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Electrometallurgy

Brine is elec...

Question

Brine is electrolysed by using inert electrodes. The reaction at anode is _____.

{C l}_{(a q)}^{-} \to \frac{1}{2} C l_{2 (g)} + e^{-}; E_{c e l l}^{0} = 1.36 V

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2 H_{2} O_{(l)} \to O_{2 (g)} + 4 H^{+} + 4 e^{-}; E_{c e l l}^{0} = 1.23 V

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N a_{(a q)}^{+} + e^{-} \to N a_{(s)}; E_{c e l l}^{0} = 2.71 V

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H_{(a q)}^{+} + e^{-} \to \frac{1}{2} H_{2 (g)}; E_{c e l l}^{0} = 0.00 V

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Solution

The correct option is A ${C l}_{(a q)}^{-} \to \frac{1}{2} C l_{2 (g)} + e^{-}; E_{c e l l}^{0} = 1.36 V$
Although reduction potential of ${C l}^{-}$ ion is higher than that of $H_{2} O$ , yet oxidation of ${C l}^{-}$ ions occurs in preference to $H_{2} O$ due to over voltage of $O_{2}$ (i.e., water needs greater voltage for oxidation of $O_{2}$ as it is a kinetically slow process.

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Similar questions

△_{r} G^{0}

M g (s) + C u^{2 +} (a q) \to M g^{2 +} (a q) + C u (s)

E_{c e l l}^{0} = + 2.71 V, 1 F = 96500 C m o l^{- 1}

O_{2} (g)

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} (c o r r e c t) C u (s) ⟶ & C u^{2 +} (a q) + 2 e^{-} E_{C u | C u^{2 +}}^{⊝} = - 0.34 V (w r o n g) 2 H_{2} O (l) ⟶ & O_{2} (g) + 4 H^{\oplus} (a q) + 4 e^{-} E_{H_{2} O | O_{2}}^{⊝} = - 1.23 V \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

\begin{matrix} I n (d) \end{matrix}

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} (c o r r e c t) N i (s) ⟶ & N i^{2 +} (a q) + 2 e^{-} E_{N i | N i^{2 +}}^{⊝} = - 0.25 V (w r o n g) 4 ⊝ O H (a q) ⟶ & O_{2} (g) + 2 H_{2} O (l) + 4 e^{-} E_{⊝ O H | O_{2}}^{\oplus} = - 0.4 V \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

(E_{c e l l}^{0})

M g^{2 +} + 2 e^{-} \to M g (s); E^{0} = - 2.37 V

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

log K_{e q}

298 K

K_{e q} is the equilibrium constant for the given reaction

C u (s) + 2 A g^{+} (a q) \to C u^{2 +} (a q) + 2 A g (s); E_{c e l l}^{0} = 0.46 V

2 H_{2} (g) + O_{2} (g) ⟶ 2 H_{2} O (l); △_{r} S_{298}^{\circ} = - 0.32 k J / K

△_{f} H_{298}^{\circ} (H_{2} O, l)

O_{2} (g) + 4 H^{+} (a q .) + 4 e^{-} ⟶ 2 H_{2} O (l); E^{\circ} = 1.23 V

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