Derive nermst equation relating the e.m.f.of a cell with the concentration of the reactants and the products of the reaction.
Answers
Explanation:
Consider a metal in contact with its own salt aqueous solution. Reactions of metal losing an electron to become an ion and the ion gaining electron to return to the atomic state are equally feasible and are in an equilibrium state.
Mn+ + ne– → nM
In the reduction reaction, ‘n’ moles of an electron is taken up by the ion against a reduction potential of Ered.
1. The work done in the movement of electron
Wred = nFEred
Where,
F is Faraday = 96487 coulomb = electrical charge carried by one mole of electrons
2. Change in the Gibbs free energy is an indication of the spontaneity and it is also equal to the maximum useful work (other than volume expansion) done in a process.
Combining work done and Gibbs free energy change:
Wred = nFEred = – ∆G or ∆G = – nFEred
3. Change in the free energy at standard conditions of 298K and one molar /one atmospheric pressure conditions is ∆G°. From the above relation, it can be written that
∆G° = – nFE°red
Where,
E°red is the reduction potential measured at standard conditions.
4. During the reaction, concentration keeps changing and the potential also will decrease with the rate of reaction.
To get the maximum work or maximum free energy change, the concentrations have to be maintained the same. This is possible only by carrying out the reaction under a reversible equilibrium condition.
For a reversible equilibrium reaction, vant Hoff isotherm says:
∆G = ∆G° + RT ln K
Where,
K is the equilibrium constant
K = Product/Reactant = [M]n/[M]n+
R is the Gas constant =8 .314J/K mole
T is the temperature in Kelvin scale.
5. Substituting for free energy changes in ant Hoff equation,
– nFEred = – nFE°red + RT ln [M]/[Mn+] = – nFE°red + 2.303 RT log [M]n/[Mn+]
Dividing both sides by – nF,
Ered = E°red – \frac{2.303 RT}{nF1}\log\frac{[M]^n}{[M^{n+}]}
nF1
2.303RT
log
[M
n+
]
[M]
n
or, EM^{n+}/M =E^o M^{n+}/M -\frac{2.303 RT}{nF1}\log\frac{[M]^n}{[M^{n+}]}EM
n+
/M=E
o
M
n+
/M−
nF1
2.303RT
log
[M
n+
]
[M]
n
The activity of the metal is, always considered as equal to unity.
Ered = E°red – or E M^{n+}/{M}EM
n+
/M
= E^o M^{n+}/{M}E
o
M
n+
/M- \frac{2.303 RT}{nF}\log\frac{1}{[Mn^{n+}]}
nF
2.303RT
log
[Mn
n+
]
1
This relation connecting reduction potential measurable at conditions other than standard conditions to the standard electrode potential is the Nernst equation.
For reaction conducted at 298K but at different concentrations, Nernst Equation is;
E M^{n+}/{M}EM
n+
/M= E^o M^{n+}/{M}E
o
M
n+
/M- \frac{2.303\times 8.314 \times293}{n96500}\log\frac{1}{[Mn^{n+}]}
n96500
2.303×8.314×293
log
[Mn
n+
]
1
= E^o M^{n+}/{M}E
o
M
n+
/M- \frac{0.0591}{n}\log\frac{1}{[Mn^{n+}]}
n
0.0591
log [Mn n+1]
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