Question

Predict the products of electrolysis in each of the following:
A.An aqueous solution of $AgN{O_3}$with silver electrodes.
B.An aqueous solution of $AgN{O_3}$ with platinum electrodes.
C.A dilute solution of ${H_2}S{O_4}$ with platinum electrodes.
D.An aqueous solution of $CuC{l_2}$ with platinum electrodes.

Answer 1

Hint: We have to know the parts of the electrolytic cell and the process of electrolysis, and the chemical reactions that take place in anode and cathode.

Complete step by step answer:
We know that electrolysis is a process in which decomposition of ionic substances takes place when electricity is passed through them. Interchanging of ions and atoms takes place either by the addition of electrons or by removal of electrons from the external circuit.
We have to understand that the process of electrolysis takes place in electrochemical cells.
An electrochemical cell contains the cathode, which is positive, anode, which is negative, and an electrolytic solution.
Let us know that an electrochemical cell transforms chemical energy to electrical energy or vice-versa.
An aqueous solution of $AgN{O_3}$ with silver electrodes.
At cathode: The following reduction reactions compete to occur at the cathode.
$A{g^ + }_{\left( {aq} \right)} + {e^ - } \to A{g_{\left( s \right)}}\,\,\,\,\,\,\,\,\,\,\,{E^ \circ } = 0.80V$
${H^ + }_{\left( {aq} \right)} + {e^ - } \to \dfrac{1}{2}{H_{2(g)}}\,\,\,\,\,\,\,\,{E^ \circ } = 0.00\,V$
The reaction that has a higher value of ${E^ \circ }$ takes place at the cathode. Hence, deposition of silver will take place at the cathode.
At anode: The Ag anode is attacked by \[N{O_3}^ - \] ions. Hence, the silver electrode at the anode dissolves in the solution to form \[A{g^ + }\].
$Ag \to A{g^ + } + {e^ - }$
An aqueous solution of $AgN{O_3}$ with platinum electrodes.
At cathode: The reduction reactions compete to occur at the cathode.
$A{g^ + }_{\left( {aq} \right)} + {e^ - } \to A{g_{\left( s \right)}}\,\,\,\,\,\,\,\,\,\,\,{E^ \circ } = 0.80V$
${H^ + }_{\left( {aq} \right)} + {e^ - } \to \dfrac{1}{2}{H_{2(g)}}\,\,\,\,\,\,\,\,{E^ \circ } = 0.00\,V$
The reaction that contains a higher value of ${E^ \circ }$ takes place at the cathode. Hence, deposition of silver will take place at the cathode.
At anode: We know the platinum electrodes are inert, and the anode is attacked by $N{O_3}^ - $ions. Thus, $O{H^ - }$ (or) $N{O_3}^ - $ ions could be oxidized at the anode. But the hydroxide ions $O{H^ - }$ have lower discharge potential and they will decompose to release oxygen.
$O{H^ - } \to OH + {e^ - }$
$4O{H^ - } \to 2{H_2}O + {O_2}$
A dilute solution of ${H_2}S{O_4}$ with platinum electrodes.
At cathode: The reduction reactions occur to form hydrogen gas.
${H^ + }_{\left( {aq} \right)} + {e^ - } \to \dfrac{1}{2}{H_{2(g)}}\,\,\,\,\,\,\,\,{E^ \circ } = 0.00\,V$
At anode: The following two processes are possible at the anode:
The first reaction is,
$2{H_2}{O_{\left( l \right)}} \to {O_{2(g)}} + 4{H^ + }_{\left( {aq} \right)} + 4{e^ - }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{E^ \circ } = + 1.23\,V$
The second reaction is,
$2S{O_4}{^{2 - }_{\left( {aq} \right)}} \to {S_2}{O_6}{^{2 - }_{\left( {aq} \right)}} + 2{e^ - }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{E^ \circ } = + 1.96V$
Incase dilute sulphuric acid, reaction (i) is preferred to produce \[{O_2}\] gas. But for concentrated sulphuric acid, reaction (ii) takes place.
An aqueous solution of $CuC{l_2}$ with platinum electrodes.
At cathode: The reduction reactions compete to occur at the cathode.
$C{u^{2 + }}_{\left( {aq} \right)} + 2{e^ - } \to C{u_{\left( s \right)}}\,\,\,\,\,\,\,\,{E^ \circ } = 0.34V$
${H^ + }_{\left( {aq} \right)} + {e^ - } \to \dfrac{1}{2}{H_{2(g)}}\,\,\,\,\,\,\,\,{E^ \circ } = 0.00\,V$
The reaction that has a higher value of ${E^ \circ }$ takes place at the cathode. Hence, deposition of copper will take place at the cathode.
At anode: The following two oxidation reactions are possible at the anode:
The first reaction is,
$C{l^ - }_{\left( {aq} \right)} \to \dfrac{1}{2}C{l_{2(g)}} + 2{e^ - }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{E^ \circ } = + 1.36\,V$
The second reaction is,
$2{H_2}{O_{\left( l \right)}} \to {O_{2(g)}} + 4{H^ + }_{\left( {aq} \right)} + 4{e^ - }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{E^ \circ } = + 1.23\,V$
At the anode, the reaction that has a lower value of is preferred. But due to the overpotential of oxygen, \[C{l^ - }\] gets oxidized at the anode to produce \[C{l_2}\] gas.

Note:
We must know that electrolysis is one of the many methods which are widely used in industry, it is used not only for extraction of metals but also to reduce the impurities in metal. This process is used in the medical field as well like for permanent removal of hair, etc. Even though such an application has some side effects but for long run it is good.