Question

♣What will be the products of electrolysis of an aqueous solution of AgNO3 in water with :-

(i) An aqueous solution of AgNO3 with silver electrodes
(ii) An aqueous solution of AgNO3 with plantinum electrodes
(iii) A dilute solution of H2SO4 with platinum electrodes

♠Answer in detail, Thanks! ☺️​

Answer 1

Answer:

1) When the electrode of the same material is used for example Ag. The metal itself gets oxidized and reduced on an electrode.

Cathode - Reduction Ag

+

+e

−

⟶Ag

Anode - Oxidation Ag⟶Ag

+

+e

−

When platinum electrodes are used, Ag

+

from solution is reduced and deposited at cathode whereas O

2

is produced at the anode.

Cathode - Reduction Ag

+

+e

−

⟶Ag

Anode - Oxidation 2H

2

O⟶4H

+

+O

2

+4e

−

Explanation:

please mark me ad BRAINLIEST

Answer 2

Answer:

[Formed products are at anode and cathode respectively]

$(i) \ \mathbf {Ag^+_{(aq)}} \ and \ \mathbf {Ag_{(s)}}\\(ii) \ \mathbf {O_{2(g)}} \ and \ \mathbf {Ag_{(s)}}\\(iii) \ \mathbf {O_{2(g)}} \ and \ \mathbf {H_{2(g)}}$

Explanation:

(i) An aqueous solution of AgNO3 with silver electrodes:

If the metal which is used for electrolysis is used as electrodes, then they are called as active electrodes. Their chemical composition changes in electrolysis.

So, in case of this type of electrolysis, active electrodes at anode dissociates first. Here, Ag is the active electrode

We always know that, oxidation takes place anode and reduction takes place at cathode. Negative ions participate at anode(+ve electrode) and +ve ions  participate at cathode(-ve electrode). But, in case of active electrodes, metal participates at anode. Hence, electrons are released at anode and are absorbed at the cathode.

As AgNO₃ is a ionic compound, it dissociates. (But as it is aqueous H₂O also lies in the Solution)

$AgNO_3 \longrightarrow Ag^+ + NO_3^-$

$\underline {{\text{At active Ag anode(+ve)}}} : [Oxidation] \\\\=> Ag_{(s)} \longrightarrow Ag^+_{(aq)} + e^-$

We know that, at cathode, positive ion, here, Ag⁺ will take part at the electrolysis.

$\underline {{\text{At cathode(-ve)}}} : [Reduction] \\\\=> Ag^+_{(aq)} + e^-\longrightarrow Ag_{(s)}$

So, the products formed at anode and cathode are $\mathbf {Ag^+_{(aq)}}$ and $\mathbf {Ag_{(s)}}$ respectively. (Maximum we can say nothing is formed at anode as ions are not considered to be formed.)

-------------------------------

(ii) An aqueous solution of AgNO3 with platinum electrodes

Electrodes like Pt, Fe, Graphite(Carbon) are called inert electrodes. These electrodes do not change their composition at the electrodes.

Now, we need to electrolyze, aqueous solution of AgNO3 which means it contains, silver and nitrate ions along with water(aqueous sol⁻ⁿ)

As AgNO₃ is a ionic compound, it dissociates. (But as it is aqueous H₂O also lies in the solution)

$AgNO_3 \longrightarrow Ag^+ + NO_3^-$

Here, as there is no active electrode, at anode, negative ions will participate as it is a positive electrode. In the solution, we have, 2 negative ions (1) NO₃⁻ and (2) H₂O (water is a bad electrolyte. It doesn't divide like H⁺ and OH⁻ unless there is a impurity{small amount of salt, acid and base}. So, H₂O directly participates in the reaction)

[Note : In some of the books it is mentioned that water dissociates and then participate. But some of them say, water is a bad electrolyte and water directly participates in the electrolysis. I follow the second method. I will do with that method now.]

Here, the main anion NO₃⁻ should take part at anode. But, in presence of water, oxoanions (SO₄⁻², SO₄⁻³, NO₃⁻, NO₂⁻, etc => Anions with oxygen) do not take participation at cathode. The reason is the ion which has lesser discharge potential gets oxidized or reduces at the respective electrode. H₂O has more lesser discharge potential that NO₃⁻ ion. So, H₂O takes place at cathode.

$\underline {{\text{At anode(+ve)}}} : [Oxidation] \\\\=> 2H_2O \longrightarrow O_{2(g)} +4H^+ + 4e^-$

We know that, at cathode positive ion, here, Ag⁺ will take part at the electrolysis. [Electric discharge potential of Ag⁺ is less than the water. Hence Ag⁺ only participate]

$\underline {{\text{At cathode(-ve)}}} : [Reduction] \\\\=> Ag^+_{(aq)} + e^-\longrightarrow Ag_{(s)}$

So, the products formed at anode and cathode are $\mathbf {O_{2(g)}}$ and $\mathbf {Ag_{(s)}}$ respectively.

---------------------------------

(iii) A dilute solution of H2SO4 with platinum electrodes:

As H₂SO₄ is a ionic compound, it dissociates. (But as it is aqueous H₂O also lies in the solution)

$H_2SO_4 \longrightarrow 2H^+ + SO_4^{-2}$

At anode, H₂O participate as oxoanions do not participate. (Electric discharge potential of H₂O is less than the SO₄⁻². Hence H₂O only participate)

$\underline {{\text{At anode(+ve)}}} : [Oxidation] \\\\=> 2H_2O \longrightarrow O_{2(g)} +4H^+ + 4e^-$

At cathode, formed H⁺ ions at anode participate.

$\underline {{\text{At cathode(-ve)}}} : [Reduction] \\\\=> 4H^+_{(aq)} + 4e^-\longrightarrow 2H_{2(g)}$

So, the products formed at anode and cathode are $\mathbf {O_{2(g)}}$ and $\mathbf {H_{2(g)}}$ respectively.

_____________________

❖ Extra information:

⟡ The order of discharge potential at cathode :

K⁺ > Ca⁺² > Na⁺ > Mg⁺² > Al⁺³ > Zn⁺²> Fe⁺² > Ni⁺ > Sn⁺² > H⁺(H₂O) > Cu⁺² > Ag⁺ > Hg⁺² > Au⁺³

⟡ The order of discharge potential at anode :

F⁻ > SO₄⁻² > NO₃⁻ > OH⁻ (H₂O) > Cl⁻ > Br⁻ > I⁻

[Note down this order and learn it. It could be useful for other questions.]

⟡ H₂O at cathode is according to 2nd method (bad electrolyte method) participates as

$2H_2O + 2e^- \longrightarrow H_{2(g)} + 4OH^-$

Hope it helps!