Chemistry Class 11 NCERT Solutions: Chapter 8 Redox Reactions Part 14

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Q: 27. Predict the products of electrolysis in each of the following:

(i) An aqueous solution of with silver electrodes

(ii) An aqueous solution with platinum electrodes

(iii) A dilute solution of with platinum electrodes

(iv) An aqueous solution of with platinum electrodes.

Answer:

(i) ionizes in aqueous solutions to form and ions.

On electrolysis, either ions or molecules can be reduced at the cathode. But the reduction potential of ions is higher than that of .

Hence, ions are reduced at the cathode. Similarly, metal or molecules can be oxidized at the anode. But the oxidation potential of is higher than that of molecules.

Therefore, Ag metal gets oxidized at the anode.

(ii) Pt cannot be oxidized easily. Hence, at the anode, oxidation of water occurs to liberate . At the cathode, ions are reduced and get deposited.

(iii) ionizes in aqueous solutions to give and ions

On electrolysis, either of ions or molecules can get reduced at the cathode. But

the reduction potential of ions is higher than that of molecules.

Hence, at the cathode, ions are reduced to liberate gas.

On the other hand, at the anode, either of ions or molecules can get oxidized. But the oxidation of involves breaking of more bonds than that of molecules. Hence, ions have a lower oxidation potential than . Thus, is oxidized at the anode to liberate molecules.

(iv) In aqueous solutions, ionizes to give and ions as:

On electrolysis, either of ions or molecules can get reduced at the cathode. But the reduction potential of is more than that of molecules

Hence, ions are reduced at the cathode and get deposited. Similarly, at the anode, either of or is oxidized. The oxidation potential of is higher than that of .

But oxidation of molecules occurs at a lower electrode potential than that of ions because of over-voltage (extra voltage required to liberate gas). As a result, ions are oxidized at the anode to liberate gas.

Q: 28. Arrange the following metals in the order in which they displace each other from the solution of their salts.

Answer:

A metal of stronger reducing power displaces another metal of weaker reducing power from its solution of salt.

The order of the increasing reducing power of the given metals is

Hence, we can say that Mg can displace Al from its salt solution, but Al cannot displace .

Thus, the order in which the given metals displace each other from the solution of their salts is given below: