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)?

Short Answer

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During the electrolysis of an aqueous solution of \(\mathrm{Na_2SO_4}\) (assuming standard conditions), the following reactions occur: Anode (Oxidation): \(\mathrm{OH^- (aq) \rightarrow \frac{1}{2}O_2(g) + H_2O(l) + e^-}\) Cathode (Reduction): \(\mathrm{2H^+ (aq) + 2e^- \rightarrow H_2(g)}\) The oxidation of hydroxide ions (\(\mathrm{OH^-}\)) takes place at the anode, and the reduction of hydrogen ions (\(\mathrm{H^+}\)) occurs at the cathode.

Step by step solution

01

Identify the possible reactions at the anode and the cathode

In the case of an aqueous solution of \(\mathrm{Na_2SO_4}\), there can be several ions present in solution. The principal ions will be: 1. Cations: \(\mathrm{Na^+}\) and \(\mathrm{H^+}\) (from water) 2. Anions: \(\mathrm{SO_4^{2-}}\) and \(\mathrm{OH^-}\) (from water) From these, we can list the possible oxidation reactions at the anode and reduction reactions at the cathode. For Anode (Oxidation): a) \(\mathrm{OH^- (aq) \rightarrow \frac{1}{2}O_2(g) + H_2O(l) + e^-}\); Standard electrode potential = +0.40 V b) \(\mathrm{SO_4^{2-} (aq) \rightarrow SO_2(g) + O_2(g) + 2e^-}\); Unable to give its standard electrode potential due to some limitations For Cathode (Reduction): c) \(\mathrm{2H^+ (aq) + 2e^- \rightarrow H_2(g)}\); Standard electrode potential = 0.00 V d) \(\mathrm{Na^+ (aq) + e^- \rightarrow Na(s)}\); Standard electrode potential = -2.71 V
02

Determine the favorable reactions at the anode and the cathode

To determine which reaction is favorable at each electrode, we need to compare their standard electrode potentials. For Anode (Oxidation): The \(\mathrm{OH^-}\) has a higher standard electrode potential (+0.40 V) than the reaction involving \(\mathrm{SO_4^{2-}}\), which suggests that the \(\mathrm{OH^-}\) reaction is more likely to occur at the anode. For Cathode (Reduction): The \(\mathrm{H^+}\) reaction has a higher standard electrode potential (0.00 V) than the reaction involving \(\mathrm{Na^+}\) (-2.71 V), which means that the \(\mathrm{H^+}\) reaction is more favorable.
03

Present the final reactions at the electrodes

Based on our analysis, we can determine the most likely reactions at each electrode during the electrolysis of aqueous \(\mathrm{Na_2SO_4}\): Anode (Oxidation): \(\mathrm{OH^- (aq) \rightarrow \frac{1}{2}O_2(g) + H_2O(l) + e^-}\) Cathode (Reduction): \(\mathrm{2H^+ (aq) + 2e^- \rightarrow H_2(g)}\) So the reactions that occur during the electrolysis of an aqueous solution of \(\mathrm{Na_2SO_4}\) (assuming standard conditions) are the oxidation of hydroxide ions (\(\mathrm{OH^-}\)) at the anode and the reduction of hydrogen ions (\(\mathrm{H^+}\)) at the cathode.

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Most popular questions from this chapter

Balance the following equations by the half-reaction method. a. \(\mathrm{Fe}(s)+\mathrm{HCl}(a q) \longrightarrow \mathrm{HFeCl}_{4}(a q)+\mathrm{H}_{2}(g)\) b. \(\mathrm{IO}_{3}^{-}(a q)+\mathrm{I}^{-}(a q) \stackrel{\text { Acid }}{\longrightarrow} \mathrm{I}_{3}^{-}(a q)\) c. \(\mathrm{Cr}(\mathrm{NCS})_{6}^{4-}(a q)+\mathrm{Ce}^{4+}(a q) \stackrel{\text { Acid }}{\longrightarrow}\) \(\mathrm{Cr}^{3+}(a q)+\mathrm{Ce}^{3+}(a q)+\mathrm{NO}_{3}^{-}(a q)+\mathrm{CO}_{2}(g)+\mathrm{SO}_{4}{ }^{2-}(a q)\) d. \(\operatorname{CrI}_{3}(s)+\mathrm{Cl}_{2}(g) \stackrel{\text { Base }}{\longrightarrow}\) \(\operatorname{CrO}_{4}{ }^{2-}(a q)+\mathrm{IO}_{4}^{-}(a q)+\mathrm{Cl}^{-}(a q)\) e. \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}(a q)+\mathrm{Ce}^{4+}(a q) \stackrel{\text { Base }}{\longrightarrow}\) \(\mathrm{Ce}(\mathrm{OH})_{3}(s)+\mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{CO}_{3}^{2-}(a q)+\mathrm{NO}_{3}^{-}(a q)\)

When balancing equations in Chapter 3, we did not mention that reactions must be charge balanced as well as mass

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A chemist wishes to determine the concentration of \(\mathrm{CrO}_{4}^{2}\) electrochemically. A cell is constructed consisting of a saturated calomel electrode (SCE; see Exercise 115 ) and a silver wire coated with \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\). The \(\mathscr{E}^{\circ}\) value for the following half-reaction is \(0.446 \mathrm{~V}\) relative to the standard hydrogen electrode: $$ \mathrm{Ag}_{2} \mathrm{CrO}_{4}+2 \mathrm{c}^{-} \longrightarrow 2 \mathrm{Ag}+\mathrm{CrO}_{4}{ }^{2-} $$ a. Calculate \(\mathscr{E}_{\text {cell }}\) and \(\Delta G\) at \(25^{\circ} \mathrm{C}\) for the cell reaction when \(\left[\mathrm{CrO}_{4}{ }^{2-}\right]=1.00 \mathrm{~mol} / \mathrm{L}\) b. Write the Nernst equation for the cell. Assume that the SCE concentrations are constant. c. If the coated silver wire is placed in a solution (at \(25^{\circ} \mathrm{C}\) ) in which \(\left[\mathrm{CrO}_{4}{ }^{2-}\right]=1.00 \times 10^{-5} M\), what is the expected cell potential? d. The measured cell potential at \(25^{\circ} \mathrm{C}\) is \(0.504 \mathrm{~V}\) when the coated wire is dipped into a solution of unknown \(\left[\mathrm{CrO}_{4}{ }^{2-}\right]\). What is \(\left[\mathrm{CrO}_{4}{ }^{2-}\right]\) for this solution? e. Using data from this problem and from Table \(18.1\), calculate the solubility product \(\left(K_{\mathrm{sp}}\right)\) for \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\)

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