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Electrochemical Series

What will the...

Question

What will the equations at the cathode and anode, when acidified water is electrolysed?

At cathode:

2 H^{+} + 2 e^{-} \to 2 H \to H_{2}

At anode:

2 O H^{-} - 2 e^{-} \to 2 O H \to H_{2} O + O

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At anode:

2 H^{+} + 2 e^{-} \to 2 H \to H_{2}

At cathode:

2 O H^{-} - 2 e - \to 2 O H \to H_{2} O + O

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At cathode:

2 H^{+} + 2 e^{-} \to H_{2}

At anode:

2 O H^{-} - 2 e^{-} \to 2 O H \to H_{2} O + O

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None of the above

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Solution

The correct option is D None of the above
During electrolysis of water,
At anode: $2 H_{2} O \to O_{2} + 4 H^{+} + 4 e^{-}$
At cathode: $4 H_{2} O + 4 e^{-} \to 2 H_{2} + 4 O H^{-}$

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Similar questions

{2 H}_{2} O + {2 e}^{-} \to H_{2} + 2 {O H}^{-}

{x O H}^{-} \to O_{2} + {2 H}_{2} O + {y e}^{-}

{2 H}_{2} O \to 2 H_{2} + O_{2}

Given are the reactions occuring during the electrolysis of sodium chloride solution:

$2 C l^{-} \to C l_{2} + 2 e^{-}$
$2 H_{2} O + 2 e^{-} \to H_{2} + 2 O H^{-}$

Which of the following is correct regarding liberation of electrolytic products?

(a) O_{3} + 2 H^{+} + 2 e^{-} ⇌ O_{2} + H_{2} O

(b) O_{3} + H_{2} O + 2 e^{-} ⇌ O_{2} + 2 O H^{-}

O_{3}

H_{2} O_{2}

H_{2} O_{2} + 2 H^{+} + 2 e^{-} \to 2 H_{2} O

H_{2} O_{2} - 2 e^{-} \to O_{2} + 2 H^{+}

H_{2} O_{2} + 2 e^{-} \to 2 O H^{-}

H_{2} O_{2} + 2 O H^{-} 2 e^{-} \to O_{2} + 2 H_{2} O

H_{2} O_{2}

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