Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 105

In the electrolysis of an aqueous solution of $\mathrm{Na}_{2} \mathrm{SO}_{4},$ what reactions occur at the anode and the cathode (assuming standard conditions)? $$\begin{array}{ll} {\text{}} & \quad{ \mathscr{E}^{\circ} } \\ \hline {\mathrm{S}_{2} \mathrm{O}_{8}^{2-}+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{SO}_{4}^{2-}} & {2.01 \mathrm{V}} \\ {\mathrm{O}_{2}+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}} & {1.23 \mathrm{V}} \\ {2 \mathrm{H}_{2} \mathrm{O}+2 \mathrm{e}^{-} \longrightarrow \mathrm{H}_{2}+2 \mathrm{OH}^{-}} & {-0.83 \mathrm{V}} \\\ {\mathrm{Na}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{Na}} & {-2.71 \mathrm{V}}\end{array}$$

Short Answer

Expert verified
During the electrolysis of an aqueous solution of \(Na_2SO_4\), the reaction at the anode (oxidation) is \(S_2O_{8}^{2-} \longrightarrow 2SO_{4}^{2-} + 2e^-\), and the reaction at the cathode (reduction) is \(O_2 + 4H^+ + 4e^- \longrightarrow 2H_2O\).

Step by step solution

01

Identify possible half-reactions.

From the given table, we have four possible half-reactions: 1. \(S_2O_{8}^{2-} + 2e^- \longrightarrow 2SO_{4}^{2-}\) , E°=2.01 V 2. \(O_2 + 4H^+ + 4e^- \longrightarrow 2H_2O\) , E°=1.23 V 3. \(2H_2O + 2e^- \longrightarrow H_2 + 2OH^-\) , E°=-0.83 V 4. \(Na^+ + e^- \longrightarrow Na\) , E°=-2.71 V
02

Determine the reaction occurring at the anode.

The anode is the electrode where oxidation occurs. Therefore, we need to consider the most positive oxidation potential. In this case, the half-reaction \(S_2O_{8}^{2-} + 2e^- \longrightarrow 2SO_{4}^{2-}\) has the highest E° value (2.01 V). Thus, at the anode, the reaction is as follows: \(S_2O_{8}^{2-} \longrightarrow 2SO_{4}^{2-} + 2e^-\)
03

Determine the reaction occurring at the cathode.

The cathode is the electrode where reduction occurs. Therefore, we need to consider the most positive reduction potential. From the table, the half-reaction \(O_2 + 4H^+ + 4e^- \longrightarrow 2H_2O\) has the highest E° value (1.23 V). Thus, at the cathode, the reaction is as follows: \(O_2 + 4H^+ + 4e^- \longrightarrow 2H_2O\)
04

Write the reactions at the anode and cathode.

The reactions at the anode and cathode are: Anode (oxidation): \(S_2O_{8}^{2-} \longrightarrow 2SO_{4}^{2-} + 2e^-\) Cathode (reduction): \(O_2 + 4H^+ + 4e^- \longrightarrow 2H_2O\) These are the reactions occurring during the electrolysis of an aqueous solution of Na2SO4 under standard conditions.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Hydrogen peroxide can function either as an oxidizing agent or as a reducing agent. At standard conditions, is \(\mathrm{H}_{2} \mathrm{O}_{2}\) a better oxidizing agent or reducing agent? Explain.

The free energy change for a reaction, \(\Delta G,\) is an extensive property. What is an extensive property? Surprisingly, one can calculate \(\Delta G\) from the cell potential, \(\mathscr{E}\), for the reaction. This is surprising because \(\mathscr{E}\) is an intensive property. How can the extensive property \(\Delta G\) be calculated from the intensive property \(\mathscr{E}\) ?

Calculate \(\mathscr{E}^{\circ}\) for the following half-reaction: $$\mathrm{AgI}(s)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)+\mathrm{I}^{-}(a q)$$ (Hint: Reference the \(K_{\mathrm{sp}}\) value for AgI and the standard reduction potential for \(\mathrm{Ag}^{+}\).)

Specify which of the following equations represent oxidation–reduction reactions, and indicate the oxidizing agent, the reducing agent, the species being oxidized, and the species being reduced. a. $\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightarrow \mathrm{CO}(g)+3 \mathrm{H}_{2}(g)$ b. $2 \mathrm{AgNO}_{3}(a q)+\mathrm{Cu}(s) \rightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}(a q)+2 \mathrm{Ag}(s)$ c. $\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \rightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)$ d. $2 \mathrm{H}^{+}(a q)+2 \mathrm{CrO}_{4}^{2-}(a q) \rightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)$

Consider the following galvanic cell: Calculate the \(K_{s p}\) value for \(\mathrm{Ag}_{2} \mathrm{SO}_{4}(s) .\) Note that to obtain silver ions in the right compartment (the cathode compartment), excess solid \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\) was added and some of the salt dissolved.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

Read Explanation

The Earths Atmosphere

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free