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Electrolysis and Electrolytes

In Dow's meth...

Question

In Dow's method, sodium is prepared by the electrolysis of molten $N a C l$ . The reaction at the cathode is :

2 C l^{-} \to C l_{2} (g) + 2 e

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N a^{+} + e \to N a (s)

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N a^{+} (a q) + e \to N a^{+} + e

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N a^{+} (a q) + e \to N a (s)

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Solution

The correct option is B $N a^{+} + e \to N a (s)$
In Dow's process,
$N a^{+}$ ions migrate to cathode where they are reduced to Na.
${N a}^{+} + e^{-} \to N a$

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Similar questions

N a^{+} (a q) + e^{-} \to N a (s) E^{0} = - 2.71 V

H^{+} (a q) + e^{-} \to \frac{1}{2} H_{2} (g) E^{0} = 0.00 V

(E^{0})

N a^{+} (a q) + e^{-} \to N a (s) E^{0} = - 2.71 V

H^{+} (a q) + e^{-} \to \frac{1}{2} H_{2} (g) E^{0} = 0.00 V

(E^{0})

(a) Following reactions occur at cathode during the electrolysis of aqueous sodium chloride solution:

$N a^{+} (a q) + e^{-} \to N a (s) E^{\circ} = - 2.71 V$
$H^{+} (a q) + e^{-} \to \frac{1}{2} H_{2} (g) E^{\circ} = 0.00 V$

On the basis of their standard reduction electrode potential $(E^{\circ})$ values, which reaction is feasible at the cathode and why?

(b) Why does the cell potential of mercury cell remain constant throughout its life?

N a (s) \to N a^{+} (a q) + e^{-}

{C o}^{3 +} (a q) + e^{-} \to {C o}^{2 +} (a q)

E^{o} = 1.82 V

{N a}^{+} (a q) + e^{-} \to N a (s)

E^{o} = - 2.71 V

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