Standard reduction electrode potentials of a few elements at 298 K
|a* = increasing tendency||b* = increasing tendency|
|for oxidation||for reduction|
|to lose electrons||to gain electrons|
|as strong reducing agent||as strong oxidizing agent|
The E° value of an electrode gives the relative tendency for the reduction reaction to occur at the electrode as compared to that of reduction of H+ ion under standard conditions. Therefore, the electrodes with positive electrode potentials show greater tendency towards reduction than the reduction tendency shown by H+. On the other hand, the electrodes with negative electrode potentials show lesser tendency towards reduction than the reduction tendency of H+ ions. This difference in the reduction tendencies of the two electrodes in a cell finds many applications in chemistry.
Applications of electrochemical series
Oxidizing and Reducing Strengths
The electrochemical series helps to pick out substances that are good oxidizing agents and those which are good reducing agents. For example, a very high negative reduction potential of lithium electrode indicates that it is very difficult to reduce Li+ ions to Li atoms. Therefore, Li+ cannot accept electrons easily and so loses electrons to behave as a reducing agent. As the reduction potential increases (negative value decreases), the tendency of the electrode to behave as reducing agent decreases. Thus, all the substances appearing on the top of the series behave as good reducing agents. For example Li and K are good reducing agents while F- and Au are the poorest reducing agents.Similarly, substances at the bottom of the table have high reduction potential and they can be easily reduced. Therefore, they act as strong oxidizing agents. From the table we can conclude that H+ is a better oxidizing agent than Zn2+ while Cu2+ is a better oxidizing agent than H+; Fe is a better oxidizing agent than Cl2 and so on. All the substances appearing at the bottom of the table are good oxidizing agents.
Comparison of Reactivities of Metals
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.
Calculation of the EMF of the Cell
The following steps determine the reduction potential of the cathode and anode:
The two half-cell reactions are written in such a way that the reaction taking place at the left hand electrode is written as an oxidation reaction and that taking place at the right electrode is written as reduction reaction.
The number of electrons in the two equations are made equal by multiplying one of the equations if necessary by a suitable number. However, electrode potential values (E°) are not multiplied.
The electrode potentials of both the electrodes are taken to be reduction potentials and so the EMF of the cell is equal to the difference between the standard potential of the right hand side and the left hand side electrode.Ecell = ER - EL
If the EMF of the cell is +ve, the reaction is feasible in the given direction and the cell is correctly represented, i.e., oxidation occurs at left electrode (anode) and reduction occurs at the right electrode (cathode). If it is -ve, the cell reaction is not feasible in the given direction and the cell is wrongly represented. Thus, to get positive value for the EMF the electrodes must be reversed.
Predicting the Liberation of Hydrogen Gas from Acids by Metals
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.
Predicting Feasibility of a 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.
6. 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.
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:
7. 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:
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°.
8. What will be the spontaneous reaction between the following half-cell reactions?
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