On electrolysis of dil.sulphuric acid using Platinum (Pt) electrode, the product obtained at anode will be:

Question

On electrolysis of dil.sulphuric acid using Platinum (Pt) electrode, the product obtained at anode will be: (A) Hydrogen gas (B) Oxygen gas (C) H2S gas (D) SO2 gas

Tardigrade Admin · Accepted Answer

Dissociation of sulfuric acid, H 2 SO 4 arrow 2 H +( aq )+ SO 4-2( aq ) Disssociation of water, H 2 O arrow H +( aq )+ OH -( aq ) At Anode: 4 OH - arrow 2 H 2 O +( l )+ O 2( g )+4 e - At Cathode: 2 H +( aq )+2 e - arrow H 2( g )

Q. On electrolysis of dil.sulphuric acid using Platinum (Pt) electrode, the product obtained at anode will be:

Hydrogen gas

Oxygen gas

$H_{2} S$ gas

$S O_{2}$ gas

Dissociation of sulfuric acid,
$H_{2} S O_{4} \to 2 H^{+} (a q) + S O_{4}^{- 2} (a q)$
Disssociation of water,
$H_{2} O \to H^{+} (a q) + O H^{-} (a q)$
At Anode : $4 O H^{-} \to 2 H_{2} O^{+} (l) + O_{2} (g) + 4 e^{-}$
At Cathode: $2 H^{+} (a q) + 2 e^{-} \to H_{2} (g)$

Q. On electrolysis of dil.sulphuric acid using Platinum (Pt) electrode, the product obtained at anode will be:

Hydrogen gas

Oxygen gas

H2​S gas

SO2​ gas

Dissociation of sulfuric acid, H2​SO4​→2H+(aq)+SO4−2​(aq) Disssociation of water, H2​O→H+(aq)+OH−(aq) At Anode : 4OH−→2H2​O+(l)+O2​(g)+4e− At Cathode: 2H+(aq)+2e−→H2​(g)

$H_{2} S$ gas

$S O_{2}$ gas

Dissociation of sulfuric acid,
$H_{2} S O_{4} \to 2 H^{+} (a q) + S O_{4}^{- 2} (a q)$
Disssociation of water,
$H_{2} O \to H^{+} (a q) + O H^{-} (a q)$
At Anode : $4 O H^{-} \to 2 H_{2} O^{+} (l) + O_{2} (g) + 4 e^{-}$
At Cathode: $2 H^{+} (a q) + 2 e^{-} \to H_{2} (g)$