Ch.20 - Electrochemistry

Problem 52c

Chapter 20, Problem 52c

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.

Verified step by step guidance

Step 1: Write the balanced redox reaction.

Step 2: Calculate the standard cell potential (E°cell) using standard reduction potentials.

Step 3: Use the Nernst equation to calculate the standard Gibbs free energy change (∆G°) at 298 K.

Step 4: Use the relationship between ∆G° and the equilibrium constant (K) to calculate K at 298 K.

Step 5: Verify the balanced equation and calculations for consistency and accuracy.

Verified Solution

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Key Concepts

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

Balancing Redox Reactions

Balancing redox reactions involves ensuring that the number of atoms and the charge are equal on both sides of the equation. This process often requires identifying the oxidation and reduction half-reactions, balancing them separately, and then combining them to form a complete balanced equation. In this case, understanding the oxidation states of chromium in Cr(OH)2 and CrO4^2- is essential.

Standard Electromotive Force (emf)

The standard electromotive force (emf) of a reaction is the voltage produced by an electrochemical cell under standard conditions. It can be calculated using the standard reduction potentials of the half-reactions involved. A positive emf indicates a spontaneous reaction, while a negative emf suggests non-spontaneity. This concept is crucial for determining the feasibility of the redox reaction in question.

Gibbs Free Energy and Equilibrium Constant

Gibbs free energy (∆G°) is a thermodynamic quantity that indicates the spontaneity of a reaction at constant temperature and pressure. The relationship between ∆G° and the equilibrium constant (K) is given by the equation ∆G° = -RT ln(K), where R is the gas constant and T is the temperature in Kelvin. This relationship allows for the calculation of K from ∆G° and vice versa, providing insight into the position of equilibrium for the reaction.

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Gibbs Free Energy of Reactions

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

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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Open Question

If the equilibrium constant for a two-electron redox reaction at 298 K is 1.5 * 10⁻⁴, calculate the corresponding ∆G° and E°.

Open Question

If the equilibrium constant for a one-electron redox reaction at 298 K is 8.7 * 10^4, calculate the corresponding ∆G° and E°.

Textbook Question

Using the standard reduction potentials listed in Appendix E, calculate the equilibrium constant for each of the following reactions at 298 K:

(a) Fe(s) + Ni²⁺(aq) → Fe²⁺(aq) + Ni(s)

(b) Co(s) + 2 H⁺(aq) → Co²⁺(aq) + H₂(g)

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