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Chapter 8: Problem 27

Predict the products of electrolysis in each of the following: (i) An aqueous solution of \(\mathrm{AgNO}_{3}\) with silver electrodes (ii) An aqueous solution \(\mathrm{AgNO}_{3}\) with platinum electrodes (iii) A dilute solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) with platinum electrodes (iv) An aqueous solution of \(\mathrm{CuCl}_{2}\) with platinum electrodes.

Short Answer

Expert verified
(i) Ag deposited at cathode, silver oxidizes at anode. (ii) Ag deposited at cathode, O鈧� at anode. (iii) H鈧� at cathode, O鈧� at anode. (iv) Cu deposited at cathode, Cl鈧� at anode.

Step by step solution

01

Analyze electrolysis with silver electrodes in AgNO鈧� solution

In an aqueous solution of \( \mathrm{AgNO}_3 \) with silver electrodes, the products of electrolysis depend on the material of the electrode and the ions present in the solution. At the anode, silver will oxidize: \( \mathrm{Ag(s) \rightarrow Ag^+(aq) + e^-} \). At the cathode, the \( \mathrm{Ag^+} \) ions from the solution will reduce: \( \mathrm{Ag^+(aq) + e^- \rightarrow Ag(s)} \). The silver electrode acts as both the source and deposition site of silver atoms.
02

Examine electrolysis with platinum electrodes in AgNO鈧� solution

For an aqueous solution of \( \mathrm{AgNO}_3 \) with platinum electrodes, the reaction will differ from that with silver electrodes. At the anode, water is oxidized because \ \( \mathrm{Pt} \) electrodes are inert, forming \( \mathrm{O_2(g)} \) and \( \mathrm{H^+(aq)} \). At the cathode, \( \mathrm{Ag^+} \) ions are reduced to form silver metal: \( \mathrm{Ag^+(aq) + e^- \rightarrow Ag(s)} \). Thus, silver deposits on the cathode, and oxygen gas evolves from the anode.
03

Predict electrolysis products for dilute H鈧係O鈧� with platinum electrodes

In the electrolysis of a dilute solution of \( \mathrm{H_2SO_4} \) with platinum electrodes, the main ions in solution are \( \mathrm{H^+} \) and \( \mathrm{SO_4^{2-}} \). At the cathode, \( \mathrm{H^+(aq)} \) ions are reduced to form hydrogen gas: \( \mathrm{2H^+(aq) + 2e^- \rightarrow H_2(g)} \). At the anode, water is oxidized to produce oxygen gas: \( \mathrm{2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^-} \). Thus, hydrogen gas forms at the cathode and oxygen gas forms at the anode, with no deposition on the electrodes.
04

