Electrolysis of 600 mL aqueous solution of N | vedclass

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(C) The electrolysis reaction of $NaCl$ is: $NaCl_{(aq)} + H_2O_{(l)} \rightarrow NaOH_{(aq)} + \frac{1}{2} Cl_{2(g)} + \frac{1}{2} H_{2(g)}$ Since th

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$2$

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(C) The electrolysis reaction of $NaCl$ is: $NaCl_{(aq)} + H_2O_{(l)} ightarrow NaOH_{(aq)} + \frac{1}{2} Cl_{2(g)} + \frac{1}{2} H_{2(g)}$ Since the final $pH$ is $12$,the concentration of $OH^{\ominus}$ ions is $[OH^{\ominus}] = 10^{-(14-12)} = 10^{-2} \ M$. The total moles of $OH^{\ominus}$ produced in $600 \ mL$ $(0.6 \ L)$ is $n = 10^{-2} \ mol/L 	imes 0.6 \ L = 6 	imes 10^{-3} \ mol$. According to Faraday's law,the charge $Q$ required is $Q = n 	imes F$,where $F = 96500 \ C/mol$. $Q = 6 	imes 10^{-3} 	imes 96500 = 579 \ C$. Since $Q = I 	imes t$,where $t = 5 \ min = 300 \ s$: $I = \frac{579}{300} = 1.93 \ A$. The nearest integer value for the current is $2 \ A$.

Electrolysis of $600 \ mL$ aqueous solution of $NaCl$ for $5 \ min$ changes the $pH$ of the solution to $12$. The current in Amperes used for the given electrolysis is $...........$. $(Nearest \ integer)$.

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