In the process of obtaining chlorine from bri | vedclass

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Question

(B) The electrolysis of brine ($NaCl$ solution) involves the oxidation of chloride ions to chlorine gas at the anode: $2Cl^- (aq) \rightarrow Cl_2 (g)

Vedclass · Accepted Answer

$2.18$

Vedclass · Answer

(B) The electrolysis of brine ($NaCl$ solution) involves the oxidation of chloride ions to chlorine gas at the anode: $2Cl^- (aq) \rightarrow Cl_2 (g) + 2e^-$ $(E^\circ = -1.36 \ V)$.At the cathode,water is reduced: $2H_2O (l) + 2e^- \rightarrow H_2 (g) + 2OH^- (aq)$ $(E^\circ = -0.83 \ V)$.The overall cell reaction is $2Cl^- (aq) + 2H_2O (l) \rightarrow Cl_2 (g) + H_2 (g) + 2OH^- (aq)$.The standard cell potential is $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = -0.83 \ V - 1.36 \ V = -2.19 \ V$.Since the reaction is non-spontaneous,an external potential of at least $2.19 \ V$ (often approximated as $2.2 \ V$ or $2.18 \ V$ in textbooks) is required to drive the reaction.

Electrolyte	$\Lambda_{m}^{o} \, (S \, cm^{2} \, mol^{-1})$
$NaCl$	$126.5$
$NaBr$	$128.2$
$KCl$	$149.5$

In the process of obtaining chlorine from brine,the minimum value of $E_{cell}$ should be ....... $V$.

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Electrolyte $\Lambda_{m}^{o} \, (S \, cm^{2} \, mol^{-1})$

$NaCl$ $126.5$

$NaBr$ $128.2$

$KCl$ $149.5$

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