and the chemical species in the electrolyte. The stoichiometry of an electrode reaction is important because it dictates the proportion of reactants and products, as well as the number of electrons exchanged, which in turn influences the cell's electrical output. For example, in the given exercise, the half-cell reaction involves oxygen, hydrogen ions, and electrons combining to form water. When we double the stoichiometric coefficients, effectively we're doubling the scale of the reaction. The number of molecules of each reactant and product involved, as well as the electrons transferred, are all doubled. While this change impacts the electron flow (i.e., current) and the reaction quotient, it doesn't alter the intrinsic nature of the chemical species or their propensity to undergo the reaction, which is reflected in the standard cell potential remaining the same. Understanding the stoichiometry is crucial in predicting the outcome of changes in the reaction and designing efficient electrochemical cells.

It's important in solving problems like these to recognize that the underlying chemistry doesn't change with the scaling of the reaction; oxygen is still reduced, and water is still produced. The main difference is in quantitative amounts, not qualitative behavior.
Nernst Equation
The Nernst Equation is a fundamental relationship used to calculate the cell potential of an electrochemical cell under non-standard conditions. It's defined as:
\[E = E° - \frac{RT}{nF} \ln Q\]
Here, \(E\) is the cell potential, \(E°\) is the standard cell potential, \(R\) is the universal gas constant, \(T\) is the temperature in Kelvin, \(n\) is the number of moles of electrons transferred in the electrochemical reaction, \(F\) is the Faraday constant, and \(Q\) is the reaction quotient. By adjusting for the actual concentration of reactants and products, the Nernst equation gives us a way to account for real-world conditions. In practice, if we change the stoichiometry of a half-cell reaction—like doubling it as in our exercise—the Nernst equation needs to be re-evaluated because of the change in \(n\) and the reaction quotient \(Q\). However, the standard electrode potential \(E°\) remains constant since it's an intrinsic property of the materials involved, independent of reaction stoichiometry or concentration of reactants. Understanding how to apply the Nernst equation is crucial in electrochemistry, as it enables us to predict cell potentials under various conditions, which is vital for the operation of batteries, sensors, and other electrochemical devices.
Reaction Quotient
The reaction quotient, denoted by \(Q\), is a measure of the relative amounts of products and reactants present during a reaction at a given point in time. It is calculated using the same expression as the equilibrium constant \(K\), but unlike \(K\), it isn't limited to conditions of equilibrium. For the given reaction:
\[Q = \frac{[H_2O]^2}{[O_2][H^+]^4}\]
When we alter the coefficients of a balanced equation, the reaction quotient expression changes accordingly. In the improved equation with doubled stoichiometric coefficients, the powers to which each concentration is raised in the quotient formula also double. As a result, the magnitude of \(Q\) depends on the concentrations of the reactants and products at a specific moment and thus has to be calculated based on the current state of the system. This is particularly important for understanding electrochemical processes, where dynamic changes in concentration can significantly affect cell behavior. It's essential not to assume that doubling the coefficients will just double the reaction quotient; the change is more complex due to the exponents in the quotient's formula. Comprehending this nuanced interplay is important for both theoretical analysis and practical application in designing and monitoring electrochemical systems.
Cell Potential
Cell potential, expressed as \(E\), is the measure of the driving force behind an electrochemical reaction and is measured in volts. It quantifies the energy difference per charge between the reactions occurring at the two electrodes of an electrochemical cell. The standard cell potential \(E°\) is the potential difference when all the reactants and products are at standard conditions, typically 1M concentration for solutions, 1 bar pressure for gases, and 25°C for temperature. As observed in the exercise, changing the stoichiometry of the half-cell reaction doesn't influence the standard cell potential since it is a property determined by the specific redox couple involved in the electrode reaction.

However, the actual cell potential does vary because it is influenced by the reaction quotient and the number of electrons transferred, both of which are affected by the stoichiometry. When stoichiometric coefficients are doubled, while \(E°\) remains unchanged, the term involving \(n\) in the Nernst equation will be affected because the number of electrons has doubled, and the reaction quotient \(Q\) has changed due to different powers in its formula. These dependencies make interpreting the changes to cell potential more nuanced. Understanding cell potential is fundamental for the practical use of electrochemical cells in applications such as batteries, where it dictates how much electrical energy can be produced or stored.

