2.10 Experimental Determination of Avogadro Constant By Electrolytic Method | Electrochemistry | Chemistry 12

Electrolysis provides a precise experimental method to determine the Avogadro Constant ( $N_{A}$ ). This technique involves passing a known quantity of electric current through an electrolytic cell containing a specific electrolyte. By measuring the mass of metal deposited at the cathode over a known period, $N_{A}$ can be calculated. This process is a practical application of Oxidation Reduction Concepts→.

Principle of the Method

The core principle relies on Faraday's laws of electrolysis, which state that the amount of chemical change produced by electric current is proportional to the quantity of electricity passed.

Measure Quantity of Charge ( $Q$ ): By passing a known current ( $I$ ) for a known time ( $t$ ), the total charge can be calculated using $Q = I \times t$ .
Measure Mass Deposited: The mass of the metal deposited at the cathode is accurately measured.
Relate Charge to Moles: Using the molar mass of the deposited metal and the stoichiometric relationship from the half-reaction at the cathode, the charge required to deposit one mole of the metal can be determined.
Calculate Avogadro's Constant: Since the charge on a single electron is known, dividing the total charge for one mole of electrons (Faraday's constant) by the charge of a single electron yields the number of electrons in one mole, which is Avogadro's Constant.

Cathode Reaction for Silver Deposition

In the electrolysis of an aqueous silver nitrate ( $AgN O_{3}$ ) solution using silver electrodes, silver ions ( $A g^{+}$ ) from the solution gain electrons at the cathode to form solid silver metal ( $A g_{(s)}$ ).

$A g_{(aq)}^{+} + e^{-} \to A g_{(s)}$

This reaction indicates that one mole of electrons is required to deposit one mole of silver atoms. For more complex reactions, see Balancing Of Equations→.

Worked Examples

Example: Calculating Avogadro Constant from Silver Deposition

Let's use the provided experimental data for the electrolysis of $AgN O_{3}$ to calculate the Avogadro Constant.

Given Values:
- Current ( $I$ ) = $0.1 A$
- Time ( $t$ ) = $30 min$
- Mass of Ag deposited = $0.201 g$
- Molar Mass of Ag ( $M_{A g}$ ) = $107.868 g/mol$
- Charge on one electron ( $e$ ) = $1.602 \times 1 0^{- 19} C$
Convert Time to Seconds: $t = 30 min \times \frac{60 s}{1 min} = 1800 s$
Calculate Total Charge ( $Q$ ) Passed: $Q = I \times t$ $Q = 0.1 A \times 1800 s = 180 C$

This means $0.201 g$ of Ag is produced by $180 C$ of electricity.
Calculate Charge Required for 1 Mole of Ag: If $0.201 g$ of Ag requires $180 C$ , then $1 mol$ (or $107.868 g$ ) of Ag would require: $Charge per mole of Ag = \frac{180 C}{0.201 g Ag} \times 107.868 g Ag/mol$ $Charge per mole of Ag = 96598.209 C/mol$

This charge must be present on the electrons responsible for producing $1 mol$ of Ag.
Relate Charge to Moles of Electrons (Faraday's Constant): From the cathode reaction, $A g_{(aq)}^{+} + 1 e^{-} \to A g_{(s)}$ , we know that $1 mole$ of electrons is required to deposit $1 mole$ of silver.

Therefore, the charge on $1 mole$ of electrons (Faraday's Constant, $F$ ) is approximately $96598.209 C/mol$ .
Calculate Avogadro's Constant ( $N_{A}$ ): Since we know the charge on one electron ( $e$ ) and the charge on one mole of electrons ( $F$ ), we can calculate the number of electrons in one mole: $N_{A} = \frac{Charge on 1 mole of electrons}{Charge on one electron} = \frac{F}{e}$ $N_{A} = \frac{96598.209 C}{1.602 \times 1 0 ^{- 19} C/electron}$ $N_{A} = 6.0298 \times 1 0^{23} electrons/mol$

This calculated value ( $6.0298 \times 1 0^{23}$ ) is very close to the accepted value of $6.02214 \times 1 0^{23}$ .

Possible Questions/Answers

CONCEPT ASSESSMENT EXERCISE 2.10

Q1: An electrolytic cell is connected to a power source for two hours. If the current flowing through the cell is $0.5 A$ during this time, find the mass of the substance liberated during this time interval. (The molar mass of the substance is $107.9 g/mol$ .)

A1:
1. Given values:
  - Time ( $t$ ) = $2 hours$
  - Current ( $I$ ) = $0.5 A$
  - Molar Mass ( $M$ ) = $107.9 g/mol$
  - Faraday's constant ( $F$ ) $\approx 96485 C/mol of e^{-}$ (assuming a 1-electron transfer process).
2. Convert time to seconds: $t = 2 hours \times \frac{3600 s}{1 hour} = 7200 s$
3. Calculate total charge ( $Q$ ): $Q = I \times t = 0.5 A \times 7200 s = 3600 C$
4. Calculate moles of electrons transferred: $Moles of e^{-} = \frac{Q}{F} = \frac{3600 C}{96485 C/mol e ^{-}} \approx 0.03731 mol e^{-}$
5. Calculate moles of substance liberated (assuming 1 mole of e- per mole of substance): $Moles of substance = Moles of e^{-} = 0.03731 mol$
6. Calculate mass of substance liberated: $Mass = Moles \times M = 0.03731 mol \times 107.9 g/mol \approx 4.026 g$
Q2: Calculate the charge in coulombs when $3 moles$ of electrons flow through a circuit.

