Name the processes from which chlorine is obtained as a by-product. What will happen if an aqueous solution of $NaCl$ is subjected to electrolysis?
(N/A) Chlorine is obtained as a by-product in the electrolysis of molten $NaCl$ and in the electrolysis of brine $(NaCl_{(aq)})$ in the Chlor-alkali process.
In the electrolysis of molten $NaCl$:
At anode: $2Cl^-_{(melt)} \longrightarrow Cl_{2(g)} + 2e^-$
At cathode: $2Na^+_{(melt)} + 2e^- \longrightarrow 2Na_{(s)}$
If an aqueous solution of $NaCl$ (brine) is subjected to electrolysis,$Cl_2$ gas is obtained at the anode,but at the cathode,$H_2$ gas is evolved instead of $Na$ metal. This is because the standard reduction potential of $Na^+$ $(E^{\circ} = -2.71 \ V)$ is much more negative than that of $H_2O$ $(E^{\circ} = -0.83 \ V)$. Therefore,$H_2O$ is preferentially reduced at the cathode.
At cathode: $2H_2O_{(l)} + 2e^- \longrightarrow H_{2(g)} + 2OH^-_{(aq)}$
At anode: $2Cl^-_{(aq)} \longrightarrow Cl_{2(g)} + 2e^-$
In the electrolysis of molten $NaCl$:
At anode: $2Cl^-_{(melt)} \longrightarrow Cl_{2(g)} + 2e^-$
At cathode: $2Na^+_{(melt)} + 2e^- \longrightarrow 2Na_{(s)}$
If an aqueous solution of $NaCl$ (brine) is subjected to electrolysis,$Cl_2$ gas is obtained at the anode,but at the cathode,$H_2$ gas is evolved instead of $Na$ metal. This is because the standard reduction potential of $Na^+$ $(E^{\circ} = -2.71 \ V)$ is much more negative than that of $H_2O$ $(E^{\circ} = -0.83 \ V)$. Therefore,$H_2O$ is preferentially reduced at the cathode.
At cathode: $2H_2O_{(l)} + 2e^- \longrightarrow H_{2(g)} + 2OH^-_{(aq)}$
At anode: $2Cl^-_{(aq)} \longrightarrow Cl_{2(g)} + 2e^-$
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View SolutionIf electrolysis of aqueous $CuSO_4$ solution is carried out using $Cu$ electrodes, the reaction taking place at the anode is:
Given below are two statements $:$
$1 \ M$ aqueous solution of each of $Cu(NO_3)_2, AgNO_3, Hg_2(NO_3)_2$ and $Mg(NO_3)_2$ are electrolysed using inert electrodes.
Given $: E_{Ag^{+}/Ag}^{\theta} = 0.80 \ V, E_{Hg_2^{2+}/Hg}^{\theta} = 0.79 \ V, E_{Cu^{2+}/Cu}^{\theta} = 0.24 \ V$ and $E_{Mg^{2+}/Mg}^{\theta} = -2.37 \ V$.
Statement $(I) :$ With increasing voltage,the sequence of deposition of metals on the cathode will be $Ag, Hg$ and $Cu$.
Statement $(II) :$ Magnesium will not be deposited at cathode instead oxygen gas will be evolved at the cathode.
In the light of the above statement,choose the most appropriate answer from the options given below.
$1 \ M$ aqueous solution of each of $Cu(NO_3)_2, AgNO_3, Hg_2(NO_3)_2$ and $Mg(NO_3)_2$ are electrolysed using inert electrodes.
Given $: E_{Ag^{+}/Ag}^{\theta} = 0.80 \ V, E_{Hg_2^{2+}/Hg}^{\theta} = 0.79 \ V, E_{Cu^{2+}/Cu}^{\theta} = 0.24 \ V$ and $E_{Mg^{2+}/Mg}^{\theta} = -2.37 \ V$.
Statement $(I) :$ With increasing voltage,the sequence of deposition of metals on the cathode will be $Ag, Hg$ and $Cu$.
Statement $(II) :$ Magnesium will not be deposited at cathode instead oxygen gas will be evolved at the cathode.
In the light of the above statement,choose the most appropriate answer from the options given below.
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