Derivation of Nernst Equation

Derivation of Nernst Equation explained step-by-step for 1st semester VTU chemistry students.
Detailed Notes: http://www.vturesource.com/post/1558/...

Let Us consider a reversible equation

Mn+ + ne- ＜---＞ M

K = ([𝐏𝐫𝐨𝐝𝐮𝐜𝐭𝐬])/([𝐑𝐞𝐚𝐜𝐭𝐚𝐧𝐭𝐬]) = ([𝐌])/([𝐌𝐧+]) --------(1)

For above reaction
Wmax = -nFE

But ΔG = -nFE & ΔG° = -nFE ° ------------------(2)



Where,
ΔG = Gibbs Free Energy
ΔG° = Standard Gibbs Free Energy
n = No. of e-
F = Faraday’s constant
E = Electrode Potential


From Vantoff’s Equation
We have,

ΔG = ΔG° + R T ln k
Where,
R = Gas Constant
T = Temperature
K = Equilibrium Constant

Substituting (1) & (2) in above equation we get,

-nFE = -nFE° + RT ln([M])/([Mn+])

Divide the Above Equation by -nF

(−nFE)/(−nF) = (−nFE°)/(−nF) + RT/(−nF) ln([M])/([Mn+])


E = E° - 𝑹𝑻/𝒏𝑭 ln([𝑴])/([𝐌𝐧+])


Converting Natural Log(ln) to Log & changing sign

E = E° + 𝐑𝐓/𝐧𝐅 2.303 log([𝐌𝐧+])/([𝐌])
Where,
ln = 2.303 log
log A/𝐁 = - log B/𝐀

Substituting for R, T & F
We get,
E = E° + (𝟎.𝟎𝟓𝟗𝟏)/𝐧 log([𝐌𝐧+])/([𝐌])


When [M] = 1
E = E° + (𝟎.𝟎𝟓𝟗𝟏)/𝐧 log[Mn+]

This is the required Nernst Equation for a single Electrode