Updated By SaraNextGen

On March 11, 2024, 11:35 AM

Question 8.27:

Predict the products of electrolysis in each of the following:

(i) An aqueous solution of AgNO₃ with silver electrodes

(ii) An aqueous solution AgNO₃ with platinum electrodes

(iii) A dilute solution of H₂SO₄ with platinum electrodes

(iv) An aqueous solution of CuCl₂ with platinum electrodes.

Answer:

(i) AgNO₃ ionizes in aqueous solutions to form Ag⁺ and   ions.

On electrolysis, either Ag⁺ ions or H₂O molecules can be reduced at the cathode. But the reduction potential of Ag⁺ ions is higher than that of H₂O.

Hence, Ag⁺ ions are reduced at the cathode. Similarly, Ag metal or H₂O molecules can be oxidized at the anode. But the oxidation potential of Ag is higher than that of H₂O 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 O₂. At the cathode, Ag⁺ ions are reduced and get deposited.

(iii) H₂SO₄ ionizes in aqueous solutions to give H⁺ and  ions.

On electrolysis, either of H⁺ ions or H₂O molecules can get reduced at the cathode. But the reduction potential of H⁺ ions is higher than that of H₂O molecules.

Hence, at the cathode, H⁺ ions are reduced to liberate H₂ gas.

On the other hand, at the anode, either of  ions or H₂O molecules can get oxidized. But the oxidation of  involves breaking of more bonds than that of H₂O molecules. Hence,  ions have a lower oxidation potential than H₂O. Thus, H₂O is oxidized at the anode to liberate O₂ molecules.

(iv) In aqueous solutions, CuCl₂ ionizes to give Cu²⁺ and Cl^– ions as:

On electrolysis, either of Cu²⁺ ions or H₂O molecules can get reduced at the cathode. But the reduction potential of Cu²⁺ is more than that of H₂O molecules.

Hence, Cu²⁺ ions are reduced at the cathode and get deposited.

Similarly, at the anode, either of Cl^– or H₂O is oxidized. The oxidation potential of H₂O is higher than that of Cl^–.

But oxidation of H₂O molecules occurs at a lower electrode potential than that of Cl^– ions because of over-voltage (extra voltage required to liberate gas). As a result, Cl^– ions are oxidized at the anode to liberate Cl₂ gas.

Question 8.28:

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.

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 Cu < Fe < Zn < Al < Mg.

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

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

Mg>Al> Zn> Fe,>Cu

Question 8.29:

Given the standard electrode potentials,

K⁺/K = –2.93V, Ag⁺/Ag = 0.80V,

Hg²⁺/Hg = 0.79V

Mg²⁺/Mg = –2.37V. Cr³⁺/Cr = –0.74V

Arrange these metals in their increasing order of reducing power.

Answer:

The lower the electrode potential, the stronger is the reducing agent. Therefore, the increasing order of the reducing power of the given metals is Ag < Hg < Cr < Mg < K.

Question 8.30:

Depict the galvanic cell in which the reaction Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s) takes place, further show:

(i) which of the electrode is negatively charged,

(ii) the carriers of the current in the cell, and

(iii) individual reaction at each electrode.

Answer:

The galvanic cell corresponding to the given redox reaction can be represented as:

(i) Zn electrode is negatively charged because at this electrode, Zn oxidizes to Zn²⁺ and the leaving electrons accumulate on this electrode.

(ii) Ions are the carriers of current in the cell.

(iii) The reaction taking place at Zn electrode can be represented as:

And the reaction taking place at Ag electrode can be represented as:

 .