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Faraday's Second Law

In the electr...

Question

In the electrolysis of aqueous sodium chloride solution which of the half-cell reaction will occur at the anode?

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

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2 H_{2} O_{(l)} \to O_{2 (g)} + 4 H_{(a q)}^{+} + 4 e^{-}; E_{c e l l}^{o} = 1.23

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

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

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Solution

The correct option is D $C l_{(a q)}^{-} \to \frac{1}{2} C l_{2 (g)} + e^{-}; E_{c e l l}^{o} = 1.36$ V
At anode, oxidation takes place.
According to preferential discharge theory, more is the $ε_{c e l l}$ of anode potential more is the tendency for reaction to take place.
hence, $C l_{2}$ gas is evolved.
$C l^{⊝} ⟶ \frac{1}{2} C l_{2} + e^{-} ε_{c e l l}^{o} = 1.36$ $V$

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Similar questions

C u_{(s)} + 2 A g_{(a q)}^{+} \to C u_{(a q)}^{2 +} + 2 A g_{(s)}

E_{c e l l}^{o} = + 0.46

C u^{2 +}

E_{c e l l}^{o}

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?

E_{C e l l}^{o}

Z n (s) + C u^{2 +} (a q) \to Z n^{2 +} (a q) + C u (s)

E_{C u^{2 +} / C u}^{\circ} = 0.34 V, E_{Z n^{2 +} / Z n}^{\circ} = - 0.76 V

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