Skip to main content
Ch.20 - Electrochemistry
Chapter 20, Problem 49

The standard reduction potential of Eu2+(aq) is -0.43 V. Using Appendix E, which of the following substances is capable of reducing Eu3+(aq) to Eu2+(aq) under standard conditions: Al, Co, H2O2, N2H5+, H2C2O4?

Verified step by step guidance
1
Step 1: Identify the half-reaction for the reduction of Eu^{3+} to Eu^{2+}. The half-reaction is Eu^{3+} + e^- \rightarrow Eu^{2+} with a standard reduction potential of -0.43 V.
Step 2: To determine which substances can reduce Eu^{3+} to Eu^{2+}, we need to find substances with standard reduction potentials less positive (or more negative) than -0.43 V, as they will act as reducing agents.
Step 3: Look up the standard reduction potentials for the given substances (Al, Co, H_2O_2, N_2H_5^+, H_2C_2O_4) in Appendix E. Note that the more negative the reduction potential, the stronger the reducing agent.
Step 4: Compare the standard reduction potentials of the given substances to -0.43 V. Any substance with a more negative standard reduction potential than -0.43 V can reduce Eu^{3+} to Eu^{2+}.
Step 5: Identify the substances from the list that have standard reduction potentials more negative than -0.43 V and conclude which can act as reducing agents for the reaction.

Key Concepts

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

Standard Reduction Potential

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 more positive E° indicates a greater likelihood of reduction, while a negative E° suggests a lesser tendency. In this question, the E° of Eu<sup>2+</sup>/Eu<sup>3+</sup> is crucial for determining which substances can reduce Eu<sup>3+</sup> to Eu<sup>2+</sup>.
Recommended video:
Guided course
01:10
Standard Reduction Potentials

Reduction and Oxidation Reactions

Reduction and oxidation (redox) reactions involve the transfer of electrons between species. Reduction refers to the gain of electrons, while oxidation refers to the loss of electrons. In this context, to reduce Eu<sup>3+</sup> to Eu<sup>2+</sup>, another substance must be oxidized, meaning it loses electrons. Understanding which substances can act as reducing agents is essential for solving the question.
Recommended video:
Guided course
01:53
Oxidation and Reduction Reactions

Reducing Agents

A reducing agent is a substance that donates electrons to another species, causing that species to be reduced. The effectiveness of a reducing agent can be assessed by comparing its standard reduction potential to that of the species being reduced. In this case, the reducing agents must have a more negative E° than -0.43 V to successfully reduce Eu<sup>3+</sup> to Eu<sup>2+</sup>. Identifying which of the listed substances can act as reducing agents is key to answering the question.
Recommended video:
Guided course
01:01
Oxidizing and Reducing Agents
Related Practice
Textbook Question

(a) Assuming standard conditions, arrange the following in order of increasing strength as oxidizing agents in acidic solution: Cr2O72-, H2O2, Cu2+, Cl2, O2.

688
views
Textbook Question

(b) Arrange the following in order of increasing strength as reducing agents in acidic solution: Zn, I-, Sn2+, H2O2, Al.

Open Question
Based on the data in Appendix E, (a) which of the following is the strongest oxidizing agent, and which is the weakest in acidic solution: Br2, H2O2, Zn, Cr2O72-?
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.