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Electrolysis and Electrolytes

In the electr...

Question

In the electrolysis of $N i S O_{4}$ using Nickel electrodes the reaction that takes place at anode is:

N i^{+ 2} + 2 e^{-} \to N i

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

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N i^{+ 2} \to N i^{+} + e^{-}

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N i \to N i^{+ 2} + 2 e^{-}

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Solution

The correct option is B $N i \to N i^{+ 2} + 2 e^{-}$
During electrolysis of nickel sulphate using nickel electrode, at anode, oxidation of nickel occurs. Nickel loses two electrons to form ${N i}^{2 +}$ ion.

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Similar questions

E

E^{\circ}

N i | N i^{2 +} | | C u^{2 +} | C u

N i^{2 +} + 2 e^{-} \to N i; E^{\circ} = - 0.25 v o l t

C u^{2 +} + 2 e^{-} \to C u; E^{\circ} = + 0.34 v o l t

[N i^{2 +}] = 1 M

[C u^{2 +}] = 10^{- 3} M)

If the reduction potential $N i \to N i^{2 +} + 2 e^{-}$ is $- 0.25 V$ . ___ V is its oxidation potential.

N i^{+ 2} + 2 e^{-} ⟶ N i; E^{0} = - 0.25 V

Z n^{+ 2} + 2 e^{-} ⟶ Z n; E^{0} = - 0.77 V

F e^{2 +} + 2 e^{-} \to F e (- 0.44 V)

N i^{2 +} + 2 e^{-} \to N i (- 0.25 V)

S n^{2 +} + 2 e^{-} \to S n (- 0.14 V)

F e^{3 +} + e^{-} \to F e^{2 +} (+ 0.77 V)

Z n^{2 +} (a q) + 2 e^{-} \to Z n (s); E^{0} = - 0.76 V

C u^{2 +} (a q) + 2 e^{-} \to C u (s); E^{0} = 0.34 V

A g^{+} (a q) + e^{-} \to A g (s); E^{0} = 0.80 V

2 H^{+} (a q) + 2 e^{-} \to H_{2} (g); E^{0} = 0 V

N i^{2 +} (a q) + 2 e^{-} \to N i (s); E^{0} = - 0.25 V

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