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Predict the products of electrolysis in each of the following.
(i) An aqueous solution of \(\mathrm{AgNO}_3\) with silver electrodes.
(ii) An aqueous solution of \(\mathrm{AgNO}_3\) with platinum electrodes.
(iii) A dilute solution of \(\mathrm{H}_2 \mathrm{SO}_4\) with platinum electrodes.
(iv) An aqueous solution of \(\mathrm{CuCl}_2\) with platinum electrodes.
(i) An aqueous solution of \(\mathrm{AgNO}_3\) with silver electrodes.
(ii) An aqueous solution of \(\mathrm{AgNO}_3\) with platinum electrodes.
(iii) A dilute solution of \(\mathrm{H}_2 \mathrm{SO}_4\) with platinum electrodes.
(iv) An aqueous solution of \(\mathrm{CuCl}_2\) with platinum electrodes.
Solution:
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Verified Answer
(i) \(\mathrm{AgNO}_3(s) \longrightarrow \mathrm{Ag}^{+}(a q)+\mathrm{NO}_3^{-}(a q)\)
At cathode: \(\mathrm{Ag}^{+}\)ions have lower discharge potential than \(\mathrm{H}^{+}\)ions. Hence, \(\mathrm{Ag}^{+}\)ions will be deposited as Ag in preference to \(\mathrm{H}^{+}\)ions.
\(\mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)\)
At anode: \(\quad\) As Ag anode is attacked by \(\mathrm{NO}_3{ }^{-}\)ions, \(\mathrm{Ag}\) of the anode will dissolve to form \(\mathrm{Ag}^{+}\)ions in the solution.
\(\mathrm{Ag}(s) \longrightarrow \mathrm{Ag}^{+}(a q)+e^{-}\)
(ii) At cathode: \(\mathrm{Ag}^{+}\)ions have lower discharge potential than \(\mathrm{H}^{+}\)ions. Hence, \(\mathrm{Ag}^{+}\)ions will be deposited as Ag in preference to \(\mathrm{H}^{+}\)ions.
At anode: Since Pt electrode are inert, the anode is not attacked by \(\mathrm{NO}_3^{-}\)ions. Out of \(\mathrm{OH}^{-}\)and \(\mathrm{NO}_3^{-}\) ions, \(\mathrm{OH}^{-}\)ions have lower discharge potential. Hence, \(\mathrm{OH}^{-}\)ions will be discharged in preference to \(\mathrm{NO}_3^{-}\)ions, which then decompose to give out \(\mathrm{O}_2\).
\(\begin{aligned}
&\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{OH}+\mathrm{e}^{-} \\
&4 \mathrm{OH} \longrightarrow 2 \mathrm{H}_2 \mathrm{O}(\ell)+\mathrm{O}_2(g)
\end{aligned}\)
\(\begin{aligned}
\text { (iii) } \mathrm{H}_2 \mathrm{SO}_4(a q) \longrightarrow 2 \mathrm{H}^{+}(a q)+\mathrm{SO}_4^{2-(a q)} \\
& \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-} \\
\text {At cathode: } & \mathrm{H}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{H} \\
& \mathrm{H}+\mathrm{H} \longrightarrow \mathrm{H}_2(g) \\
\text { At Anode: } & \mathrm{OH}^{-} \longrightarrow \mathrm{OH}^{-} \\
& 4 \mathrm{OH} \longrightarrow 2 \mathrm{e}_2^{-} \\
& 2 \mathrm{O}_2+\mathrm{O}_2(g)
\end{aligned}\)
Thus, \(\mathrm{H}_2\) gas is liberated at the cathode and \(\mathrm{O}_2\) gas at the anode.
(iv) \(\mathrm{CuCl}_2(s) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q)\)
At cathode: \(\mathrm{Cu}^{2+}\) ions will be reduced in preference to \(\mathrm{H}^{+}\)ions and copper will be deposited at cathode.
\(\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}\)
At anode: \(\mathrm{Cl}^{-}\)ions will be discharged in preference to \(\mathrm{OH}^{-}\)ions which remains in solution.
Thus, \(\mathrm{Cu}\) will be deposited on the cathode and \(\mathrm{Cl}_2\) gas will be liberated at the anode.
At cathode: \(\mathrm{Ag}^{+}\)ions have lower discharge potential than \(\mathrm{H}^{+}\)ions. Hence, \(\mathrm{Ag}^{+}\)ions will be deposited as Ag in preference to \(\mathrm{H}^{+}\)ions.
\(\mathrm{Ag}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Ag}(s)\)
At anode: \(\quad\) As Ag anode is attacked by \(\mathrm{NO}_3{ }^{-}\)ions, \(\mathrm{Ag}\) of the anode will dissolve to form \(\mathrm{Ag}^{+}\)ions in the solution.
\(\mathrm{Ag}(s) \longrightarrow \mathrm{Ag}^{+}(a q)+e^{-}\)
(ii) At cathode: \(\mathrm{Ag}^{+}\)ions have lower discharge potential than \(\mathrm{H}^{+}\)ions. Hence, \(\mathrm{Ag}^{+}\)ions will be deposited as Ag in preference to \(\mathrm{H}^{+}\)ions.
At anode: Since Pt electrode are inert, the anode is not attacked by \(\mathrm{NO}_3^{-}\)ions. Out of \(\mathrm{OH}^{-}\)and \(\mathrm{NO}_3^{-}\) ions, \(\mathrm{OH}^{-}\)ions have lower discharge potential. Hence, \(\mathrm{OH}^{-}\)ions will be discharged in preference to \(\mathrm{NO}_3^{-}\)ions, which then decompose to give out \(\mathrm{O}_2\).
\(\begin{aligned}
&\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{OH}+\mathrm{e}^{-} \\
&4 \mathrm{OH} \longrightarrow 2 \mathrm{H}_2 \mathrm{O}(\ell)+\mathrm{O}_2(g)
\end{aligned}\)
\(\begin{aligned}
\text { (iii) } \mathrm{H}_2 \mathrm{SO}_4(a q) \longrightarrow 2 \mathrm{H}^{+}(a q)+\mathrm{SO}_4^{2-(a q)} \\
& \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-} \\
\text {At cathode: } & \mathrm{H}^{+}+\mathrm{e}^{-} \longrightarrow \mathrm{H} \\
& \mathrm{H}+\mathrm{H} \longrightarrow \mathrm{H}_2(g) \\
\text { At Anode: } & \mathrm{OH}^{-} \longrightarrow \mathrm{OH}^{-} \\
& 4 \mathrm{OH} \longrightarrow 2 \mathrm{e}_2^{-} \\
& 2 \mathrm{O}_2+\mathrm{O}_2(g)
\end{aligned}\)
Thus, \(\mathrm{H}_2\) gas is liberated at the cathode and \(\mathrm{O}_2\) gas at the anode.
(iv) \(\mathrm{CuCl}_2(s) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q)\)
At cathode: \(\mathrm{Cu}^{2+}\) ions will be reduced in preference to \(\mathrm{H}^{+}\)ions and copper will be deposited at cathode.
\(\mathrm{Cu}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Cu}\)
At anode: \(\mathrm{Cl}^{-}\)ions will be discharged in preference to \(\mathrm{OH}^{-}\)ions which remains in solution.
Thus, \(\mathrm{Cu}\) will be deposited on the cathode and \(\mathrm{Cl}_2\) gas will be liberated at the anode.
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