Given the following reduction half-reactions: Fe 3+ (aq) + e - → Fe 2+ (aq) E° red = +0.77 V S 2 O 6 2- (aq) + 4 H + (aq) + 2 e - → 2 H 2 SO 3 (aq) E° red = +0.60 V N 2 O(g) + 2 H + (aq) + 2 e - → N 2 (g) + H 2 O(l) E° red = -1.77 V VO 2+ (aq) + 2 H + (aq) + e - → VO 2+ + H 2 O(l) E° red = +1.00 V (c) Calculate the equilibrium constant K for each reaction at 298 K.

Hello. In this problem, we are asked to refer to the following half reactions and their standard reduction potential values for the combination of reactions that produces the largest voltage of 25°C were asked by the overall cell reaction to determine the standard cell potential. The change in Gibbs, free energy and the equilibrium constant. The reactions that will then provide the largest voltage are those that have the greatest difference in their standard reduction potentials. That will be the first and the third. So the one that has the largest most positive standard reduction potential is the one that has the greatest potential to be reduced. So that reaction will occur at the cathode, the one that has the smallest most negative standard reduction potential. Then this reaction in reverse will take place at the an ode. So writing our reactions then at the cathode, we have bismuth being reduced. The standard reduction potential is 0.240V. Yeah, node then have Valium being oxidized. The standard oxidation potential is equal to the negative of the standard reduction potential. This is equal to 0.549. So we'll combine these two half reactions to get an overall reaction. In order to do that. The moles of electrons being gained has to equal those being lost. So we'll take the reaction occurring at the cathode and multiply through by three. This will then give us an overall reaction of the modes of business ions reacting with thallium. The Form three moles of Bismuth plus Valium irons brand standard cell potential then will be equal to the standard reduction potential plus the standard oxidation potential. This is 0.240V plus 0.549V. This works out to then 0.789V. So we have our overall fast reaction and we have the standard cell potential for this overall reaction. The next thing we're asked to find then these are changing, gives free energy, so the change and gives energy then is related to the standard cell potential. This is equal to then we have three moles of electrons being exchanged. We have the charge on one mole of electrons equal to a fair day, which is 96,004 and 85 columns. And we have our standard cell potential which is 0.789V the call that and is equal to a jewel. And we can convert our jewels kill jules. So our moles of electrons cancels, cancels, bolts, cancels jewels cancels. And we're left with kill jules. So this works out to -288. Kill jules. So this is our changing gives you energy for the overall reaction? We can then make use of change and gives for energy for the overall reaction to find our equilibrium constant, rearranging equation. And solving for equilibrium constant. Are you equilibrium constant? Then have the exponent of the negative of changing good for energy. Goodbye to buy. R gas constant times are temperature. And so our units of joules will cancel Kelvin cancels. We find then that our equilibrium constant is equal to 8.83 times 10 to the 39. So overall reaction, the standard cell potential, the change in gibbs, free energy and our equilibrium constant. All correspondent to answer B. Thanks for watching. Hope. This helped.

Ch.20 - Electrochemistry

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Brown 14th Edition

Ch.20 - Electrochemistry

Problem 51c

Chapter 20, Problem 51c

Given the following reduction half-reactions:
Fe³⁺(aq) + e^- → Fe²⁺(aq) E°_red = +0.77 V
S₂O₆^2-(aq) + 4 H⁺(aq) + 2 e^- → 2 H₂SO₃(aq) E°_red = +0.60 V
N₂O(g) + 2 H⁺(aq) + 2 e^- → N₂(g) + H₂O(l) E°_red = -1.77 V
VO²⁺(aq) + 2 H⁺(aq) + e^- → VO²⁺ + H₂O(l) E°_red = +1.00 V
(c) Calculate the equilibrium constant K for each reaction at 298 K.

