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(3) To predict spontancity of redox rxn.:
For redox rxn. to be spotaneous, EMF of cell must be +ve.
Illustration: Cen Nickel vessel used to store CuSO4 sol ?
 = - 0.25 v
 = 0.34 v
Ans: NI2+/Ni || Cu2+/Cu
E0cell = - = 0.34 - (- 0.25) = 0.59 v
EMF comes out to be +ve. So, CuSO4 will reacts with Ni. So, cannot store in nickel vessel.
Nerst's Equation:
Mn+ + ne-  M
Nerst eq. as follows
E = E0 -  
E electrode potential at given conc. of Mn+ ions & Temp. T.
R gas constant.
T TEmp. in K.
F 1 Faraday.
n no. of electrons involved in electrode rxn.
For pure solids or liquids or gases, at 1 atm, molar conc. taken as unity.
i.e. [M) = 1
So, E = E0 -
Putting value of R, T & F R = 8.314, T = 298 K, F = 965000 coulom
 ......................................... (i) At 298 K Temp.
Nerts eq. for EMF of cell:
Zn(s) + Cu2+(ag) Zn2+(ag) + Cu(s)
For Zn(s) | Zn2+ (ag.) electrode,
Zn2+(ag) + 2e- Zn(s)
 =  +  [Zn2+(ag.)]
For Cu/Cu2+ electrode
Cu2+ + 2e- Cu(s)
 = + ln[Cu2+(ag.)]
Since oxidation takes place at Zn & reduction takes place at Cu electrode.
Ecell = Ecathode - Eanode
= -
= { + ln[Cu2+(ag.)]} - { + ln[Zn2+(ag.)]}
= ( - ) + n
= E0cell - ln
For cell rxn,
aA + bB  xX + yY
Ecell = E0cell - ln
At 296 K
Illustration: Calculate emf of cell
Cr|Cr3+(0.1 M) || Fe2+(0.01 M) | Fe
 = - 0.75 v
 = - 0.45 v
Ans: 2Cr + 3Fe2+ 2Cr2+ + 3Fe n = 6
Ecell = E0cell - 
= ( - ) - 
= (- 0.45 + 0.75 v) - 
Ecell = 0.2606 v
Eqm. constant form Nerst Eq.:
Zn + Cu2+  Zn2+ + Cu
eqn. const =  = KC
Eqn. occurs when Ecell = 0
0 = E0cell -  log
E0cell = logKc
At 296 k,
E0cell =  logKc |
Illustration
: Calculate eqm. constant for rxn.
Cu(s) + 2Ag+ Cu2++ 2Ag
Ans: Here n = 2 E0cell=
logKc  -=logKc
0.8 - 0.34 =logKc Kc= 3.688 x 1015
Dibb's Free Energy & Cell Potential
: - G0= nFE0cell But for rxn. at eqm E0cell=ln Kc
G0= - nF xln Kc
G0 = - 2.303 RT ln Kc |
Illustration
: Calculate standard free energy change
Zn(s)|Zn2+|| Cu2+(1M)|CU(s)
 = - 0.76 v = + 0.34 v F = 96500 C mol-1
Also calculate eqm constant for rxn.
Ans: mrxn. n = 2 G0= - nFE0cell E0cell= 0.34 - (- 0.76) = 1.1 v
G0= - 2 x 96500 x 1.1 = - 212.3 KJ/mol
G0= - 2.303 RT logKc - 212300 = - 2.303 xn 8.314 x 298 x logKc Kc= 1.6 x 1037
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E-Store
