Question

Predict the products of electrolysis in each of the following:
(i) An aqueous solution of $AgNO_{3}$ with silver electrodes.
(ii) An aqueous solution of $AgNO_{3}$ with platinum electrodes.
(iii) A dilute solution of $H_{2}SO_{4}$ with platinum electrodes.
(iv) An aqueous solution of $CuCl_{2}$ with platinum electrodes.

Answer 1

"At cathode:

The following reduction reactions supposed to occur at the cathode

$\quad { Ag+ }_{ aq }{ e }^{ - }\rightarrow { Ag }_{ s };\quad { E }^{ \circ }=0.00V$

${ H }^{ + }+{ e }^{ - }\rightarrow \frac { 1 }{ 2 } { H }_{ 2(g) };\quad { E }^{ \circ }=0.00V$

The reaction with a "higher value" of${ E }^{ \circ }$occurs at the "cathode".

Therefore, "deposition of silver" will occur at the "cathode".

At anode: The "Ag anode" is "attacked" by ${ NO }_{ 3 }^{ - }$ ions.

Therefore, the "silver electrode" at the "anode" dissolves on to form ${ Ag }^{ + }.$

At cathode: The following reduction reactions supposed to occur at the cathode.

${ Ag }_{ aq }^{ + }+{ e }^{ - }\rightarrow { Ag }_{ s) };\quad { E }^{ \circ }=0.80V$

${ H }_{ (aq) }^{ + }+{ e }^{ - }\rightarrow \frac { 1 }{ 2 } { H }_{ 2(g) };\quad { E }^{ \circ }=0.00V$

The reaction with a "higher value" of ${ E }^{ \circ }$ occurs at the "cathode".

Therefore, "deposition of silver" will occur at the "cathode".

At anode:

Since platinum electrodes are inert, the anode is not attacked by ${ { NO }_{ 3 }^{ - } }$ions

Therefore,${ OH }^{ - } or { NO }_{ 3 }^{ - }$ ions can be oxidized at the anode.

But ${ OH }^{ - }$ions having a lower discharge potential and get preference and decompose to liberate${ O }_{ 2 }$

${ OH }^{ - }\rightarrow OH\quad +{ e }^{ -\quad  }\\ 4{ OH }^{ - }\rightarrow 2{ H }_{ 2 }O+{ O }_{ 2 }$

At cathode, the following reduction reaction occurs to produce ${ H }_{ 2 }$ gas.

${ H }_{ aq }^{ + }\quad +\quad { e }^{ \_  }\rightarrow \frac { 1 }{ 2 } { H }_{ 2(g) }$

At the anode, the following process are possible

$2{ H }_{ 2 }{ O }_{ l }\quad \longrightarrow \quad { O }_{ 2(g) }\quad +\quad 4{ H }_{ (aq) }^{ + }\quad +\quad 4{ e }^{ - };\quad  { E }^{ \circ }\quad =\quad +1.23\quad V..................(i)$

$2S{ O }_{ 4(aq) }^{ 2- }\quad \longrightarrow \quad { S }_{ 2 }{ O }_{ 6(aq) }^{ 2- }\quad +\quad 2{ e }^{ - };\quad  { E }^{ \circ }\quad =\quad +1.96\quad V............(ii)$

For the "dilute sulphuric acid" reaction (i) is preferred to produce ${ O }_{ 2 }$gas.

But, "concentrated sulphuric acid" reaction (ii) occurs.

At cathode: The following reactions occur at at the cathode.

${ Cu }_{ (aq) }^{ 2+ }\quad +\quad 2{ e }^{ - }\quad \longrightarrow { Cu }_{ (s) };\quad  { E }^{ \circ }\quad =\quad 0.34\quad V$

${ H }_{ (aq) }^{ + }\quad +\quad { e }^{ - }\quad \longrightarrow \quad \frac { 1 }{ 2 } { H }_{ 2(g) };\quad  { E }^{ \circ }\quad =\quad 0.00\quad V$

The reaction with a "higher value" of ${ E }^{ \circ }$ takes place at the "cathode".

Therefore, "deposition of copper" will take place at the "cathode".

The following oxidation reactions are possible at the anode.

${ Cl }_{ (aq) }^{ - }\quad \longrightarrow \quad \frac { 1 }{ 2 } { Cl }_{ 2 }\quad +\quad { e }^{ - };\quad  { E }^{ \circ }\quad =\quad 1.36\quad V$

$2{ H }_{ 2 }{ O }(l)\quad \longrightarrow \quad { O }_{ 2(g) }\quad +\quad 4{ H }_{ (aq) }^{ + }\quad +\quad 4e^{ - };\quad  { E }^{ \circ }\quad =\quad 1.23\quad V$

At the "anode", the reaction with a "lower value" of${ E }^{ \circ }$is preferred.

But due to the "over-potential" of "oxygen", ${ Cl }^{ - }$gets "oxidized" at the "anode" to produce ${ Cl }_{ 2 }$ gas."