Chapter 1:- Concept of Electrode Potential 2nd year Book
Advance Inorganic Chemistry
(Page 6)

Electrochemical Series

‘Electrochemical Series’ is the series that represents an arrangement of electrodes in the increasing order of their standard reduction potentials(or decreasing order of standard oxidation potentials) as compared to that of standard hydrogen electrode potential from top to bottom along with their half-cell reactions. (see table 1.0).

Important Characteristics of the Electrochemical Series :

(a) Most active metals are placed at the top of the series. They are readily oxidized to give cations and are good reducing agents. Most active metals displace hydrogen gas from water and acids. For example, if lithium is added to water or a solution containing H+ ion, the H+ ions are reduced to H2, and lithium is oxidized to Li+ ions.

(b) The more negative the reduction potential, the more reactive the metal will be in displacing other metals from their salts. For example, Fe displaces Cu from CuSO4 solution, Cu displaces Ag from silver salt solutions and Ag displaces Au from gold salt solutions as in photography during gold toning processes. It is clear from the following representation:

(c) Metals above hydrogen displace H2 gas from dilute acids. e.g.

Dil. H2SO4————> ZnSO4  +  H2
2 HCI  +  Mg———> MgCl2 +  H2

(d) Metals lying above hydrogen are easily rusted in contact with water, moisture, or acids while those situated below are not rusted. e.g. Iron is rusted easily whereas gold does not. It is because Fe is placed above and Au below the hydrogen in the electrochemical series.

(e) Oxidants (oxidizing agents) are arranged in increasing order and reductants (reducing agents) are arranged in decreasing order in the series. i.e., Li situated at the top is the weakest oxidizing agent but the strongest reducing agent.

(f) Oxides of iron and other metals lying below hydrogen can be reduced easily by the reaction with active metals or H2 gas.

electrochemical serieselectrochemical series electrochemical series news

From the above table following points may be noted-

(i) The species written on the left-hand side(LHS) of the reduction half-cell reactions in an oxidized form are called ‘oxidizing agents’ or ‘oxidants’ while the species written on the right-hand side(RHS) are in a reduced form called ‘reducing agents’ or ‘reductants’. i. e. oxidizing agents accept electrons and are reduced into reducing agents.

(ii) The value of standard reduction electrode potential (E °red) and standard oxidation potential (E °ox)are equal to each other but they have opposite signs. i.e. if one has a positive sign, the other will have a negative sign, and vice-versa.

(iii) As we proceed from top to bottom in the series, the value of (E °red) goes on increasing and becomes more and more positive(-ve to + ve including zero). The negative value of E °red of a given electrode indicates that this electrode will constitute a negative terminal(anode) of the galvanic cell which is obtained by joining the given electrode with an NHE. On the other hand, the positive value red of a given electrode implies, that this electrode will constitute a positive terminal(cathode) of the galvanic cell, which is obtained by connecting the given electrode with an NHE.

(iv) The electrodes, having negative values forred are placed above the hydrogen electrode, while the electrodes, having positive values of red are placed below the hydrogen electrode.

Application of Electrochemical Series:

red values given in the electrochemical series play the most important role in the following-

(a) To calculate the standard EMF (Eocell) of any electrochemical cell:

From the standard electrode potentials of two electrodes involve in constituting an electrochemical cell, the Eocell is given by:

From the standard electrode potentials of two electrodes involve in constituting an electrochemical cell, the Eocell is given by:



Example: Calculate the standard EMF of a cell which involves the following cell reaction
Zn  +  2Ag+————————-> Zn2+  +  2Ag

given that E0Zn2+/Zn = -0.76 V and E0Ag+/Ag = +0.80 V

Solution: The above redox reaction can also be shown as-


Thus, E0cell is determined by any one of the following ways-

(1) E0Cell = [ E0Zn/Zn2+ + E0Ag+/Ag]  = [(+0.76) + (+0.80)] = +1.56 V

(2)E0Cell = [(E0red)right(E0red)Left = [(E0Ag+/Ag)- (E0 Zn2+/Zn)]
= [(+0.80) – (- 0.76) ]
= [(+0.80) + 0.76) ] = +1.56 V

