ELECTROCHEMISTRY. References: Engg.Chemistry by Jain and Jain Engg.Chemistry by Dr. R.V.Gadag and Dr. A.Nithyananda Shetty Principles of Physical Chemistry by Puri and Sharma.
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A galvanic cell is an electrochemical cell that produces electricity as a result of the spontaneous reaction occurring inside it. Galvanic cell generally consists of two electrodes dipped in two electrolyte solutions which are separated by a porous diaphragm or connected through a salt bridge. To illustrate a typical galvanic cell, we can take the example of Daniel cell.
At the cathode: Cu 2+ + 2e- → Cu
Net reaction: Zn(s)+Cu 2+ (aq)→ Zn 2+ (aq)+ Cu(s)
An electrolytic cell is an electro –chemical cell in which a non- spontaneous reaction is driven by an external source of current although the cathode is still the site of reduction, it is now the negative electrode whereas the anode, the site of oxidation is positive.
Zn│Zn2+ or Zn ; Zn2+
Zn │ ZnSO4 (1M) or Zn ; ZnSO4 (1M)
Cu2+/Cu or Cu2+ ;Cu
Cu2+(1M) ; Cu or CuSO4(1M)/Cu
Zn │ ZnSO4 (1M)║ CuSO4(1M)/Cu
Ej = Ø soln, R - Ø soln,L
Eg: *Contact between two different electrolytes (ZnSO4/ CuSO4).
*Contact between same electrolyte of different concentrations(0.1M HCl / 1.0 M HCl).
Ecell = Ecathode- Eanode
Ex α AD
Es α AD1
Ex ═ AD
Ex = AD x Es
Eg: Weston Cadmium Cell. The emf of the cell is 1.0183 V at 293 K and 1.0181 V at 298 K.
Soturated solution of
Paste of Hg2SO4
At the anode:
Cd (s) → Cd2+ + 2e-
At the cathode:
Hg2SO4(s) + 2e- → 2 Hg (l)+ SO42-(aq)
Cd + Hg22+ → Cd2+ + 2Hg
Anodic process:Zn(s) → Zn2+(aq)
Cathodic process: Zn2+(aq) → Zn(s)
Metal has net negative charge and solution has equal positive charge leading to the formation of an Helmholtz electrical layer.
Electric layer on the metal has a potential Ø (M).
Electric layer on the solution has a potential Ø (aq)
Is positive: If the electrode reaction is reduction when coupled with the standard hydrogen electrode
Is negative: If the electrode reaction is oxidation when coupled with standard hydrogen electrode. According to latest accepted conventions, all single electrode potential values represent reduction tendency of electrodes.
H2(g) → 2H+ +2e- (oxidation)
Cu2+ +2e- → Cu (reduction)
Hence electrode potential of copper is assigned a positive sign. Its standard electrode potential is 0.34 V.
Zn → Zn2+ +2e-
2H+ + 2e-→ H2.
Hence, electrode potential of zinc is negative. The standard electrode potential of zinc electrode is -0.74 V.
Mn+(aq) + ne- → M(s) ----(1)
We know that, ΔG =-nFE ----- (2)
ΔG =ΔGo +RT ln K
ΔG =ΔGo +RT ln[M]/[Mn+]-----(4)
-nFE= -nFEo + RT ln [M]/[Mn+]----(5)
E= Eo – RT/nF ln 1/[Mn+]------(6)
E=Eo- 2.303 RT/nF log 1/[Mn+]---(7)
E= Eo-0.0592/n log 1/[Mn+]-------(8)
Fe + Cu2+↔ Fe2+ + Cu
ECell= E0Cell-0.0592/2 log [Fe2+ ]/[Cu2+]
ECell= 0.78 + 0.0296 log 10/1
Zn + 2Ag+↔ Zn2+ + 2Ag
ECell= E0Cell-0.0592/2 log [Zn2+ ]/[Ag+]2
ECell= 1.56 + 0.0592 log 10/1.0
Mg│ Mg 2+ (0.01M)║ Cu 2+ /Cu is measured to be 2.78 V at 298K. The standard eletrode potential of magnesium electrode is -2.37 V. Calculate the electrode potential of copper electrode
Mg + Cu2+↔ Mg2+ + Cu
E= Eo-0.0592/n log 1/[Mn+]
EMg= EoMg-0.0592/2 log 1/[Mg2+]
2.78 = ECu-[-2.429]
Cu│ Cu 2+ (0.02M)║ Ag+ /Ag is measured to be 0.46 V at 298K. The standard eletrode potential of copper electrode is 0.34 V. Calculate the electrode potential of silver. electrode
W= QE ------(1)
Charge on 1mol electrons is F(96,500)Coulombs.
