During the electrolysis of dilute sulphuric acid, the following process is possible at anode.

Question

During the electrolysis of dilute sulphuric acid, the following process is possible at anode. (A) 2H2O(l) → O2(g)+4H+(aq)+4e- (B) 2SO2-4(aq) → S2O2-8(aq)+2e- (C) H2O(l) → H+(aq)+OH-(aq) (D) H2O(l)+e- →(1/2) H2(g)+OH-(aq)

Tardigrade Admin · Accepted Answer

During electrolysis of H 2 SO 4 H 2 is evolved at cathode. 2 H ++2 e - longrightarrow H 2 O 2 is evolved at anode, 2 H 2 O longrightarrow O 2+4 H ++4 e -

Q. During the electrolysis of dilute sulphuric acid, the following process is possible at anode.

$2 H_{2} O_{(l)} \to O_{2 (g)} + 4 H_{(a q)}^{+} + 4 e^{-}$

$2 S O_{4 (a q)}^{2 -} \to S_{2} O_{8 (a q)}^{2 -} + 2 e^{-}$

$H_{2} O_{(l)} \to H_{(a q)}^{+} + O H_{(a q)}^{-}$

$H_{2} O_{(l)} + e^{-} \to \frac{1}{2} H_{2 (g)} + O H_{(a q)}^{-}$

During electrolysis of $H_{2} S O_{4}$
$H_{2}$ is evolved at cathode.
$2 H^{+} + 2 e^{-} ⟶ H_{2}$
$O_{2}$ is evolved at anode,
$2 H_{2} O ⟶ O_{2} + 4 H^{+} + 4 e^{-}$

Q. During the electrolysis of dilute sulphuric acid, the following process is possible at anode.

2H2​O(l)​→O2(g)​+4H(aq)+​+4e−

2SO4(aq)2−​→S2​O8(aq)2−​+2e−

H2​O(l)​→H(aq)+​+OH(aq)−​

H2​O(l)​+e−→21​H2(g)​+OH(aq)−​

During electrolysis of H2​SO4​ H2​ is evolved at cathode. 2H++2e−⟶H2​ O2​ is evolved at anode, 2H2​O⟶O2​+4H++4e−

$2 H_{2} O_{(l)} \to O_{2 (g)} + 4 H_{(a q)}^{+} + 4 e^{-}$

$2 S O_{4 (a q)}^{2 -} \to S_{2} O_{8 (a q)}^{2 -} + 2 e^{-}$

$H_{2} O_{(l)} \to H_{(a q)}^{+} + O H_{(a q)}^{-}$

$H_{2} O_{(l)} + e^{-} \to \frac{1}{2} H_{2 (g)} + O H_{(a q)}^{-}$

During electrolysis of $H_{2} S O_{4}$
$H_{2}$ is evolved at cathode.
$2 H^{+} + 2 e^{-} ⟶ H_{2}$
$O_{2}$ is evolved at anode,
$2 H_{2} O ⟶ O_{2} + 4 H^{+} + 4 e^{-}$