(3) E0Cell = (E0ox)Left – (E0Ox)Right =  [( E0Zn/Zn2+)(E0 Ag/Ag+)]
                                                         = [+ 0.76) – (-0.80) ]
= [(+0.76) + 0.80)] =+ 1.56 V

(b) To predict whether a given cell representation is correct or wrong:

If the difference (E0red)cathode –( E0red)anode which is equal to E0cell is positive, the cell representation is correct and if the potential difference between the electrodes will have a negative sign, the cell representation is wrong.

Example: For a cell representation of Daniel cell, Zn, Zn2+ (1M) || Cu2+ (1M), Cu

Given that E0zn2+ / zn= -0.76 V and E0Cu2+ / Cu = +0.34 V

E0cell = (Eo red)cathode – (Eo red)anode = +0.34-(0.76)V
= + [0.34 + 0.76)]V
= + 1.10 V

Thus, the cell representation, Zn, Zn2+(1M) || Cu2+ (1M), Cu is correct.

(c) To predict which half-cell will constitute the cathode and which half-cell constitutes anode when we construct a galvanic cell:

The half-cell having higher values of Eored (less negative or more positive) will form the cathode while the half-cell having lower values of Eored (more negative less positive) will act as anode of the galvanic cell.

For example, when a galvanic cell is constructed by the coupling of the electrodes

−Al3+ /Al and Fe2+ /Fe;

given that
Eo Al3+/Al=−1.67 V and E0Fe2+/Fe=−0.41 V.

It is evident that E0Al3+/Al < E0Fe2+/Fe=−0.41 V.
Thus, Al3+/Al electrode will act as an anode and Fe2+/Fe electrode will act as a cathode.

(d) To predict the spontaneity of the redox reaction :
To see whether the redox reaction is feasible or not, the EMF of the cell based on the given redox reaction is calculated. For a redox reaction to be spontaneous, the EMF of the cell must be positive. If the EMF of the cell comes out to be negative, the direct reaction, as given, can not take place.
Example: Give an answer with a suitable reason that the following reactions are feasible are not.
(1) 2Ag+  +  Cu⟶Cu2+  +  2Ag
(2) Ni2+    +  Cu⟶Cu2+  +  Ni

Given that E0 Ag+/Ag = +0.80 V, E0Cu2+/Cu= +0.34 V, E0 Ni2+/Ni = -0.25 V

Solution :
(1)  EMF of the redox reaction (Ecell=[E0 Ag+/Ag – E0 Cu2+/Cu]
= [0.80-(+0.34)]
=[ 0.80 – 0.34]
= + 0.4 6 V

The positive value of EMF of the redox reaction shows Cu can reduce Agion to Ag and the forward reaction is feasible.

(2) EMF of the redox reaction (Ecell[E0 Ni2+/Ni-E0 Cu2+/Cu]
= -0.5 9 V

The negative value of EMF of the redox reaction shows Cu can reduce Ni2+ ion to Ni and the forward reaction is not feasible.

(e) To compare the relative activities of metals:
The greater the oxidation potential of a metal, the more easily it can lose electrons and hence greater its reactivity. As a result, a metal with greater oxidation potential can displace metals with lower oxidation potentials from their salt solutions. For example, oxidation potentials of Na, K, Mg, Ca, Fe, Co, Ni, Cu, Ag, Au, and Zn are in the order:

K > Na > Ca > Mg > Zn > Fe > Co > Ni > Cu > Ag > Au
More active ———————————-> Less active
Decreasing the Order of Reactivity

It is clear from their standard oxidation electrode potentials as shown below:-


Hence, each metal can displace metals on its right from the salt solutions.
Thus, we can say on moving down the electrochemical series reactivities of the metals decreases with the increase of reduction electrode potential.

(f) To construct different galvanic cells of specified EMF.
(g) To select the suitable reducing or oxidizing agent.
(h) To compare the reducing /or oxidizing powers.


  Previous PageNext Page  

Spread The Love