The electrochemical series helps to pick out substances that are good oxidizing agents and those which are good reducing agents. In an electrochemical series the species which are placed above hydrogen are more difficult to be reduced and their standard reduction potential values are negative. The Li : Li+ (aq) electrode has the least Eo value and therefore, it is reduced with more difficulty. Therefore, Li+ cannot accept electrons easily and so loses electrons to behave as a reducing agent. Li is the strongest reducing agent.

The species which are easily reduced than hydrogen are palced below it in electrochemical series and their Eo value are positive. The F2 : 2F-(aq) electrode has the highest Eo value and therefore, F2 has the greatest tendency to get reduced, it is consequently the strongest oxidizing agent. In general, oxidizing agents have + Eo values.

Higher the positive value, stronger will be the oxidizing agent and reducing agents have -Eo values, higher the negative value, stronger will be the reducing agent.

For Example:

Increasing order of reducing power of metal is

Ag+/Ag(+0. 80V) < Cr3+/Cr(-074V) < K+/K (-2. 93V)

To predict whether a metal will react with acids to give H2 gas:-

Metals above hydrogen in Electrochemical series have great tendency for oxidation, so they displace hydrogen from acids. All metals having negative electrode potentials (negative E° values) show greater tendency of losing electrons as compared to hydrogen. So, when such a metal is placed in an acid solution, the metal gets oxidized, and H+ (hydrogen) ions get reduced to form hydrogen gas. Thus, the metals having negative E° values liberate hydrogen from acids.

metal having negative E° value

For example, metals such as Mg (E (Mg2+ Mg) = – 2. 37 V),

Zn (E (Zn2+ Zn) = – 0. 76 V), Iron (E (Fe2+ Fe) = – 0. 44 V) etc. , can displace hydrogen from acids such as HCl and HSO4. But metals such as Copper, (E (Cu2+ Cu) = + 0. 34V), silver (E (Ag+ Ag) = + 0. 80V) and gold (E (Au3+ Au) = +1. 42 V) cannot displace hydrogen from acids because of their positive reduction potential value.

To predict the Feasibility of Redox Reaction:-

From the E° values of the two electrodes one can find out whether a given redox reaction is feasible or not. A redox reaction is feasible only if the species which has higher potential is reduced i. e. , accepts the electrons and the species which has lower reduction potential is oxidized i. e. , loses electrons.

The electrochemical series gives the increasing order of electrode potentials (reduction) of different electrodes on moving down the table. This means that the species, which accept the electrons (reduced) must be lower in the electrochemical series as compared to the other which is to lose electrons. (oxidized). For example,

From the electrochemical series E° value of Cu = +0. 34 V and that of Ag = +0. 80 V since the reduction potential of Ag is more than that of Cu, this means that silver has greater tendency to get reduced in comparison to copper. Thus, the reaction

occurs more readily than the reaction

The reduction potential of copper is less than that of Ag, this means that copper will be oxidized or will go into solution as ions in comparison to Ag. Thus, the reaction,

occurs more readily than

Therefore, silver will be reduced and copper will be oxidized and the above reaction is not feasible. Rather the reverse reaction,

can occur. Thus a metal will displace, any other metal, which occurs below it in the electrochemical series from its salt solution. When a metal having lower E° value is placed in a solution, containing ions of another metal having higher E° value, then the metal having lower E° value gets dissolved and the ions of the metal having higher E° value get precipitated.

Problems

Q:- Write the half-cell reaction and the overall cell reaction for the electrochemical cell:

Calculate the standard emf for the cell if standard electrode potentials (reduction) Pb2+ Pb and Zn2+ Zn electrodes are -0. 126V and -0. 763 V respectively.

Solution

Zn electrode acts as anode while Pb electrode acts as cathode and, therefore oxidation occurs at zinc electrode and reduction occurs at lead electrode. The half cell reactions are:

Q:- Iodine (I2) and bromine (Br2) are added to a solution containing iodide (I-) and bromide (Br-) ions. What reaction would occur if the concentration of each species is 1 M? The electrode potentials for the reactions are:

Solution

Since the reduction potential of Br2 is more than that of I2, it means that bromine can be readily reduced. Therefore, I- will be oxidized to I2 and this reaction should be written as oxidation. Therefore, the following reactions will occur:

Since for the feasibility of the reaction, the emf should be +ve, and to get + ve value for the cell reaction, subtract the equation representing lower value of E° from the equation representing the higher value of E°.

Q:-. What will be the spontaneous reaction between the following half-cell reactions?

Calculate Ecell.

Solution

Since the reduction potential of reaction (ii) is more than that of reaction (i); reaction (ii) will occur as reduction. Therefore, reaction (i) should be written as oxidation. To obtain the net reaction, we multiply the reactions by appropriate coefficients so that electrons get cancelled.

Ecell = Esubstance reduced – Esubstance oxidized

= 1. 28 – (- 0. 74) = 2. 02V

To predict the spontaneity of any redox reaction:-

For any spontaneous reaction (deltaG) should be negative. Since

deltaG = -nFE cell

Hence E cell should be positive for spontaneous reaction. E cell is the emf of the cell and is calculated from the standard redox potentials by using the reaction.

E cell = Ecathode – Eanode

If E cell is positive, the cell reaction is spontaneous, otherwise not.

To predict the Replacement tendency :-

The relative ease with which the various species of metals and ions may be oxidized or reduced is indicated by the reduction potential values. The metals with lower reduction potential are not reduced easily but are easily oxidized to their ions losing electrons. These electrons would reduce the other metals having higher reduction potentials. In other words, a metal having smaller reduction potential can displace metals having larger reduction potentials from the solution of their salt. For example, copper lies above silver in the electrochemical series, therefore, if copper metal is added to AgNO3 solution, silver is displaced from the solution. In general a metal occupying higher position in the series can displace the metals lying below it from the solutions of their salts and so are more reactive in displacing the other metals. Thus, Li is the most electropositive element in solutions and fluorine is the most electronegative element.

To predict the correct Metallurgical Methods :-

Eo values of Cu, H2O and Al are +0. 34V, -0. 83V and -1. 66V. It means Cu gets more easily reduced than water and water gets more easily reduced than aluminium. Hence copper can be produced by the electrolysis of aqueous copper sulphate but not aluminium. this is due to the fact that when Al3+(aq) is electrolysed, the H2O will be electrolysed but not Al3+(aq).