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Most popular questions from this chapter

Determine the half-cell reactions and the overall cell reaction, calculate the cell potential, and determine the equilibrium constant at \(298.15 \mathrm{K}\) for the cell \\[\begin{array}{c}\mathrm{H}_{2}(g)|\operatorname{Pt}(s)| \mathrm{H}^{+}\left(a q, a_{H^{+}}=0.250\right) | \mathrm{Cu}^{2+} \\ \left(a q, a_{C u^{2+}}=0.100\right) | \mathrm{Cu}(s)\end{array}\\] Is the cell reaction spontaneous as written?

Harnet and Hamer [J. American Chemical Society \(57(1935): 33]\) report values for the potential of the cell \(\mathrm{Pt}(s)\left|\mathrm{PbSO}_{4}(s)\right| \mathrm{H}_{2} \mathrm{SO}_{4}(a q, a)\left|\mathrm{PbSO}_{4}(s)\right| \mathrm{PbO}_{2}(s) | \mathrm{Pt}(s)\) over a wide range of temperature and \(\mathrm{H}_{2} \mathrm{SO}_{4}\) concentrations. In \(1.00 m \mathrm{H}_{2} \mathrm{SO}_{4}\), their results were described by \(E(V)=\) \(1.91737+56.1 \times 10^{-6} t+108 \times 10^{-8} t^{2}\) where \(t\) is the temperature on the Celsius scale. Calculate \(\Delta G_{R}^{\circ}, \Delta H_{R}^{\circ},\) and \(\Delta S_{R}^{\circ}\) for the cell reaction at \(11^{\circ}\) and \(35^{\circ} \mathrm{C}\).

Between \(0^{\circ}\) and \(90^{\circ} \mathrm{C}\), the potential of the cell \(\operatorname{Pt}(s)\left|\mathrm{H}_{2}(g, f=1.00 \mathrm{atm})\right| \mathrm{HCl}(a q, m=0.100)|\mathrm{AgCl}(s)|\) \(\mathrm{Ag}(s)\) is described by the equation \(E(V)=0.35510-\) \(0.3422 \times 10^{-4} t-3.2347 \times 10^{-6} t^{2}+6.314 \times 10^{-9} t^{3}\) where \(t\) is the temperature on the Celsius scale. Write the cell reaction and calculate \(\Delta G_{R}^{\circ}, \Delta H_{R}^{\circ},\) and \(\Delta S_{R}^{\circ}\) for the cell reaction at \(50 .^{\circ} \mathrm{C}\).

The standard half-cell potential for the reaction \\[\mathrm{O}_{2}(g)+4 \mathrm{H}^{+}(a q)+4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \text { is }+1.229 \mathrm{V} \text { at }\\] \(298.15 \mathrm{K} .\) Calculate \(E\) for a 0.300 -molal solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) for \(a_{\mathrm{O}_{2}}=1.00\) (a) assuming that the \(a_{H^{+}}\) is equal to the molality and (b) using the measured mean ionic activity coefficient for this concentration from the data tables. How large is the relative error if the concentrations, rather than the activities, are used?

You are given the following half-cell reactions: $$\begin{array}{ll}\mathrm{Pd}^{2+}(a q)+2 \mathrm{e}^{-} \rightarrow \operatorname{Pd}(s) & E^{\circ}=0.83 \mathrm{V} \\ \mathrm{PdCl}_{4}^{2-}(a q)+2 \mathrm{e}^{-} \rightarrow \mathrm{Pd}(s)+4 \mathrm{Cl}^{-}(a q) & E^{\circ}=0.64 \mathrm{V}\end{array}$$ a. Calculate the equilibrium constant for the reaction \(\mathrm{Pd}^{2+}(a q)+4 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{PdCl}_{4}^{2-}(a q)\) b. Calculate \(\Delta G^{\circ}\) for this reaction.

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