2.9 Relation Between the Faraday Constant, Avogadro Constant, and the Charge on the Electron | Electrochemistry | Chemistry 12

This section explores the fundamental relationship between the quantity of electricity, the charge of an electron, and the Avogadro constant, leading to the definition and application of the Faraday constant in electrochemistry.

Oxidation Reduction Concepts→

2.9.1 Quantity of Electric Charge in Electrolysis

During electrolysis, the amount of substance produced at an electrode is directly proportional to the total electric charge that passes through the electrolyte. This charge is related to the current and the time for which it flows.

Definition: The quantity of charge ( $Q$ ) is the product of the constant current ( $I$ ) and the time ( $t$ ) for which the current is passed.

Formula: $Q = I \times t$

Where:

$Q$ = Quantity of charge in coulombs (C)
$I$ = Current in Amperes (A)
$t$ = Time of electrolysis in seconds (s)

2.9.2 The Faraday Constant (F)

The Faraday constant is a crucial concept in electrochemistry, representing the amount of electric charge carried by one mole of electrons.

Calculation: The Faraday constant can be derived by multiplying the charge on a single electron by the Avogadro constant.

Charge on one electron ( $e$ ) $= 1.60217662 \times 1 0^{- 19} C$
Avogadro's constant ( $N_{A}$ ) $= 6.022 \times 1 0^{23} mol^{- 1}$
Charge on one mole of electrons $= (1.60217662 \times 1 0^{- 19} C) \times (6.022 \times 1 0^{23} mol^{- 1})$ $= 96485.332 C mol^{- 1}$

Definition: This quantity of charge, $96485.332 C mol^{- 1}$ , is referred to as one Faraday (F).

Approximation: For most calculations, the value of the Faraday constant is approximated as $96500 C mo l^{- 1}$ .

2.9.3 Relationship Between Faraday Constant, Avogadro Constant, and Elementary Charge

The fundamental relationship connecting these three constants is expressed by the equation:

$F = N_{A} \cdot e$

Where:

$F$ = the Faraday constant (charge per mole of electrons, typically $96485.332 C mol^{- 1}$ or $96500 C mo l^{- 1}$ )
$N_{A}$ = the Avogadro constant (number of entities per mole, approximately $6.022141 \times 1 0^{23} mol^{- 1}$ )
$e$ = the elementary charge (charge on a single electron, approximately $1.60217662 \times 1 0^{- 19} C$ )

2.9.4 Stoichiometry of Electrolysis and Faraday's Constant

The Faraday constant provides a direct link between the amount of electricity passed and the moles of substance produced or consumed in an electrochemical reaction.

Example 1: Deposition of Sodium The reduction of a sodium ion ( $N a^{+}$ ) to sodium metal ( $N a$ ) involves the transfer of one electron per ion: $N a^{+} + e^{-} \to N a$ To deposit one mole of $N a$ , one mole of electrons is required. Therefore, the amount of electricity needed is $1 F$ ( $96500 C$ ).

Example 2: Deposition of Copper The reduction of a copper(II) ion ( $C u^{2 +}$ ) to copper metal ( $C u$ ) involves the transfer of two electrons per ion: $C u^{2 +} + 2 e^{-} \to C u$ To deposit one mole of $C u$ , two moles of electrons are required. Therefore, the amount of electricity needed is $2 F$ .

General Principle: If an electrochemical reaction requires $n$ electrons to produce or consume one mole of a substance, then $n$ Faradays of charge are needed to complete the reaction for one mole of that substance.

Experimental Determination Of Avogadro Constant→

Worked Examples

Example 2.8: Zinc Deposition

In the electrolysis of molten $ZnC l_{2}$ , how much Zn can be deposited at the cathode by the passage of a $0.01$ -ampere current for one hour?

Solution:

Given values:
- Current ( $I$ ) = $0.01 A$
- Time ( $t$ ) = $1 hour = 3600 s$
- Faraday constant ( $F$ ) = $96500 C$
- Molar mass of Zn ( $M_{Z n}$ ) = $63.37 g mo l^{- 1}$ (Textbook value)
Calculate the total charge ( $Q$ ) passed: $Q = 0.01 A \times 3600 s = 36 C$
Convert charge ( $Q$ ) to Faradays ( $F_{t o t a l}$ ): $F_{t o t a l} = \frac{36 C}{96500 C/F} = 3.73 \times 1 0^{- 4} F$
Write the cathode reaction: $Z n^{2 +} + 2 e^{-} \to Z n$ $2 F$ of charge are required to deposit $1 mol$ of $Z n$ .
Calculate moles of Zn deposited: $n_{Z n} = 3.73 \times 1 0^{- 4} F \times \frac{1}{2} = 1.865 \times 1 0^{- 4} mol Z n$
Calculate the mass of Zn deposited: $Mass = 1.85 \times 1 0^{- 4} mol \times 63.37 g mo l^{- 1} \approx 0.012 g$

