By Sunil Bhardwaj

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If a solution contains two cations, then it is possible that their simultaneous discharge may occur. It is expected that the simultaneous deposition of the two metals would come, provided the discharge ( reduction potential ) of the two ions were the same. For example, the reversible potential OR theoretical discharge potential of a metal M in a solution of its ions of activity 2* i.e. of the electrode m mo* would be given by, $${ E }_{ m } = { E }_{ m }^{ 0 } + \frac { RT }{ nF } \ln { { a }_{ m }^{ n+ } }$$When the gas is discharged for e.g hydrogen gas in order to obtain the actual discharge potential, it is necessary to include the over voltage ie. Tam so for hydrogen gas equation I will be$${ E }_{ { H }_{ 2 } } = { E }_{ { H }_{ 2 } }^{ 0 } + { \eta }_{ { H }_{ 2 } } \frac { RT }{ nF } \ln { { a }_{ { H }^{ + } } } $$if the solution contain two cations, one is metal and the other is hydrogen, then simultaneous deposition will occur when the discharge potential of the two is same equation I and II$${ E }_{ m }^{ 0 } + \frac { RT }{ nF } \ln { { a }_{ m }^{ n+ } } = { E }_{ { H }_{ 2 } }^{ 0 } + { \eta }_{ { H }_{ 2 } } \frac { RT }{ nF } \ln { { a }_{ { H }^{ + } } } $$There are the three ways in which discharge potential of the two cations may brought near each other.

1. If standard potentials are approximately equal and the over voltages are small

2. If the standard potentials are different but over voltages differ sufficiently to compensate for the standard potentials and

3. If the difference in standard potential and over voltage are compensated by the differences in the activities of the ions

These three types of behaviour are explained below

1. The standard potential of lead and tin are - 0.126V and - 0.140V rep. As they are solid over voltage is negligible in this case mall Concentration adjustment is sufficient to allow simultaneous deposition from their chloride solution

2. The discharge (i.e. reduction) potential of a zinc from a molar solution of ZnSO4 is about -0.76V$${ H }^{ + } + { e }^{ - } \longrightarrow \frac { 1 }{ 2 } { H }_{ 2 }$$ $$\therefore K = \frac { 1 }{ { a }_{ { H }^{ + } } } $$ $$\therefore { E }_{ r } = { E }^{ 0 } - \frac { RT }{ nF } \ln { K } $$ $$ { E }_{ r } = { E }^{ 0 } - \frac { 0.0592 }{ n } \ln { K } $$ $$ { E }_{ r } = { E }^{ 0 } - \frac { 0.0592 }{ 1 } \ln { \left( \frac { 1 }{ { a }_{ { H }^{ + } } } \right) } $$ $$ { E }_{ r } = { E }^{ 0 } + \frac { 0.0592 }{ 1 } \ln { \left( { a }_{ { H }^{ + } } \right) } $$ $$ { E }_{ r } = { E }^{ 0 } + 0.0592\ln { \left( { a }_{ { H }^{ + } } \right) } $$ $$ { E }_{ r } = 0 + 0.0592\ln { \left( { a }_{ { H }^{ + } } \right) } $$ $$ { E }_{ r } = 0.0592\ln { \left( { a }_{ { H }^{ + } } \right) } $$ $$ { E }_{ r } = 0.0592\ln { \left( { 10 }^{ -7 } \right) } $$ $$ { E }_{ r } = 0.0592 \times \left( -7 \right) $$ $$ { E }_{ r } = -0.4144 volt $$ Therefore hydrogen evolution will occur preference to zinc But Hydrogen over voltage on zinc is large ie. 0.7 to 0.8 Volt It adjust to bring the actual discharge potential closer together (ie. He and Za ) and simultaneous deposition of Hz and Zn occurs from slightly acid solution by using za electrode.

3. Although the normal deposition potential of Cd and Cu and of Za sad Cum far part they may be brought together by adjustment of ionic activities. If KCN is added to solutions of salts of these metals, complex oynnides have forned in each cuse But those complexes have different instability constant the concentration of simple ions se reduced to different extent so that the discharge potentials now approach one another.

The metals Cu and Zn are deposited separately from a solution containing their mixed sulphates, because the required discharge potentials from their IM solutions are 0.34 V and - 0.76 V resp. If however excess of alkali cyanido is added to the solution, both metal from complex cyanides from which deposition take place at about - 1.0 Volt in each case. The Cuprocyanide complex is much more stable than that of Zinc, due to which the concentration of free copper ions gets reduced to much greater extent than zinc ions and hence the potential of two electrodes are brought close to each other. Thus by electrolysis of copper and Zinc salt containing excess of cyanides. It is possible to prepare alloy of the metals.