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Electrochemistry

December 8, 2016
Published By : Pratik Kataria
Categorised in:

Electrical Conductance (G):
$1/R = I/V$

$R = ρ l/a$

Conductivity of Conductor (k)

$G l/a = 1/ρ$

Molar Conductivity
$⋀ = k/c = {\text"electrolytic conductivity"}/{\text"molar concentration"}$

⋀ = k V

$⋀ = ⋀_o – a √c$
c -> concentration ; $⋀_o$ -> molar conductivity at zero concentration or infinite dilution
a -> constant

Kohlrausch Law: $⋀_o = λ^o_+ + λ^o_-$

where $λ^o_+$ & $λ^o_-$ are molar conductivities of cation and anion at zero concentration

Degree of dissociation of weak electrolyte: $α = {⋀}/{⋀_o}$

Dissociation constant: $k = {α^2 C}/{1 – α}$

$k = {⋀^2 C}/{⋀_o (⋀_o – ⋀)}$

Cell Constant: $b = l/a$ = ${\text"Distance between electrodes"}/{\text"Area of cross section of electrode"}$ $m^{-1}$

[OR] b = R k

Popular Use:
$⋀ = {1000 k}/{C}$

Note: In oxidation half reactions: electron is on right side
In reduction half reactions: electron is on left side

Faraday’s 1st Law of electrolytes: Q = I t

Charge of 1 mole of electron = 96500 C i.e. 1 Faraday = 96500 C

No. of moles of electrons actually passed = ${Q (C)}/{96500 (c/mol)}$

Mole Ratio = ${\text"Moles of substance (produced) in half reaction"}/{\text"moles of electrons required in half reaction"}$

Mass of substance produced: ${I t}/{96500}$ x mole ratio x molar mass of substance
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Standard EMF of cell: $E_{cell}_^o = E^o_{cathode} (RHE) – E^o_{anode} (LHE)$

Nernst Equation

Cell/Electrode Potential
$E = E^o – {2.303 RT}/{nF} {log_{10}{Products}/{Reactants}}$

n -> moles of electrons used in reaction

At 25°C,
$E = E^o – {0.0592}/{n} {log_{10} {Products}/{Reactants}}$

Electrical Work:
$∆G = -nF E_{cell}$

Also,

$∆G^o = – RT ln K$

$E^o = {2.303 RT}/{nF} log_{10} k$
$E^o_{cell} = {0.0592}/{n} log_{10} k$ at 25°C

For spontaneous cell reaction: ∆G < 0 and so, $E_{cell} > 0$