Skip to main content
Pearson+ LogoPearson+ Logo
Ch.20 - Electrochemistry
All textbooksBrown 14th EditionCh.20 - ElectrochemistryProblem 52a
Chapter 20, Problem 52a

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.

Verified step by step guidance
1
Step 1: Write the balanced redox reaction. Identify the oxidation and reduction half-reactions. For the given reaction, iodide ions (I-) are oxidized to iodine (I2), and mercury ions (Hg22+) are reduced. Balance each half-reaction for mass and charge.
Step 2: Combine the balanced half-reactions to form the overall balanced equation. Ensure that the number of electrons lost in the oxidation half-reaction equals the number of electrons gained in the reduction half-reaction.
Step 3: Calculate the standard cell potential (E°) using standard reduction potentials from a table. Use the formula: E° = E° (cathode) - E° (anode).
Step 4: Calculate the standard Gibbs free energy change (∆G°) using the formula: ∆G° = -nFE°, where n is the number of moles of electrons transferred, F is the Faraday constant (96,485 C/mol), and E° is the standard cell potential.
Step 5: Calculate the equilibrium constant (K) at 298 K using the relationship: ∆G° = -RT ln K, where R is the universal gas constant (8.314 J/mol·K) and T is the temperature in Kelvin. Rearrange to solve for K.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
10m
Was this helpful?

Key Concepts

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

Balancing Chemical Equations

Balancing chemical equations involves ensuring that the number of atoms for each element is the same on both sides of the equation. This is crucial for accurately representing the conservation of mass during a chemical reaction. In redox reactions, it is also important to balance the charges, which may involve adding electrons to one side of the equation.
Recommended video:
Guided course
01:32
Balancing Chemical Equations

Standard Electromotive Force (emf)

The standard electromotive force (emf) of a reaction is the measure of the voltage produced by an electrochemical cell under standard conditions. It is calculated using standard reduction potentials from a table of half-reactions. A positive emf indicates a spontaneous reaction, while a negative emf suggests non-spontaneity.
Recommended video:
Guided course
01:10
Standard Reduction Potentials

Gibbs Free Energy (∆G°) and Equilibrium Constant (K)

Gibbs Free Energy (∆G°) is a thermodynamic quantity that indicates the spontaneity of a reaction at standard conditions. It is related to the equilibrium constant (K) by the equation ∆G° = -RT ln(K), where R is the gas constant and T is the temperature in Kelvin. A negative ∆G° corresponds to a reaction that favors products, while K > 1 indicates a product-favored equilibrium.
Recommended video:
Guided course
01:19
∆G° and K
Related Practice
Textbook Question

Given the following reduction half-reactions:

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

S2O62-(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 S2O62-(aq), by N2O(aq), and by VO2+(aq).

728
views
Textbook Question

Given the following reduction half-reactions:

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

S2O62-(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.

369
views
Textbook Question

Given the following reduction half-reactions:

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

S2O62-(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.

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. (b) In acidic solution, copper(I) ion is oxidized to copper(II) ion by nitrate ion.

380
views
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.

311
views