Verified step by step guidance

Identify the relationship between the standard reduction potential (E°) and the equilibrium constant (K) using the Nernst equation: E° = (RT/nF) * ln(K), where R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin (298 K), n is the number of moles of electrons transferred, and F is the Faraday constant (96485 C/mol).

Rearrange the Nernst equation to solve for the equilibrium constant (K): ln(K) = (nF/R) * E° / T.

For each half-reaction, determine the number of electrons transferred (n) from the balanced equation. For example, in the reaction Fe3+(aq) + e- → Fe2+(aq), n = 1.

Substitute the values of R, T, F, n, and E° into the rearranged Nernst equation for each reaction to calculate ln(K).

Exponentiate the result from the previous step to solve for K: K = e^(ln(K)). This will give you the equilibrium constant for each reaction at 298 K.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Standard Reduction Potential (E°)

Standard reduction potential (E°) is a measure of the tendency of a chemical species to gain electrons and be reduced. It is measured in volts and is determined under standard conditions (1 M concentration, 1 atm pressure, and 25°C). A higher E° value indicates a greater likelihood of reduction occurring, which is crucial for predicting the direction of redox reactions.

Nernst Equation

The Nernst equation relates the standard reduction potential of a half-reaction to the concentrations of the reactants and products. It allows for the calculation of the cell potential under non-standard conditions. This equation is essential for determining the equilibrium constant (K) for a reaction, as it connects thermodynamic properties with electrochemical behavior.

Equilibrium Constant (K)

The equilibrium constant (K) quantifies the ratio of the concentrations of products to reactants at equilibrium for a given reaction. It is derived from the Gibbs free energy change and can be calculated using the standard reduction potentials of the half-reactions involved. Understanding K is vital for predicting the extent of a reaction and its favorability under specific conditions.

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Related Practice

Textbook Question

Given the following reduction half-reactions:

Fe³⁺(aq) + e^- → Fe²⁺(aq) E°_red = +0.77 V

S₂O₆^2-(aq) + 4 H⁺(aq) + 2 e^- → 2 H₂SO₃(aq) E°_red = +0.60 V

N₂O(g) + 2 H⁺(aq) + 2 e^- → N₂(g) + H₂O(l) E°_red = -1.77 V

VO²⁺(aq) + 2 H⁺(aq) + e^- → VO²⁺ + H₂O(l) E°_red = +1.00 V

(a) Write balanced chemical equations for the oxidation of Fe²⁺(aq) by S₂O₆^2-(aq), by N₂O(aq), and by VO²⁺(aq).

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Textbook Question

Given the following reduction half-reactions:

Fe³⁺(aq) + e^- → Fe²⁺(aq) E°_red = +0.77 V

S₂O₆^2-(aq) + 4 H⁺(aq) + 2 e^- → 2 H₂SO₃(aq) E°_red = +0.60 V

N₂O(g) + 2 H⁺(aq) + 2 e^- → N₂(g) + H₂O(l) E°_red = -1.77 V

VO²⁺(aq) + 2 H⁺(aq) + e^- → VO²⁺ + H₂O(l) E°_red = +1.00 V

(b) Calculate ∆G° for each reaction at 298 K.

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Textbook Question

For each of the following reactions, write a balanced equation, calculate the standard emf, calculate ∆G° at 298 K, and calculate the equilibrium constant K at 298 K. (a) Aqueous iodide ion is oxidized to I21s2 by Hg22+1aq2.

For each of the following reactions, write a balanced equation, calculate the standard emf, calculate ∆G° at 298 K, and calculate the equilibrium constant K at 298 K. (b) In acidic solution, copper(I) ion is oxidized to copper(II) ion by nitrate ion.

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Textbook Question

For each of the following reactions, write a balanced equation, calculate the standard emf, calculate ∆G° at 298 K, and calculate the equilibrium constant K at 298 K. (c) In basic solution, Cr1OH231s2 is oxidized to CrO42-1aq2 by ClO-1aq2.

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