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ChemistryEdu Logo Electrochemistry | Introduction#

Electrochemistry can be defined as the branch of chemistry which deals with the study of relationships between electrical energy and chemical energy and interconversion of one form of energy into another.

Charge#

  • A charged object can be either electron deficient (positively charged) or electron surplus (negatively charged).
  • If 'i' amount of current flows per second in a wire, then after time 't', the charge, Q can be calculated as:
\[Q = it\]
  • Standard unit of charge is coulomb, denoted by C.
  • Other commonly used unit of charge in electrochemistry is Faraday or Farad denoted by F.

Definition of Faraday#

  • 1 Faraday can be defined as the charge on 1 mole of electrons.
  • Let us derive relation between Faraday(F) and Coulomb(C):
\[Charge\ on\ 1\ electron = 1.6 \times 10^{-19} C\]
\[Charge\ on\ 1\ mole\ electrons = 6.022 \times 10^{23} \times 1.6 \times 10^{-19} C\]
\[∴\ 1\ F = 96500\ C\]

Example

4 F charge means charge on 4 moles of electrons. Similarly, 0.982 F means charge on 0.982 moles of electrons.

  • Number of Faradays can be calculated as:
\[Number\ of\ Faradays = {Charge\ in\ C \over 96500}\]
\[Number\ of\ Faradays = {it \over 96500}\]

Where, 'i' is in Ampere and 't' in seconds.

Examples

In the given examples, let us consider 1 F of charge is used or 1 mole of electrons is used.

\[Ag_{(aq)}^+ + e^- → Ag_{(s)} ↓ (1\ mole\ Ag\ is\ deposited)\]
\[Cu_{(aq)}^{2+} + 2e^- → Cu_{(s)} ↓ ({1\over 2}\ moles\ Cu\ is\ deposited)\]
\[Al_{(aq)}^{3+} + 3e^- → Al_{(s)} ↓ ({1\over 3}\ moles\ Al\ is\ deposited)\]

Similarly:

\[M_{(aq)}^{n+} + ne^- → M_{(s)} ↓ ({1\over n}\ moles\ M\ is\ deposited)\]

Thus, if 1 Farad of charge supplied to 1 mole of Mn+, $ {1\over n} $ moles of metal, M is deposited.

  • If x Farad charge is supplied to 1 mole of Mn+, \({x\over n}\) moles of metal, M is deposited.
  • We can also say that: If x Farad charge is supplied to 1 mole of Mn+, then mass of metal deposited = \({xM \over n}\), where M = molar mass of metal.

Some Important Terms#

1. Electrolyte#

An electrolyte is a substance which can make water conductive when mixed wih it. Example: Pure water is non-conductive but when salt(NaCl) is added to it, it becomes conductive. Hence, NaCl is an electrolyte.

2. Non-Electrolyte#

A non-electrolyte is a substance which cannot make water conductive when mixed wih it. Example: A sugar solution is non-conductive. Hence, sugar is a non-electrolyte.

3. Electrolytic Solution#

An electrolytic solution is a solution which is capable of conducting current. When an electrolyte is added to water, it becomes an electrolytic solution. Example, NaCl solution is an electrolytic solution.

4. Electrode#

It is a rod which is immersed in an electrolytic solution and conducts electricity. There are two types of electrodes: (a) Cathode (b) Anode.

5. Anode#

Anode is an electrode where oxidation takes place.

6. Cathode#

Cathode is an electrode where reduction takes place.

Questions#

In an electrolytic solution, 10 gm of Ag needs to be deposited on Cu surface. Calculate the time taken in this process if current passed is 2 x 10-3 A.

\[Ag^+ + e^- → Ag ↓\]
\[Mass\ of\ Ag\ formed = 10\ g\]
\[Moles\ of\ Ag\ formed = {10 \over 108}\]
\[Moles\ of\ electrons\ required = {10 \over 108}\]
\[∴ Charge\ required(Q) = {10 \over 108} F\]
\[Q = {10 \over 108} \times 96500\ C\]
\[it = {10 \over 108} \times 96500\]
\[ {2 \times 10^{-3}} t = {10 \over 108} \times 96500\]
\[t = {10 \times 96500 \times 1000 \over 108 \times 2}\]
\[t = 4467592.593\ s = 1241\ hours\]

Hence, the time taken for deposition of 10 g Ag is 1241 hours.

In an electrolytic solution Cu gets deposited if 2 A current is passed for 3 hours. Calculate the mass of Cu deposited.

\[Cu^{2+} + 2e^- → Cu ↓\]
\[Charge\ passed,\ Q = it\]
\[Q = 2 \times 3\times 60\times 60 = 21600\ C\]
\[Q = {21600 \over 96500} F \]
\[Moles\ of\ electrons\ used = {21600 \over 96500}\]

If 2 moles electrons are used, 1 mole Cu is deposited.

If \({21600 \over 96500}\) moles electrons are used, \({21600 \over 96500 \times 2}\) moles Cu is deposited.

\[Mass\ of\ Cu\ deposited = {21600 \over 96500 \times 2} \times 63.5\ g\]
\[Mass\ of\ Cu\ deposited = 7.107\ g\]