Use the standard reduction potentials to find the equilibrium constant for each of the following reactions at \(25^{\circ} \mathrm{C}\): (a) \(\operatorname{Br}_{2}(l)+2 \mathrm{I}^{-}(a q) \rightleftharpoons 2 \mathrm{Br}^{-}(a q)+\mathrm{I}_{2}(s)\) (b) \(2 \mathrm{Ce}^{4+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons$$\mathrm{Cl}_{2}(g)+2 \mathrm{Ce}^{3+}(a q)\) (c) \(5 \mathrm{Fe}^{2+}(a q)+\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q) \rightleftharpoons\) \(\mathrm{Mn}^{2+}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l)+5 \mathrm{Fe}^{3+}(a q)\)

Short Answer

Expert verified
The equilibrium constant (K) for each reaction is calculated by applying the equation derived from the Nernst equation as stated in the steps above.

Step by step solution

01

Calculate Standard Cell Potential

Calculate the standard cell potential (E\(^{\circ}\)) for each reaction. This can be done by subtracting the standard reduction potential of the oxidation half-reaction from that of the reduction half-reaction.
02

Use Nernst Equation

Use the Nernst equation to calculate the equilibrium constant (K) for the reactions. Also remember to take into consideration the stoichiometry of the reactions. The Nernst equation is given by E = E\(^{\circ}\) – (0.0592/n)logQ, where n is the number of electron transfers. At equilibrium, E = 0, Q becomes K. This gives: 0 = E\(^{\circ}\) – (0.0592/n)logK. Solving for K:\( K = 10^{(nE°) /0.0592}\).
03

Apply Equation to Each Reaction

Apply the above equation to each of the given reactions.

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Key Concepts

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

Standard Reduction Potentials
Understanding standard reduction potentials is crucial for students delving into electrochemistry, as they provide insight into the tendencies of various species to gain electrons, known as reduction. These potentials are measured under standard conditions, meaning at a temperature of 25°C, 1 M concentration for solutions, and 1 atm for gases.

Think of it as a numeric value that represents how likely a chemical species is to be reduced. These values are tabulated relative to the standard hydrogen electrode, which has an assigned potential of zero volts. When comparing two half-reactions in an electrochemical cell, the species with the higher standard reduction potential will undergo reduction, while the other will be oxidized. This comparison helps in determining the overall cell potential for a reaction.
Nernst Equation
Moving on to the Nernst equation, this powerful formula allows students to relate the concentrations of reactants and products to the cell potential in an electrochemical cell. In essence, it allows us to predict how cell potential changes as a reaction proceeds. The Nernst equation, presented as E = E° – (0.0592/n)logQ, incorporates the standard cell potential (), the number of electrons transferred in the reaction (n), and the reaction quotient (Q).

At equilibrium, the cell potential (E) becomes zero, and Q is replaced by the equilibrium constant (K). The resulting equation, 0 = E° – (0.0592/n)logK, can then be rearranged to solve for the equilibrium constant. This forms a direct relationship between the cell potential and the equilibrium constant, demonstrating how thermodynamics and electrochemistry are intertwined.
Cell Potential
Cell potential, often symbolized as E or when referring to standard conditions, is essentially the 'driving force' of an electrochemical reaction. It represents the difference in potential energy between the two electrodes in a cell and determines the direction and magnitude of electron flow. Calculating cell potential involves using standard reduction potentials for the half-reactions occurring at the anode (oxidation) and cathode (reduction).

The standard cell potential can be found by taking the difference between the cathode's and anode's standard reduction potentials. This value is instrumental in predicting the feasibility of a reaction and calculating the equilibrium constant by using the Nernst equation. A positive cell potential indicates a spontaneous reaction under standard conditions, illustrating a fundamental connection between voltage and the thermodynamic favorability of a reaction.
Electrochemistry
Electrochemistry is a branch of chemistry that focuses on the changes between electrical and chemical energy, particularly those involving electron transfer processes. Central to this discipline are redox reactions, where oxidation and reduction occur, converting energy from chemical to electrical form and vice versa. In studying electrochemistry, concepts like standard reduction potentials, cell potential, and the Nernst equation come together to provide a comprehensive understanding of how chemical reactions can produce electrical current, and how that current can instigate chemical change.

With applications in batteries, fuel cells, corrosion prevention, and even biological systems like nerve cell operation, electrochemistry bridges the gap between physical science and practical technology. It enables us to calculate essential values such as cell potentials and equilibrium constants, which inform the design and interpretation of a myriad of electrochemical processes.

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Most popular questions from this chapter

What is the function of a salt bridge? What kind of electrolyte should be used in a salt bridge?

What is a cell diagram? Write the cell diagram for a galvanic cell consisting of an Al electrode placed in a \(1 \mathrm{MAl}\left(\mathrm{NO}_{3}\right)_{3}\) solution and a Ag electrode placed in a \(1 M \mathrm{AgNO}_{3}\) solution.

Balance the following redox equations by the halfreaction method: (a) \(\mathrm{Mn}^{2+}+\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{MnO}_{2}+\mathrm{H}_{2} \mathrm{O}\) (in basic solution) (b) \(\mathrm{Bi}(\mathrm{OH})_{3}+\mathrm{SnO}_{2}^{2-} \longrightarrow \mathrm{SnO}_{3}^{2-}+\mathrm{Bi}\) (in basic solution) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{C}_{2} \mathrm{O}_{4}^{2-} \longrightarrow \mathrm{Cr}^{3+}+\mathrm{CO}_{2}\) (in acidic solution) (d) \(\mathrm{ClO}_{3}^{-}+\mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_{2}+\mathrm{ClO}_{2}\) (in acidic solution)

Given that $$ \begin{array}{cl} 2 \mathrm{Hg}^{2+}(a q)+2 e^{-} \longrightarrow \mathrm{Hg}_{2}^{2+}(a q) & E^{\circ}=0.92 \mathrm{~V} \\ \mathrm{Hg}_{2}^{2+}(a q)+2 e^{-} \longrightarrow 2 \mathrm{Hg}(l) & E^{\circ}=0.85 \mathrm{~V} \end{array} $$ calculate \(\Delta G^{\circ}\) and \(K\) for the following process at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{Hg}_{2}^{2+}(a q) \longrightarrow \mathrm{Hg}^{2+}(a q)+\mathrm{Hg}(l) $$ (The preceding reaction is an example of a disproportionation reaction in which an element in one oxidation state is both oxidized and reduced.)

When \(25.0 \mathrm{~mL}\) of a solution containing both \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) ions is titrated with \(23.0 \mathrm{~mL}\) of \(0.0200 \mathrm{M}\) \(\mathrm{KMnO}_{4}\) (in dilute sulfuric acid), all of the \(\mathrm{Fe}^{2+}\) ions are oxidized to \(\mathrm{Fe}^{3+}\) ions. Next, the solution is treated with \(Z\) n metal to convert all of the \(\mathrm{Fe}^{3+}\) ions to \(\mathrm{Fe}^{2+}\) ions. Finally, \(40.0 \mathrm{~mL}\) of the same \(\mathrm{KMnO}_{4}\) solution are added to the solution in order to oxidize the \(\mathrm{Fe}^{2+}\) ions to \(\mathrm{Fe}^{3+}\). Calculate the molar concentrations of \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) in the original solution.

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