You visited us 1 times! Enjoying our articles? Unlock Full Access!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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.
${N a}^{+} + e^{-} \to N a$

Suggest Corrections

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

Join BYJU'S Learning Program