Electrochemical Series - Definition, Char and Applications

The electrochemical series, also known as the electrode potential series, was developed by the German chemist Walter Nernst. Nernst's work in the late 19th century provided a theoretical framework for understanding electrochemical reactions and electrode potentials. Walter Nernst’s development of the electrochemical series provided crucial insights into electrochemical reactions and their potential, greatly advancing the field of electrochemistry.

This Story also Contains

1. Electrochemical Series
2. Characteristics of Electrochemical Series
3. Some Solved Examples
4. Summary

Electrochemical series are discovered based on electrode potential and the Nernst equation in such a way that the electrochemical series is based on the standard electrode potentials of various half-reactions. These potentials are measured relative to a standard reference electrode, typically the standard hydrogen electrode (SHE), which is assigned a potential of 0 volts. And Nernst equation as Nernst formulated the Nernst equation, which relates the electrode potential to the concentration of ions in solution. This equation helps predict how the electrode potential changes with concentration and temperature.

Electrochemical Series

Characteristics of Electrochemical Series

Metals with greater negative E^o (reduction) are strongly electropositive and have more reactivity. It means a lower-placed element or metal in the given series is less reactive is replaced by upper placed or higher element while a higher element can be coated by a lower metal.
Example, (i) $\mathrm{Zn}+\mathrm{CuSO}_4 \rightarrow \mathrm{ZnSO}_4+\mathrm{Cu}$
Here Cu is replaced by Zn due to more oxidation potential or reactivity of Zn, while Zn is coated by Cu. Zn- Cu couple is also coated by Cu. Here, the solution turns from blue to colorless and the rod becomes Reddish-brown from Gray white.
(ii) $\mathrm{Cu}+2 \mathrm{AgNO}_3 \rightarrow \mathrm{Cu}\left(\mathrm{NO}_3\right)_2+2 \mathrm{Ag}$
Here solution becomes colorless to blue and the rod becomes reddish-brown to white.

• Metals above H₂ can easily replace H₂, from acid, bases, etc. due to their more positive E^o_op or reactivity.
  For example,
  $\mathrm{Mg}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{MgSO}_4+\mathrm{H}_2$$\mathrm{E}_{\mathrm{op}}^{\circ}$ of $\mathrm{Mg}>\mathrm{E}_{\mathrm{OP}}^{\circ}$ of $\mathrm{H}_2$
  $$
  \mathrm{R}-\mathrm{OH}+\mathrm{Na} \rightarrow \mathrm{R}-\mathrm{ONa}+\mathrm{H}^{+}
  $$
• Lower placed metals (Cu Hg Ag Pt Au) to H₂ can not do that as E^o_op of H₂ is more than their E^o_op.$\mathrm{Cu}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow$ no reaction
• Oxides of lower metals (Cu, Hg, Ag, Pt, Au) are easily reduced by H₂ or carbon. As they are thermally more unstable due to positive E_rp, they also decompose on heating.
• More E^o_OP means more ease or tendency to get oxidized that is, they act as better reducing agents while more E^o_RP means more ease to reduce that is, they act as better oxidizing agents. It means metal above hydrogen having positive E_op are reducing agents.
  Reducing property $\propto \mathrm{E}_{\mathrm{OP}}^{\circ}$
  For example, Li is the strongest reducing agent due to maximum E^o_OP
• Metals placed lower in the reactivity series (Cu Hg Ag Pt Au) having high E^o_RP are oxidizing agents and they tend to be reduced.
  
  
  For example, Oxidizing power $\propto \mathrm{E}_{\mathrm{RP}}$
  $\mathrm{F}_2>\mathrm{Cl}_2>\mathrm{Br}_2>\mathrm{I}_2$
  Reducing power decreases
  As $\mathrm{E}_{\mathrm{op}}^{\circ}$ of $\mathrm{I}^{-}>\mathrm{Br}^{-}>\mathrm{Cl}^{-}>\mathrm{F}^{-}$
• Elements with more positive E^o_RP will be discharged first at the cathode i.e., discharging order increases as reduction potential increases.

Increasing ease of deposition of some cations

$\mathrm{Li}^{+}, \mathrm{K}^{+}, \mathrm{Ca}^{+2}, \mathrm{Na}^{+}, \mathrm{Mg}^{+2}, \mathrm{Al}^{+3}, \mathrm{Zn}^{+2}, \mathrm{Fe}^{+2}, \mathrm{H}^{+}, \mathrm{Cu}^{+2}, \mathrm{Ag}^{+}, \mathrm{Au}^{+3}$

• In the case of negative ions, an anion with a stronger reducing nature is discharged first at the anode.

Increasing ease of discharge of some anion$\mathrm{SO}_4^{-2}<\mathrm{NO}_3^{-}<\mathrm{OH}^{-}<\mathrm{Cl}^{-}<\mathrm{Br}^{-}$

• Hydroxides of upper metals are strongly basic and their salts do not undergo hydrolysis while hydroxides of lower metals are weakly acidic and their salts undergo hydrolysis.

Recommended topic video on(Electrochemical Series )

Some Solved Examples

Example.1

1. Which of the following can replace hydrogen from its compounds, where it has a +1 oxidation state?

a)Na

b)Hg

c)Zn

d)Fe

1)a,b,c

2)c,d

3)a,c

4) (correct)a,c,d

Solution

The standard reduction potential of a large number of electrodes has been measured using a standard hydrogen electrode as the reference electrode. These various electrodes can be arranged in increasing electrode potential.

