(a) Following reactions occur at cathode during the electrolysis of aqueous sodium chloride solution:
${N a}^{+} (a q) + e^{-} ⟶ N a (s) E^{0} = - 2.71 V$
$H^{+} (a q) + e^{-} ⟶ \frac{1}{2} H_{2} (g) E^{0} = 0.00 V$
On the basis of their standard reduction electrode potential $(E^{0})$ values, which reaction is feasible at the cathode and why?
(b) Why does the cell potential of mercury cell remain constant throughout its life?

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Solution

$(a) E_{{N a}^{+} / N a}^{o} = - 2.71 V E_{H^{+} / H_{2}}^{o} = 0.00 V$
Since $E_{H^{+} / H_{2}}^{o} > E_{{N a}^{+} / N a}$ , so reduction of $H^{+}$ will occur and oxidation of $N a$ will occur.
At cathode $⟶$ Reduction will occur.
$H^{+} + e^{-} ⟶ \frac{1}{2} H_{2}$

$(b)$ A mercury cell consist of $Z n$ container as anode, a carbon rod as cathode and a paste of mercuric oxide with $K O H$ as electrolyte. In this cell overall cell reduction doesn't contain any ion whose concentration may change.
$∴$ cell constant remain constant.

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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?

A g^{+} (a q) + e \to A g (s) E^{\circ} = + .80 V

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

(E^{o})

C u S O_{4}

C u^{+ 2} + 2 e^{-} \to C u

C u \to C u^{+ 2} + 2 e^{-}

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