Therefore, the reactivity of metals decreases in increasing order of their electrode potentials and metals in the electrochemical series are placed in decreasing order of their reactivity.

Example: Fe is a more reactive metal than Cu.

Putting a more active metal in a solution of the cations of a less active metal frees the less active metal.

Example: Fe is located above Cu in the Electrochemical Series and is more reactive than Cu.

2Fe + 3CuSO4 → Fe2(SO4)3 + 3Cu

Therefore, by releasing Fe into the CuSO4 solution, it liberates Cu.

2) With the help of the electrochemical Series, information is available about the ability of different metals to displace hydrogen from acids.

In the electrochemical Series, metal is located above hydrogen, compared to hydrogen.

• Their electrode potentials are lower,
• Their reduction potentials are lower,
• They have higher oxidation potential.
• They are prone to oxidation in semi-reactions.
• They have a higher tendency to oxidize.
• Those metals have a higher tendency to make cations.
• Those metals can displace hydrogen from acids.

Therefore, in the electrochemical category, the metal which is located above hydrogen, displaces hydrogen from the acids.

In other words, the metals whose electron potentials are lower in the electrode potential of hydrogen displace hydrogen from the metal acids. Metals whose electrode potentials are greater than the electrode potential of hydrogen do not displace hydrogen from the metal acids.

2Al + 6HCl → 2AlCl3 + 3H2

Zn + 2HCl → ZnCl2 + H2

Example: Aluminum and zinc lie above hydrogen in the electrochemical series. The standard electrode potentials of Al, Zn and hydrogen are 1.66, -0.76 and 0 volts respectively. Hence, Al and Zn displace hydrogen from HCl. The reaction speed of Al with HCl is faster than the reaction of Zn with HCl.

Copper is located below the hydrogen in the electrochemical series. The standard electrode potential of Cu is +0.34 volts which is higher than the standard electrode potential of hydrogen. Thus Cu cannot displace hydrogen from HCl.

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3) With the help of the electrochemical series, the oxidizing capacities of different oxidants and the oxidizing capacities of different catabolists can be compared.

From the electrode potential of a semi-reaction, it is known how much it is prone to oxidation or reduction. If the electrode potential of a semi-reaction is high then –

• The reduction potential of that semi-reaction will be higher
• In that semi-reaction, the tendency of reduction will be more
• The oxidizer used in that semi-reaction will be strong oxidizer.
• The reducing agent used in that half-reaction will be a weak reducing agent.

Similarly, if the electrode potential of a semi-reaction is low, the reducing agent used in that semi-reaction will be a strong oxidizer and the oxidizer used in that semi-reaction will be a weak oxidizer.

Example:

Zn2+ + 2e– ⇌ Zn (E0= -0.76Volt)

Cu2+ + 2e– ⇌ Cu (E0= +0.34Vold)

The above semi-reactions are written for reduction. Hence the voltage E0 written with them shows their standard reduction potential. In the above semi-reactions, Zn2+ is Cu2+ oxidant and Zn is Cu oxidant.

Since the reduction potential of the first half-reaction (Zn2+ + 2e ⇌ Zn) is low, so in this half-reaction, oxidizer will be strong(Zn → Zn2+ + 2e), Zn is a strong oxidizer and Zn2+ is a weak oxidizer. Since the second half-reaction(Cu2+ + 2e ⇌ Cu) has a higher reduction potential, so this half-reaction(Cu2+ + 2e → Cu) will have a reduction, Cu2+ is a strong oxidizer and Cu is a weak oxidizer.

Following is the mixed equation of both the above semi-reactions.

Zn + Cu2+ ⇌ Zn2+ + Cu

Since Zn is a fast reducing agent and Cu2+ is a strong oxidizer, this reaction is possible in the right direction. This reaction is not possible in the left direction because Zn2+ is a weak oxidizer and Cu is a weak oxidizer.

Zn + Cu2+ → Zn2+ + Cu

When Zn rods are added to CuSO4 solution, Zn will dissolve in solution and Cu will precipitate but there will be no action if Cu rod is added to ZnZO4 solution.

4) The E.M.F. of a galvanic cell can be calculated from the electrochemical Series.

E.M.F. of a galvanic cell = potential of positive electrode – potential of negative electrode

Standard electrode potentials of various electrodes are given in the electrochemical series. Therefore, with its help the E.M.F. of a galvanic cell can be calculated.

Example: If a galvanic cell has zinc and copper rods immersed in one molar solution of ZnSO4 and CuSO4, then the zinc rod has negative charge and negative potential and the copper rod has positive charge and money potential. The device consists of a Cu positive electrode and a Zn negative electrode. Cu has E0 = +0.34 volts and Zn has E0 = -0.76 volts.

Therefore

ECell = E