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 \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.
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 \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:

(I) In fluid arrangement, AgNO3 ionizes to give Ag+(aq) and NO3–(aq) particles.

 

    \[AgN03\left( aq \right)\text{ }\to \text{ }Ag+\left( aq \right)\text{ }+\text{ }NO3\left( aq \right)\]

Hence, when power is passed, Ag+(aq) particles move towards the cathode while NO3–particles move towards the anode. As such, at the cathode, either Ag+(aq) particles or H2O atoms might be diminished. Which of these will really get released would rely on their anode possibilities which are given beneath:

 

    \[\begin{array}{*{35}{l}} Ag+\left( aq \right)\text{ }+e\text{ }\to \text{ }Ag\left( s \right);\text{ }E{}^\circ \text{ }=\text{ }+0.80\text{ }V\text{ }\ldots \text{ }\left( I \right)  \\ ~  \\ 2H2O\left( Z \right)\text{ }+\text{ }2e\text{ }\to \text{ }H2\left( g \right)\text{ }+\text{ }2OH\left( aq \right);\text{ }E{}^\circ \text{ }=\text{ }-\text{ }0.83\text{ }V\text{ }\ldots \text{ }\left( ii \right)  \\ \end{array}\]

Since the anode potential (i.e., decrease capability of Ag+(aq) particles is higher than that of H2O atoms, accordingly, at the cathode, it is the Ag+(aq) particles (as opposed to H2O particles) which are diminished.

 

Likewise, at the anode, either Ag metal of the anode or H2O atoms might be oxidized. Their terminal possibilities are:

 

    \[\begin{array}{*{35}{l}} Ag\left( s \right)\text{ }\to \text{ }Ag+\left( aq \right)\text{ }+\text{ }e;\text{ }E{}^\circ \text{ }=\text{ }-\text{ }0.80\text{ }V\text{ }\ldots \text{ }\left( iii \right)  \\ ~  \\ 2H2O\left( l \right)\text{ }\to \text{ }02\left( g \right)\text{ }+4H+\left( aq \right)+4e;\text{ }E{}^\circ \text{ }=\text{ }-\text{ }1.23\text{ }V\text{ }\ldots \text{ }\left( iv \right)  \\ \end{array}\]

Since the oxidation capability of Ag is a lot higher than that of H2O, subsequently,

 

at the anode, it is the Ag of the silver anode which gets oxidized and not the H2O particles. It might, nonetheless, be referenced here that the oxidation capability of N03–particles is even lower than that of H2O since more bonds are to broken during decrease of N03 particles than those in H2O.

 

Along these lines, when a watery arrangement 0f AgN03 is electrolysed, Ag from Ag anode breaks up while Ag+(aq) particles present in the arrangement get diminished and get kept on the cathode.

 

(ii) If, nonetheless, electrolysis of AgN03 arrangement is completed utilizing platinum terminals, rather than silver cathodes, oxidation of water happens at the anode since Pt being an honorable metal doesn’t go through oxidation without any problem. Thus, O2 is freed at the anode as per condition (iv).

 

Along these lines, when a watery arrangement of AgNO3 is electrolysed utilizing platinum terminals, Ag+ particles from the arrangement get stored on the cathode while 02 is freed at the anode.

 

(iii) In watery arrangement, H2S04ionises to give H+(aq) and SO42-(aq) particles.

 

    \[H2S04\left( aq \right)\text{ }\to \text{ }2H+\left( aq \right)\text{ }+S04\left( aq \right)\]

Hence, when power is passed, H+ (aq) particles move towards cathode while SO42-(aq) particles move towards anode. In other wode either H+(aq) particles or H2O atoms are diminished. Their terminal possibilities are:

    \[\begin{array}{*{35}{l}} 2H+\left( aq \right)2e\text{ }\to \text{ }H2\left( g \right);\text{ }E{}^\circ \text{ }=\text{ }0.0\text{ }V  \\ ~  \\ H2O\left( aq \right)\text{ }+\text{ }2e\text{ }\to \text{ }H2\left( g \right)\text{ }+\text{ }2OH(\left( aq \right);\text{ }E{}^\circ \text{ }=\text{ }-\text{ }0.83\text{ }V  \\ \end{array}\]

Since the electron potential (i.e., decrease capability) of H+(aq) particles is higher than that of H2O, accordingly, at the cathode, it is H+(aq) particles (as opposed to H2O atoms) which are diminished to develop H2 gas.

 

Additionally at the anode, either SO42-(aq) particles or H2O atoms are oxidized. Since the oxidation capability of SO4 is relied upon to be a lot of lower (since it included cleavage of many bonds when contrasted with those in H20)

 

than that of HjO atoms, consequently, at the anode, it is H2O particles (instead of SO42-particles) which are oxidized to develop O2 gas.

 

From the above conversation, it follows that during electrolysis of a fluid arrangement of H2S04 just the electrolysis of H2O happens freeing H2 at the cathode and O2 at the anode.

 

(iv) In watery arrangement, CuCl2 ionizes as follows:

 

    \[CuCl2\left( aq \right)\text{ }\to \text{ }CU2+\left( aq \right)\text{ }+\text{ }2Cl\left( aq \right)\]

On passing power, CU2+(aq) particles move towards cathode and CU2+(aq) particles move towards anode.

 

Hence, at cathode, either CU2+(aq) or H2O particles are diminished. Their anode possibilities are:

 

    \[\begin{array}{*{35}{l}} CU2++\text{ }2e\text{ }\to \text{ }Cu\left( s \right);\text{ }E{}^\circ \text{ }=\text{ }+0.34\text{ }V  \\ ~  \\ H2O\left( l \right)\text{ }+\text{ }2e\text{ }\to \text{ }H2\left( g \right)\text{ }+\text{ }2OH;\text{ }E{}^\circ \text{ }=\text{ }-\text{ }0.83\text{ }V  \\ \end{array}\]

Since the anode capability of CU2+(aq) particles is a lot higher than that of H2O, hence, at the cathode, it is CU2+(aq) particles which are diminished and not H2Omolecules.

 

Essentially, at the anode, either Cl–(aq) particles or H2O atoms are oxidized. Their oxidation possibilities

 

    \[\begin{array}{*{35}{l}} 2Cl\left( aq \right)\text{ }\to \text{ }Cl2\left( g \right)\text{ }+\text{ }2e;\text{ }AE{}^\circ \text{ }=\text{ }-\text{ }1.36\text{ }V  \\ ~  \\ 2H2O\left( l \right)\text{ }\to O2\left( g \right)\text{ }+\text{ }4H+\left( aq \right)\text{ }+\text{ }4e;\text{ }E{}^\circ \text{ }=\text{ }-\text{ }1.23\text{ }V  \\ \end{array}\]

Despite the fact that oxidation capability of H2O atoms is higher than that of Cl–particles, by the by, oxidation of Cl–(aq) particles happens in inclination to H2O since due to overvoltage much lower potential than – 1.36 V is required for the oxidation of H2O particles.

 

In this way, when a fluid arrangement of CuCl2 is electrolysed, Cu metal is freed at the cathode while Cl2 gas is advanced at the anode.