Example 2.9: Gold and Chlorine Production

A constant current was passed through a solution of $AuC l_{4}^{-}$ ions between gold electrodes. After a period of $10.0$ minutes, the cathode increased in weight by $1.314$ grams.

i) How much charge was passed? ii) What was the amount of current? iii) What volume of $C l_{2}$ was collected at the anode at $1 atm$ and $2 5^{\circ} C$ ?

Solution:

Cathode reaction: $A u C l_{4}^{-} + 3 e^{-} \to A u + 4 C l^{-}$ (3 Faradays per mole of Au)
Moles of Au: $n_{A u} = \frac{1.314 g}{197 g mo l ^{- 1}} = 6.67 \times 1 0^{- 3} mol$
Charge in Faradays: $F_{t o t a l} = 6.67 \times 1 0^{- 3} \times 3 = 2.001 \times 1 0^{- 2} F$
Current ( $I$ ): $Q = 2 \times 1 0^{- 2} \times 96500 = 1930 C$ . $I = \frac{1930}{600} = 3.22 A$
Anode reaction: $2 C l^{-} \to C l_{2} + 2 e^{-}$ (2 Faradays per mole of $C l_{2}$ )
Moles of $C l_{2}$ : $n_{C l_{2}} = \frac{2 \times 1 0 ^{- 2} F}{2} = 1 \times 1 0^{- 2} mol$
Volume of $C l_{2}$ : $V = \frac{n R T}{P} = \frac{0.01 \times 0.0821 \times 298}{1} = 0.245 dm^{3}$

Worked Examples

In the electrolysis of molten

ZnC l_{2}

, how much Zn can be deposited at the cathode by the passage of a

0.01

-ampere current for one hour?

Solution:

Given values:

Current ( $I$ ) = $0.01 A$
Time ( $t$ ) = $1 hour = 3600 s$
Faraday constant ( $F$ ) = $96500 C$
Molar mass of Zn ( $M_{Z n}$ ) = $63.37 g mo l^{- 1}$ (Textbook value)

Calculate the total charge ( $Q$ ) passed: $Q = 0.01 A \times 3600 s = 36 C$

Convert charge ( $Q$ ) to Faradays ( $F_{t o t a l}$ ): $F_{t o t a l} = \frac{36 C}{96500 C/F} = 3.73 \times 1 0^{- 4} F$

Write the cathode reaction: $Z n^{2 +} + 2 e^{-} \to Z n$ $2 F$ of charge are required to deposit $1 mol$ of $Z n$ .

Calculate moles of Zn deposited: $n_{Z n} = 3.73 \times 1 0^{- 4} F \times \frac{1}{2} = 1.865 \times 1 0^{- 4} mol Z n$

Calculate the mass of Zn deposited: $Mass = 1.85 \times 1 0^{- 4} mol \times 63.37 g mo l^{- 1} \approx 0.012 g$

A constant current was passed through a solution of

AuC l_{4}^{-}

ions between gold electrodes. After a period of

10.0

minutes, the cathode increased in weight by

1.314

grams.

i) How much charge was passed? ii) What was the amount of current? iii) What volume of $C l_{2}$ was collected at the anode at $1 atm$ and $2 5^{\circ} C$ ?

Solution:

Cathode reaction:

A u C l_{4}^{-} + 3 e^{-} \to A u + 4 C l^{-}

(3 Faradays per mole of Au)

Moles of Au:

n_{A u} = \frac{1.314 g}{197 g mo l ^{- 1}} = 6.67 \times 1 0^{- 3} mol

Charge in Faradays:

F_{t o t a l} = 6.67 \times 1 0^{- 3} \times 3 = 2.001 \times 1 0^{- 2} F

Current ( $I$ ):

Q = 2 \times 1 0^{- 2} \times 96500 = 1930 C

I = \frac{1930}{600} = 3.22 A

Anode reaction:

2 C l^{-} \to C l_{2} + 2 e^{-}

(2 Faradays per mole of

C l_{2}

)

Moles of $C l_{2}$ :

n_{C l_{2}} = \frac{2 \times 1 0 ^{- 2} F}{2} = 1 \times 1 0^{- 2} mol

Volume of $C l_{2}$ :

V = \frac{n R T}{P} = \frac{0.01 \times 0.0821 \times 298}{1} = 0.245 dm^{3}