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Chapter 18: Problem 46

Describe the electrolysis of an aqueous solution of \(\mathrm{KNO}_{3}\)

Short Answer

Expert verified
In the electrolysis of an aqueous solution of \(\mathrm{KNO}_{3}\), at the anode, \(OH^{-}\) ions are oxidized to form oxygen and water with the release of electrons. At the cathode, \(H^{+}\) ions are reduced by gaining electrons to form hydrogen gas.

Step by step solution

01

Understanding Electrolysis

Electrolysis is the process in which an electric current is passed through an electrolyte to bring about chemical changes. In this context, aqueous Potassium Nitrate (\(\mathrm{KNO}_{3}\)) is the electrolyte.
02

Determining the Ions Present

In an aqueous solution, \(\mathrm{KNO}_{3}\) will dissociate into its ions: \(K^{+}\), \(NO_{3}^{-}\), \(H^{+}\) and \(OH^{-}\). The \(H^{+}\) and \(OH^{-}\) ions come from the self-dissociation of water.
03

Determining the Anode Reaction (Oxidation)

The anode is the site of oxidation. Here, the \(OH^{-}\) ion will be oxidized, because it has the lowest oxidation potential among the ions present. It will lose electrons to form oxygen and water while also releasing electrons. The reaction can be written as: \[4OH^{-}\rightarrow2H_{2}O+O_{2}+4e^{-}\]
04

Determining the Cathode Reaction (Reduction)

The cathode is where reduction takes place. In this case, the \(H^{+}\) ion will typically be reduced, because it has the highest reduction potential among the current ions. Therefore, it will gain electrons to form hydrogen gas. The reaction can be written as: \[2H^{+} + 2e^{-}\rightarrow H_{2}\]

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Most popular questions from this chapter

A galvanic cell is constructed by immersing a piece of copper wire in \(25.0 \mathrm{~mL}\) of a \(0.20 \mathrm{M} \mathrm{CuSO}_{4}\) solution and a zinc strip in \(25.0 \mathrm{~mL}\) of a \(0.20 \mathrm{M} \mathrm{ZnSO}_{4}\) solution. (a) Calculate the emf of the cell at \(25^{\circ} \mathrm{C}\) and predict what would happen if a small amount of concentrated \(\mathrm{NH}_{3}\) solution were added to (i) the \(\mathrm{CuSO}_{4}\) solution and (ii) the \(\mathrm{ZnSO}_{4}\) solution. Assume that the volume in each compartment remains constant at \(25.0 \mathrm{~mL}\). (b) In a separate experiment, \(25.0 \mathrm{~mL}\) of \(3.00 \mathrm{M} \mathrm{NH}_{3}\) are added to the \(\mathrm{CuSO}_{4}\) so- lution. If the emf of the cell is \(0.68 \mathrm{~V},\) calculate the formation constant \(\left(K_{\mathrm{f}}\right)\) of \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\)

The ingestion of a very small quantity of mercury is not considered too harmful. Would this statement still hold if the gastric juice in your stomach were mostly nitric acid instead of hydrochloric acid?

For each of the following redox reactions, (i) write the half-reactions, (ii) write a balanced equation for the whole reaction, (iii) determine in which direction the reaction will proceed spontaneously under standard-state conditions: (a) \(\mathrm{H}_{2}(g)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{Ni}(s)\) (b) \(\mathrm{MnO}_{4}^{-}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow\) \(\quad \mathrm{Mn}^{2+}(a q)+\mathrm{Cl}_{2}(g)\) (in acid solution) \(\begin{array}{ll}\text { (c) } \mathrm{Cr}(s)+\mathrm{Zn}^{2+}(a q) & \longrightarrow \mathrm{Cr}^{3+}(a q)+\mathrm{Zn}(s)\end{array}\)

Consider a Daniell cell operating under nonstandardstate conditions. Suppose that the cell's reaction is multiplied by \(2 .\) What effect does this have on each of the following quantities in the Nernst equation: (a) \(E,(b) E^{\circ},(c) Q\) (d) \(\ln Q,\) (e) \(n ?\)

Consider the following half-reactions: $$\mathrm{MnO}_{4}^{-}(a q)+8 \mathrm{H}^{+}(a q)+5 e^{-} \overrightarrow{\mathrm{Mn}^{2+}}(a q)+4 \mathrm{H}_{2} \mathrm{O}(l)$$ $$\mathrm{NO}_{3}^{-}(a q)+4 \mathrm{H}^{+}(a q)+3 e^{-} \longrightarrow_{\mathrm{NO}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)}$$
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