Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

What will be the products of electrolysis of an aqueous solution of \(\mathrm{AgNO}_{3}\) with silver electrodes? (a) Ag from Ag anode dissolves while \(\mathrm{Ag}^{+}\)from solution gets deposited on cathode. (b) \(\mathrm{Ag}\) is liberated at cathode and \(\mathrm{O}_{2}\) is liberated at anode. (c) \(\mathrm{Ag}\) at cathode and nitric acid at anode is liberated. (d) No reaction takes place.

Short Answer

Expert verified
The correct products of electrolysis for an aqueous solution of \(\mathrm{AgNO}_{3}\) with silver electrodes are elemental silver at the cathode and the dissolution of silver at the anode. Hence, option (a) is correct.

Step by step solution

01

Understand Electrolysis

Electrolysis is a method of using a direct electric current (DC) to drive an otherwise non-spontaneous chemical reaction. During electrolysis, reduction occurs at the cathode (negative electrode) and oxidation occurs at the anode (positive electrode). In aqueous solutions, water can also undergo electrolysis.
02

Identify the Electrodes and Electrolyte

In our scenario, the electrolyte is an aqueous solution of \(\mathrm{AgNO}_{3}\), and the electrodes are made of silver (Ag).
03

Determine the Likely Reactions at Cathode and Anode

At the cathode, where reduction occurs, the positive \(\mathrm{Ag}^{+}\) ions will likely be reduced to elemental silver (Ag) due to their lower reduction potential compared to water. At the anode, silver (Ag) will likely oxidize to \(\mathrm{Ag}^{+}\) ions since the electrode itself is made of silver, which prefers to oxidize before water in this scenario.
04

Eliminate the Incorrect Options

- Option (a): This option correctly states Ag is deposited on the cathode, but the anode reaction isn't explicitly mentioned.- Option (b): Liberation of \(\mathrm{O}_{2}\) at the anode would only occur if water was being oxidized; however, with silver electrodes, the silver would oxidize preferentially.- Option (c): Nitric acid at the anode is not possible since the \(\mathrm{AgNO}_{3}\) solution provides \(\mathrm{Ag}^{+}\) ions for oxidation, not \(\mathrm{NO}_{3}^{-}\).- Option (d): A reaction will definitely take place.
05

Determine the Correct Option

Given the information and the likely reactions at both the cathode and anode, option (a) is the most accurate description of what occurs during the electrolysis of an aqueous solution of \(\mathrm{AgNO}_{3}\) with silver electrodes. Silver from the anode dissolves as \(\mathrm{Ag}^{+}\), and \(\mathrm{Ag}^{+}\) from the solution gets deposited on the cathode.

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.

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

Key Concepts

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

Electrolysis
Imagine placing two pencils (electrodes) in a glass of water and running electricity through it; you would see bubbles forming around the pencils. This is similar to the process of electrolysis, a fascinating method where we use electric current to spur a chemical reaction that wouldn't happen on its own. In the case of the aquaeous AgNO3 solution, we have silver nitrate dissolved in water, and when electric current flows through, it breaks down into its components. At the negative end, the cathode, we see the silver ions in the solution gain electrons (remember, cathodes attract cations!). This is reduction, where the silver ions become shiny silver metal, plating the electrode. At the positive end, the anode, the reverse happens. Silver atoms lose electrons (oxidation) and enter the solution as silver ions. This fascinating shuffle is constantly renewing the solution with silver ions as the reaction proceeds.

Thus, in simple terms, electrolysis is like a party where ions and electrons are the guests, and the cathode and anode are the dance floor where all the action happens!
Reduction and Oxidation Reactions
To get a good handle on reduction and oxidation reactions, affectionately called redox reactions, think of it as a cosmic dance of electrons. In every redox reaction, there is a give and take. One substance (the reducer) hands over electrons, while another (the oxidizer) eagerly grabs them.