When n electrons are involved in the cell reaction,
the charge on n mole of electrons = nF
Substituting for Q in eqn (1)
W = nFE ----------(2)
The cell does net work at the expense of
ΔG = ΔH + T [δ(ΔG)/ δT]P ------- (2)
-nFE = ΔH – nFT (δ E/ δT)P
ΔH = nFT (δ E/ δ T)P – nFE
ΔH = nF[T(δ E/ δT)P –E]
ΔS = nF (δE/ δT)P
Calculate ∆G, ΔH and ΔS of the cell reaction at 298 K.
Solution:- ∆G: ∆G = - n FE
n = 2 for the cell reaction; F = 96,500 C E= 1.0181 V at 298 K
∆G = -2 x 96,500 x 1.0181 J = -196.5 KJ
The electrode reaction is reversible as it can undergo either oxidation or reduction depending on the other half cell.
E=Eo- 2.303 RT/nF log [H2]1/2/[H+]
= 0 -0.0591 log 1/[H+]
Cell Scheme: Pt,H2,H+(x)// SHE
= 0 – (- 0.0592 pH)
E (cell) = 0.0592 pH
pH = E(cell)/ 0.0592
Eg: Calomel electrode.
It can act as anode or cathode depending on the nature of the other electrode.
= Eo -0.0591 log [Cl-] at 298 K
Its electrode potential depends on the concentration of KCl.
Conc. of Cl- Electrode potential
0.1M 0.3335 V
1.0 M 0.2810 V
Saturated 0.2422 V
Pt,H2/H+(X) // KCl,Hg2Cl2,Hg
pH = E(cell) – 0.2422/ 0.0592
*This type of electrode is called indicator electrode.
H+ + Na+ Na+ + H+
Soln. glass soln. glass
Eg = Eog – 0.0592 pH
The overall potential of the glass electrode has three components:
Eg = Eb + Eref. + Easy.
= RT/nF ln C1 – RT/nF ln C2
= L + RT/nF ln C1
Eb depends upon [H+]
Eg = Eb + EAg/AgCl + Easy.
= L + RT/nF ln C1 + EAg/AgCl + Easy.
= Eog + RT/nF ln C1
= Eog + 0.0592 log [H+]
Eg = Eog – 0.0592 pH.
6. If recently calibrated, the glass electrode gives an accurate response.
7. The glass electrode is much more convenient to handle than the inconvenient hydrogen gas electrode.
Cell: SCE ║Test solution / GE
E cell = Eg – Ecal.
E cell = Eog – 0.0592 pH – 0.2422
pH = Eog -Ecell – Ecal. / 0.0592
The cell SCE ΙΙ (0.1M) HCl Ι AgCl(s) /Ag
gave emf of 0.24 V and 0.26 V with buffer having pH value 2.8 and unknown pH value respectively. Calculate the pH value of unknown buffer solution. Given ESCE= 0.2422 V
= 0.0592x2.8 +0.24 + 0.2422
pH = Eog -Ecell – Ecal. / 0.0592
= 0.648 -0.26-0.2422/0.0592
M/ Mn+[C1] ║ Mn+/M[C2]
= E0 + (2.303RT/ nF)logC2- [E0+(2.303RT/nF)logC1]
Ecell is positive only if C2 > C1
0.09 =(0.0592/2) log ( x / 0.001)
with the help of the following cell.
Hg/ Mercurous || Mercurous /Hg
nitrate (0.001N) nitrate (0.01N) when the emf observed at 18˚ C is 0.029 V
Ecell=(2.303 RT/nF) log C2/C1
0.029 = 2.303RT/n) log (0.01/0.001)
0.029 =0.057 x 1/ n
n = 0.057/0.029 =̃ 2
Valency of mercurous ions is 2, Hg2 2+
Answer the following questions:
1.Distinguish between electrolytic and galvanic cells.
2.Explain the origin of electrode potential. What are the sign conventions for electrode potential?
3.Give reasons for the following.
i) The glass electrode changes its emf over a period
ii) KCl is preferred instead of NaCl as an electrolyte
in the preparation of salt bridge
4. What is meant by a standard cell? Give an example
6.Define liquid junction potential. How it can be eliminated or minimized?
7.Derive Nernst equation for the single electrode potential.
8.Describe potentiometric determination of emf of a cell.
9.Writ e construction and working of Calomel Electrode
10.What are concentration cells? Show that emf of concentration cell becomes zero at a certain point of its working.