According to the electrochemical Series -

$\mathrm{Na}^{+} / \mathrm{Na} \mathrm{Zn}{ }^{2+} / \mathrm{Zn}, \mathrm{Fe}^{2+} / \mathrm{Fe}, \mathrm{H}^{+} / \mathrm{H}_2, \mathrm{Hg}_2^{2+} / \mathrm{Hg}$

Hg lies below hydrogen in the electrochemical series and is, therefore, less reactive

Hence, the answer is the option (4).

Example.2

Which of the following metals is the least reactive?

1)Al

2) (correct)Cu

3)Fe

4)Zn

Solution

According to the electrochemical Series -

The order of the reduction potential is:

$\mathrm{Al}^{3+} / \mathrm{Al} \mathrm{Zn}^{2+} / \mathrm{Zn}, \mathrm{Fe}^{2+} / \mathrm{Fe}^{2+}, \mathrm{Cu}^{2+} / \mathrm{Cu}$

When we arrange these metals in order of their electropositive character, then Al>Zn>Fe>Cu
Hence, the answer is the option (2).

Example.3

3. Is it possible to store, copper sulphate solution in a zinc vessel?

1) (correct)No

2)yes

3)yes, only above 25⁰C

4)can't say

Solution

It is not possible to store, copper sulfate solution in a zinc vessel because Cu will be deposited on zinc.

Hence, the answer is the option (1).

Example.4

4. Given :

$\mathrm{Co}^{3+}+e^{-} \rightarrow \mathrm{Co}^{2+}+; \mathrm{E}^0=+1.81 V$

$\mathrm{Pb}^{4+}+2 e^{-} \rightarrow \mathrm{Pb}^{2+}+; E^0=+1.67 \mathrm{~V}$

$C e^{4+}+e^{-} \rightarrow C e^{3+}+; E^0=+1.61 V$

$B i^{3+}+3 e^{-} \rightarrow B i ; E^0=+0.20 \mathrm{~V}$

Oxidizing power of the species will increase in the order:

1) (correct)$\mathrm{Bi}^{3+}<\mathrm{Ce}^{4+}<\mathrm{Pb}^{4+}<\mathrm{Co}^{3+}$

2)$\mathrm{Ce}^{4+}<\mathrm{Pb}^{4+}<\mathrm{Bi}^{3+}<\mathrm{Co}^{3+}$

3)$\mathrm{Co}^{3+}<\mathrm{Ce}^{4+}<\mathrm{Bi}^{3+}<\mathrm{Pb}^{4+}$

4)$\mathrm{Co}^{3+}<\mathrm{Pb}^{4+}<\mathrm{Ce}^{4+}<\mathrm{Bi}^{3+}$

Solution

The greater the Standard Reduction Potential, the more will be its oxidizing power.

$\therefore$ The correct sequence will be :

$\mathrm{Co}^{3+}>\mathrm{Pb}^{4+}>\mathrm{Ce}^{4+}>\mathrm{Bi}^{3+}$

Hence, the answer is the option (1).

Example.5

5. The correct order of reduction potentials of the following pairs is

A. $\mathrm{Cl}_2 \mid \mathrm{Cl}^{-}$

B. $\mathrm{I}_2 \mid \mathrm{I}^{-}$

C. $\mathrm{Ag}^{+} \mid \mathrm{Ag}$

D. $\mathrm{Na}^{+} \mid \mathrm{Na}$

E. $\mathrm{Li}^{+} \mid \mathrm{Li}$

1) (correct)$A>C>B>D>E$

2)$A>B>C>D>E$

3)$A>C>B>E>D$

4)$A>B>C>E>D$

Solution

Fact-based on the reduction potential values.

The correct order will be

$\mathrm{Cl}_2\left|\mathrm{Cl}^{\ominus}>\mathrm{Ag}^{+}\right| \mathrm{Ag}>\mathrm{I}_2\left|\mathrm{I}^{\ominus}>\mathrm{Na}^{\oplus}\right| \mathrm{Na}>\mathrm{Li}^{\oplus} \mid \mathrm{Li}$

Thus, the given electrode couple can be arranged in order of their reduction potential values as

$\mathrm{A}>\mathrm{C}>\mathrm{B}>\mathrm{D}>\mathrm{E}$

Hence, the answer is the option (1).

Example.6

6. In van der Waals equation of state of the gas law, the constant b is a measure of

1)intermolecular repulsions

2)intermolecular attraction

3) (correct)volume occupied by the molecules

4)intermolecular collisions per unit volume.

Solution

As we learned in

Vander Waal equation for real gas -

$\left(p+\frac{a n^2}{v}\right)(V-n b)=n R T$

- wherein

a, b : Vander waal Constants, P- Pressure, V- Volume, n- No. of moles, R- Gas Constant, T- Temperature

The constant b is a measure of the volume of the molecule

$b=4 N_A \times \frac{4}{3} \pi r^3$

The correct option is 3.

Summary

Electrochemical cells are devices that convert chemical energy into electrical energy through redox reactions. There are two main types: galvanic (or voltaic) cells, which produce electrical energy from spontaneous chemical reactions, and electrolytic cells, which require an external power source to drive non-spontaneous reactions. A typical electrochemical cell consists of two electrodes (anode and cathode) and an electrolyte. The anode is where oxidation occurs, and the cathode is where reduction takes place. The flow of electrons between the electrodes through an external circuit generates electrical energy.