Determine electrolysis products for CuCl鈧� with platinum electrodes

In an aqueous solution of \( \mathrm{CuCl_2} \) with platinum electrodes, the ions present are \( \mathrm{Cu^{2+}} \) and \( \mathrm{Cl^-} \). At the cathode, \( \mathrm{Cu^{2+}} \) ions are reduced to copper metal: \( \mathrm{Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)} \). At the anode, \( \mathrm{Cl^-} \) ions are oxidized to chlorine gas: \( \mathrm{2Cl^-(aq) \rightarrow Cl_2(g) + 2e^-} \). Therefore, copper metal deposits at the cathode, and chlorine gas is released at the anode.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Silver Electrodes
Silver electrodes play a crucial role in the electrolysis of silver nitrate solutions. They not only conduct electricity but also participate actively in the electrochemical process. Here's how:
  • Anode Reaction: At the anode, silver from the electrode oxidizes, losing electrons to form silver ions in the solution. The reaction can be represented as \( \mathrm{Ag(s) \rightarrow Ag^+(aq) + e^-} \). This means the silver metal is transitioning into solution.
  • Cathode Reaction: Simultaneously, at the cathode, silver ions in the solution gain electrons and reduce to form solid silver. The reaction is \( \mathrm{Ag^+(aq) + e^- \rightarrow Ag(s)} \). Here, ions in the solution become solid and coat the cathode.
This cycle of dissolution and deposition allows the silver to continuously flow to the cathode, eventually leading to a change in the mass and morphology of the electrodes.
Platinum Electrodes
Platinum electrodes are often chosen for electrolysis due to their inert nature. Unlike silver electrodes, platinum does not react with the ions in the solution, making it a versatile choice for different electrolytic solutions.
  • Inert Nature: Platinum does not oxidize or reduce itself during electrolysis. This property means it does not interfere with the reactions of the ions in the solution, allowing for a clear study of the electrolytes involved.
  • Electrolysis in AgNO鈧� Solution: For an aqueous solution of \( \mathrm{AgNO}_{3} \), platinum serves as an efficient platform where \( \mathrm{Ag^+} \) ions reduce at the cathode to form solid silver, while at the anode, water oxidizes to produce oxygen gas, thanks to its non-reactive nature.
Platinum electrodes therefore serve as a passive medium critical for carrying out electrolysis without contributing to any side reactions.
Electrolysis of Aqueous Solutions
Electrolysis of aqueous solutions involves using electricity to drive chemical reactions that would not otherwise occur naturally in water. The setup typically involves an electrolytic cell containing a liquid (or aqueous) solution of ions.
  • Cathode and Anode Reactions: During electrolysis, cations (positive ions) are attracted to the cathode (negative electrode), where they gain electrons in a reduction reaction. Conversely, anions (negative ions) migrate to the anode (positive electrode), where they lose electrons in an oxidation reaction.
  • Example with \( \mathrm{H_2SO_4} \): For a dilute \( \mathrm{H_2SO_4} \) solution with platinum electrodes, the reactions involve the formation of hydrogen gas at the cathode as \( \mathrm{H^+(aq) + e^- \rightarrow H_2(g)} \) and oxygen gas at the anode as \( \mathrm{2H_2O(l) \rightarrow O_2(g) + 4H^+(aq) + 4e^-} \). This showcases how water itself can participate in electrolysis.
These processes highlight the ability to drive chemical changes through electrical energy, leading to the formation of new substances.
Redox Reactions
Redox reactions are fundamental to the process of electrolysis. The term 'redox' is short for reduction-oxidation, comprising two complementary processes:
  • Reduction: This is the gain of electrons by ions or molecules, decreasing their oxidation state. In the electrolysis of \( \mathrm{AgNO_3} \), for example, \( \mathrm{Ag^+} \) ions are reduced at the cathode to metallic silver, \( \mathrm{Ag(s)} \).
  • Oxidation: This involves the loss of electrons, increasing the oxidation state of the element. An example is the oxidation of \( \mathrm{Cl^-} \) ions to \( \mathrm{Cl_2(g)} \) at the anode in the electrolysis of \( \mathrm{CuCl_2} \).
Understanding redox reactions allows us to predict the products of electrolysis. The movement of electrons between the anode and cathode drives these reactions, transforming the chemical species involved and leading to diverse applications in industrial processes.

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Most popular questions from this chapter

Write formulas for the following compounds: (a) Mercury(II) chloride (b) Nickel(II) sulphate (c) Tin(IV) oxide (d) Thallium(I) sulphate (e) Iron(III) sulphate (f) Chromium(III) oxide

Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn.

Consider the reactions: (a) \(\mathrm{H}_{3} \mathrm{PO}_{2}(\mathrm{aq})+4 \mathrm{AgNO}_{3}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq})+4 \mathrm{Ag}(\mathrm{s})+4 \mathrm{HNO}_{3}(\mathrm{aq})\) (b) \(\mathrm{H}_{3} \mathrm{PO}_{2}(\mathrm{aq})+2 \mathrm{CuSO}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq})+2 \mathrm{Cu}(\mathrm{s})+\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) (c) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}(\mathrm{l})+2\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}\right]^{+}(\mathrm{aq})+3 \mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}^{-}(\mathrm{aq})+2 \mathrm{Ag}(\mathrm{s})+\) \(4 \mathrm{NH}_{3}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) (d) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CHO}(\mathrm{l})+2 \mathrm{Cu}^{2+}(\mathrm{aq})+5 \mathrm{OH}^{-}(\mathrm{aq}) \rightarrow\) No change observed. What inference do you draw about the behaviour of \(\mathrm{Ag}^{+}\) and \(\mathrm{Cu}^{2+}\) from these reactions?

Assign oxidation number to the underlined elements in each of the following species: (a) \(\mathrm{NaH}_{2} \mathrm{PO}_{4}\) (b) \(\mathrm{NaH} \underline{\mathrm{S} O}_{4}\) (c) \(\mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{7}\) (d) \(\mathrm{K}_{2} \underline{\mathrm{Mn}} \mathrm{O}_{4}\) (e) \(\mathrm{Ca} \underline{\mathrm{O}}_{2}\) (f) \(\mathrm{NaB} \mathrm{H}_{4}\) (g) \(\mathrm{H}_{2} \underline{\mathrm{S}}_{2} \mathrm{O}_{7}\) (h) \(\operatorname{KAl}\left(\underline{\mathrm{S}} \mathrm{O}_{4}\right)_{2} \cdot 12 \mathrm{H}_{2} \mathrm{O}\)

Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.
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