By Sunil Bhardwaj

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As we know that during the electrolysis of 1 N sulphuric acid, H2 and O2 gases are liberated at Cathode and Anode respectively. Which converts Pt electrode into H2 and O2 gas electrode and there is generation of electricity which opposes the applied potential from the battery. This process is known as Polarisation.

And due to the polarisation current does not increases even if we increase the applied potential. But after certain value of applied potential current increase rapidly. This voltage above which the current increase rapidly is known as Decomposition Potential (Ed). And the value of Ed for all aqueous acids and alkali is same which is 1.7 volt except HCl which is 1.3 Volts.

This decomposition potential of 1.7 Volt is the observed value and can be determined experimentally so lets try to find the decomposition value of the cell or emf of a galvanic cell with H2 and O2 gas electrodes dipped in 1N H2SO4.

This decomposition potential of 1.7 Volt is the observed value and can be determined experimentally so lets try to find the decomposition value of the cell or emf of a galvanic cell with H2 and O2 gas electrodes dipped in 1N H2SO4.

So the galvanic cell fored with H2 and O2 gas electrodes will be represented as, $$\circleddash \underset { 1 atm }{ Pt ,{ H }_{ 2 }(g) } | \underset { aq soln }{ 1N { H }_{ 2 }{ SO }_{ 4 } } | \underset { 1 atm }{ { O }_{ 2 }(g) ,Pt } \oplus$$ The cell reactions are:

At LHE oxidation $$\qquad { H }_{ 2 }(g) \longrightarrow 2{ H }^{ + } + 2{ e }^{ - }$$

At RHE reduction $$\qquad \frac { 1 }{ 2 } { O }_{ 2 }(g) + { H }_{ 2 }O + 2{ e }^{ - }\longrightarrow 2{ OH }^{ - }$$

$$\overline { Net cell reaction\qquad { H }_{ 2 }(g) + \frac { 1 }{ 2 } { O }_{ 2 }(g) + { H }_{ 2 }O \longrightarrow 2{ H }^{ + } + 2{ OH }^{ - } }$$ On applying the law of mass action to above cell reaction, $$K = \frac { { \left( { a }_{ { H }^{ + } } \right) }^{ 2 } \times { \left( { a }_{ { OH }^{ - } } \right) }^{ 2 } }{ \left( { a }_{ { H }_{ 2 }(g) } \right) \times { \left( { a }_{ { O }_{ 2 }(g) } \right) }^{ 2 } \times \left( { a }_{ { H }_{ 2 }O } \right) } \qquad ...(1)$$ In 1N H2SO4, neglecting activity coefficient, $${ { a }_{ { H }^{ + } } } = { { C }_{ { H }^{ + } } } = 1;$$ $${ { a }_{ { OH }^{ - } } } = { { C }_{ { OH }^{ - } } } = { 10 }^{ -14 };$$ $${ a }_{ { H }_{ 2 }(g) } = { C }_{ { H }_{ 2 }(g) } = 1 atm;$$ $${ a }_{ { O }_{ 2 }(g) } = C_{ { O }_{ 2 }(g) } = 1 atm;$$ $${ a }_{ { H }_{ 2 }O } = { C }_{ { H }_{ 2 }O } = 1;$$ Substituting all the values in eq (1)$$K = \frac { { \left( 1 \right) }^{ 2 } \times { \left( { 10 }^{ -14 } \right) }^{ 2 } }{ \left( 1 \right) \times { \left( 1 \right) }^{ 2 } \times \left( 1 \right) } = { \left( { 10 }^{ -14 } \right) }^{ 2 }$$ On applying the Nernst expression for emf of the cell, $${ E }_{ cell } = { E }_{ cell }^{ 0 } - \frac { RT }{ nF } \ln { K }$$ $${ E }_{ cell } = \left[ { E }_{ { { O }_{ 2 } }|{ OH }^{ - } }^{ 0 } - { E }_{ { { H }_{ 2 } }|{ H }^{ + } }^{ 0 } \right] - \frac { RT }{ 2F } \ln { \left[ { \left( { 10 }^{ -14 } \right) }^{ 2 } \right] }$$ Standard reduction potential of $${ E }_{ { { O }_{ 2 } }|{ OH }^{ - } }^{ 0 } = 0.4V$$ and $${ E }_{ { { H }_{ 2 } }|{ H }^{ + } }^{ 0 } = 0.0V$$ $${ E }_{ cell } = \left[ 0.4 - 0.0 \right] - \frac { 0.0591 }{ 2 } \ln { \left[ { \left( { 10 }^{ -14 } \right) }^{ 2 } \right] }$$ $${ E }_{ cell } = 0.4 - \frac { 0.0591 }{ 2 } \left[ 2 \times \ln { \left( { 10 }^{ -14 } \right) } \right]$$ $${ E }_{ cell } = 0.4 - 0.0591\left[ \ln { \left( { 10 }^{ -14 } \right) } \right]$$ $${ E }_{ cell } = 0.4 - 0.0591\left[ -14 \times \ln { \left( 10 \right) } \right]$$ $${ E }_{ cell } = 0.4 - 0.0591 \times \left( -14 \right)$$ $${ E }_{ cell } = 0.4 + 0.8274$$ $${ E }_{ cell } = 1.2274 \cong 1.23 Volts.$$

So practically we found the value of 1.7 Volt while theoretically it comes 1.23 Volt. This difference between the actual and theoretical voltage for electrolysis to happen when gas evolution is observed is known as Over Voltage or Over Potential.

If we represent observed decomposition potential by Ed

And theoretical reversible potential with Er.

Then Over Voltage $$(\eta)$$ = Ed - Er.

So in this case $$(\eta)$$ = 1.7 - 1.23 = 0.47 Volt.

Thus, the over voltage can be defined as �The difference between the potential of the electrode when gas evolution is actually observed and the theoretical reversible potential of the involved galvanic cell.�

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