Here's a friendly way to remember: 'LEO the lion says GER.' That means 'Loss of Electrons is Oxidation' and 'Gain of Electrons is Reduction.' In our example with the aquaeous AgNO3, the silver ions are like someone who has lost their hat in the wind (the electrons), and when they reach the cathode, they pick up their hat. That's reduction! At the same time, the silver atoms at the anode are like people giving away their hats to the wind. That's oxidation! Through redox reactions, electrons are the currency that keeps the chemical world spinning.
Electrode Reactions
Let's dive a bit deeper into the realm of electrode reactions. These are the redox reactions that happen at the two poles of an electrolysis setup. At the cathode, we observe the 'reduction' part of the redox, where species gain electrons. You can imagine this as a touchdown in football where the player finally gets the ball in the endzone – a gain!

At the anode, the opposite occurs with 'oxidation.' Think of this as a player losing the ball – an unexpected loss. So, when we look at the silver nitrate solution, we see this clear: at the cathode, silver ions (Ag+) get their electrons and form silver metal (Ag), and at the anode, silver metal gives away electrons and turns into silver ions (Ag+). These transformations are not just abstract ideas; they have real, visible effects like the formation of new substances or the deposition of metals. It's like watching the matter rearrange itself right before our eyes, all thanks to the flow of current through the electrodes.

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

Which of the following is not an example of disproportionation reaction? (a) \(4 \mathrm{ClO}_{3}^{-} \rightarrow \mathrm{Cl}^{-}+3 \mathrm{ClO}_{4}^{-}\) (b) \(2 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}\) (c) \(2 \mathrm{NO}_{2}+2 \mathrm{OH}^{-} \rightarrow \mathrm{NO}_{2}^{-}+\mathrm{NO}_{3}^{-}+\mathrm{H}_{2} \mathrm{O}\) (d) \(\mathrm{TiCl}_{4}+2 \mathrm{Mg} \rightarrow \mathrm{Ti}+2 \mathrm{MgCl}_{2}\)

Oxidation numbers of \(\mathrm{Mn}\) in its compounds \(\mathrm{MnCl}_{2}\) \(\mathrm{Mn}(\mathrm{OH})_{3}, \mathrm{MnO}_{2}\) and \(\mathrm{KMnO}_{4}\) respectively are (a) \(+2,+4,+7,+3\) (b) \(+2,+3,+4,+7\) (c) \(+7,+3,+2,+4\) (d) \(+7,+4,+3,+2\)

Which of the following is a disproportionation reaction? \((c): 2 \mathrm{Fe}_{(6)}+3 \mathrm{H}_{2} \mathrm{O}_{(D} \stackrel{\Delta}{\longrightarrow} \mathrm{Fe}_{2} \mathrm{O}_{3(j)}+3 \mathrm{H}_{2(g)}\) d) \(2 \mathrm{H}_{2} \mathrm{O}_{(\bar{D}}+2 \mathrm{~F}_{2(\mathrm{e})} \rightarrow 4 \mathrm{HF}_{\left(a_{q}\right)}+\mathrm{O}_{2(\mathrm{~g})}\)

When a piece of sodium metal is dropped in water, hydrogen gas evolved because (a) sodium is reduced and acts as an oxidising agent (b) water is oxidised and acts as a reducing agent (c) sodium loses electrons and is oxidised while water in reduced (d) water loses electrons and is oxidised to hydrogen.

Given below are few statements regarding electrode potentials. Mark the correct statements. (i) The potential associated with each electrode is known as electrode potential. (ii) A negative \(E^{\circ}\) means that the redox couple is a stronger reducing agent than \(\mathrm{H}^{+} / \mathrm{H}_{2}\) couple. (iii) A positive \(E^{\circ}\) means that the redox couple is a weaker reducing agent than \(\mathrm{H}^{+} / \mathrm{H}_{2}\) couple.

See all solutions

Recommended explanations on Chemistry Textbooks

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