By Sunil Bhardwaj

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Consider a general reaction, $$aA + bB \rightleftarrows cC + dD$$ The equilibrium constant K is given by, $$\therefore K = \frac { { a }_{ C }^{ c } \times { a }_{ D }^{ d } }{ { a }_{ A }^{ a } \times { a }_{ B }^{ b } } \qquad ....(1)$$ As the reaction is reversible, the equilibrium is established and this equilibrium depends upon the concentration of species present. The free energy change associated for above cell reaction is given by Vant Hoff isotherm. i.e. $$\Delta G = \Delta { G }^{ o } + RT\ln { K }$$ lets put the value of K from equation (1) $$\Delta G = \Delta { G }^{ o } + RT\ln { \left( \frac { { a }_{ C }^{ c } \times { a }_{ D }^{ d } }{ { a }_{ A }^{ a } \times { a }_{ B }^{ b } } \right) } \qquad ....(2)$$ But $$\Delta G = -nFE \ and \ \Delta { G }^{ o } = -nFE^{ o }$$ Where $$E$$ and $$E^{ o }$$ are the emf and standard emf of cell, on putting these values in above equation (2). $$-nFE = -nFE^{ o } + RT\ln { \left( \frac { { a }_{ C }^{ c } \times { a }_{ D }^{ d } }{ { a }_{ A }^{ a } \times { a }_{ B }^{ b } } \right) }$$ Lets cancel the minus sign, $$nFE = nFE^{ o } - RT\ln { \left( \frac { { a }_{ C }^{ c } \times { a }_{ D }^{ d } }{ { a }_{ A }^{ a } \times { a }_{ B }^{ b } } \right) }$$ Divide all the terms with nF, $$E = E^{ o } - \frac { RT }{ nF } \ln { \left( \frac { { a }_{ C }^{ c } \times { a }_{ D }^{ d } }{ { a }_{ A }^{ a } \times { a }_{ B }^{ b } } \right) }$$ Or $$E = E^{ o } - \frac { 2.303RT }{ nF } \log { \left( \frac { { a }_{ C }^{ c } \times { a }_{ D }^{ d } }{ { a }_{ A }^{ a } \times { a }_{ B }^{ b } } \right) }$$ The ratio of activities of product to reactant $$\left( \frac { { a }_{ C }^{ c } \times { a }_{ D }^{ d } }{ { a }_{ A }^{ a } \times { a }_{ B }^{ b } } \right)$$ is known are reaction quotient and can be represented as $${ a }_{ Q }$$ thus, $$\boxed { E = E^{ o } - \frac { 2.303RT }{ nF } \log { { a }_{ Q } } } \qquad ....(3)$$ This is Nernst Equation.

At standard conditions, R = 8.314J, T = 20? = 293K and F = 96500C.

Lets put these values in Nernst Equation (3), $$E = E^{ o } - \frac { 2.303 \times 8.314 \times 293 }{ n \times 96500 } \log { { a }_{ Q } }$$ $$\boxed { E = E^{ o } - \frac { 0.059 }{ n } \log { { a }_{ Q } } } \qquad ....(4)$$ This is Nernst Equation at standard conditions.

Numerical Problems: Find the cell potential of a galvanic cell based on the following reduction reactions at 25°C $${ Cd }^{ +2 } + 2{ e }^{ - } \longrightarrow Cd\qquad { E }^{ 0 } = -0.403V$$ $${ Pb }^{ +2 } + 2{ e }^{ - } \longrightarrow Pb\qquad { E }^{ 0 } = -0.126V$$ where $$\left[ { Cd }^{ +2 } \right] = 0.020M$$ and $$\left[ { Pb }^{ +2 } \right] = 0.200M$$.

Solution:

The first step is to determine the cell reaction, In order for the cell to be galvanic, $${ E }_{ cell }^{ 0 } > 0.$$ i.e. $${ E }_{ cell }^{ 0 }$$ must be positive. That is only possible when Cd electrode undergoes oxidation reaction. i.e. $$Cd \longrightarrow { Cd }^{ +2 } + 2{ e }^{ - }\qquad (Oxidation)\qquad { E }^{ 0 } = +0.403V$$ $${ Pb }^{ +2 } + 2{ e }^{ - } \longrightarrow Pb\qquad (Reduction)\qquad { E }^{ 0 } = -0.126V$$ $$\boxed { Cd + { Pb }^{ +2 }(aq) \longrightarrow { Cd }^{ +2 }(aq) + Pb }$$ $$\therefore { E }_{ cell }^{ 0 } = \left( 0.403V \right) + \left( -0.126V \right) = 0.277V$$ The Nernst equation is: $${ E }_{ cell } = { E }_{ cell }^{ 0 } - \frac { 2.303RT }{ nF } \log { { a }_{ Q } }$$ Here, $${ E }_{ cell }^{ 0 } = 0.277V$$ R is gas constant = 8.314 J/mol.K

T is absolute temperature = 25? = 300K

n is number of moles of electrons transfered = 2

F is faradays constant = 96485 C/mol

and $${ a }_{ Q }$$ is reaction quotient = $$\frac { { a }_{ { Cd }^{ +2 } } \times { a }_{ Pb } }{ { a }_{ { Cd } } \times { a }_{ { Pb }^{ +2 } } } =\frac { { a }_{ { Cd }^{ +2 } } }{ { a }_{ { Pb }^{ +2 } } } =\frac { \left[ 0.020M \right] }{ \left[ 0.200M \right] } = 0.1$$ Lets substitute all the values in nernst equation, we get $${ E }_{ cell } = 0.277V - \frac { 2.303 \times 8.314 J/mol.K \times 300K }{ 2 \times 96485 C/mol } \log { \left( 0.1 \right) }$$ $$= 0.277V - \frac { 5744 }{ 192970 } \log { \left( 0.1 \right) }$$ $$= 0.277V - 0.0298 \left( -1 \right)$$ $$= 0.277V + 0.0298$$ $$= 0.3068V$$

Numerical Problems:

Find the cell potential of a galvanic cell based on the following reduction reactions at 25°C $$Cd | { Cd }^{ +2 } (0.002M) \parallel { Zn }^{ +2 } (0.075M) | Zn$$ where reduction potential of $${ E }_{ Cd }^{ 0 } = -0.403V$$ and $${ E }_{ Zn }^{ 0 } = -0.762V$$

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