A2:
1. Given values:
  - Moles of electrons ( $n_{e}$ ) = $3 mol$
  - Faraday's constant ( $F$ ) $\approx 96485 C/mol e^{-}$
2. Calculate total charge ( $Q$ ): $Q = n_{e} \times F = 3 mol \times 96485 C/mol e^{-} = 289455 C$
Q3: Calculate the mass of silver deposited at the cathode during electrolysis of $AgN O_{3}$ solution, if you use a current of $0.1 A$ for $20 minutes$ .

A3:
1. Given values:
  - Current ( $I$ ) = $0.1 A$
  - Time ( $t$ ) = $20 min$
  - Molar Mass of Ag ( $M_{A g}$ ) = $107.868 g/mol$
  - Faraday's constant ( $F$ ) $\approx 96485 C/mol e^{-}$ (for $A g^{+} + e^{-} \to Ag$ ).
2. Convert time to seconds: $t = 20 min \times \frac{60 s}{1 min} = 1200 s$
3. Calculate total charge ( $Q$ ): $Q = I \times t = 0.1 A \times 1200 s = 120 C$
4. Calculate moles of electrons transferred: $Moles of e^{-} = \frac{Q}{F} = \frac{120 C}{96485 C/mol e ^{-}} \approx 0.0012437 mol e^{-}$
5. Calculate moles of silver deposited: From the reaction $A g_{(aq)}^{+} + e^{-} \to A g_{(s)}$ , $1 mol$ of $e^{-}$ deposits $1 mol$ of Ag. $Moles of Ag = Moles of e^{-} = 0.0012437 mol$
6. Calculate mass of silver deposited: $Mass of Ag = Moles of Ag \times M_{A g} = 0.0012437 mol \times 107.868 g/mol \approx 0.134 g$

Worked Examples

Let's use the provided experimental data for the electrolysis of

AgN O_{3}

to calculate the Avogadro Constant.

Given Values:

Current ( $I$ ) = $0.1 A$
Time ( $t$ ) = $30 min$
Mass of Ag deposited = $0.201 g$
Molar Mass of Ag ( $M_{A g}$ ) = $107.868 g/mol$
Charge on one electron ( $e$ ) = $1.602 \times 1 0^{- 19} C$

Convert Time to Seconds: $t = 30 min \times \frac{60 s}{1 min} = 1800 s$

Calculate Total Charge ( $Q$ ) Passed: $Q = I \times t$ $Q = 0.1 A \times 1800 s = 180 C$

This means $0.201 g$ of Ag is produced by $180 C$ of electricity.

Calculate Charge Required for 1 Mole of Ag: If $0.201 g$ of Ag requires $180 C$ , then $1 mol$ (or $107.868 g$ ) of Ag would require: $Charge per mole of Ag = \frac{180 C}{0.201 g Ag} \times 107.868 g Ag/mol$ $Charge per mole of Ag = 96598.209 C/mol$

This charge must be present on the electrons responsible for producing $1 mol$ of Ag.

Relate Charge to Moles of Electrons (Faraday's Constant): From the cathode reaction, $A g_{(aq)}^{+} + 1 e^{-} \to A g_{(s)}$ , we know that $1 mole$ of electrons is required to deposit $1 mole$ of silver.

Therefore, the charge on $1 mole$ of electrons (Faraday's Constant, $F$ ) is approximately $96598.209 C/mol$ .

Calculate Avogadro's Constant ( $N_{A}$ ): Since we know the charge on one electron ( $e$ ) and the charge on one mole of electrons ( $F$ ), we can calculate the number of electrons in one mole: $N_{A} = \frac{Charge on 1 mole of electrons}{Charge on one electron} = \frac{F}{e}$ $N_{A} = \frac{96598.209 C}{1.602 \times 1 0 ^{- 19} C/electron}$ $N_{A} = 6.0298 \times 1 0^{23} electrons/mol$

This calculated value ( $6.0298 \times 1 0^{23}$ ) is very close to the accepted value of $6.02214 \times 1 0^{23}$ .

Possible Questions/Answers

Q1: An electrolytic cell is connected to a power source for two hours. If the current flowing through the cell is $0.5 A$ during this time, find the mass of the substance liberated during this time interval. (The molar mass of the substance is $107.9 g/mol$ .)

A1:

Given values:
- Time ( $t$ ) = $2 hours$
- Current ( $I$ ) = $0.5 A$
- Molar Mass ( $M$ ) = $107.9 g/mol$
- Faraday's constant ( $F$ ) $\approx 96485 C/mol of e^{-}$ (assuming a 1-electron transfer process).
Convert time to seconds: $t = 2 hours \times \frac{3600 s}{1 hour} = 7200 s$
Calculate total charge ( $Q$ ): $Q = I \times t = 0.5 A \times 7200 s = 3600 C$
Calculate moles of electrons transferred: $Moles of e^{-} = \frac{Q}{F} = \frac{3600 C}{96485 C/mol e ^{-}} \approx 0.03731 mol e^{-}$
Calculate moles of substance liberated (assuming 1 mole of e- per mole of substance): $Moles of substance = Moles of e^{-} = 0.03731 mol$
Calculate mass of substance liberated: $Mass = Moles \times M = 0.03731 mol \times 107.9 g/mol \approx 4.026 g$

Q2: Calculate the charge in coulombs when $3 moles$ of electrons flow through a circuit.

A2:

Given values:
- Moles of electrons ( $n_{e}$ ) = $3 mol$
- Faraday's constant ( $F$ ) $\approx 96485 C/mol e^{-}$
Calculate total charge ( $Q$ ): $Q = n_{e} \times F = 3 mol \times 96485 C/mol e^{-} = 289455 C$

Q3: Calculate the mass of silver deposited at the cathode during electrolysis of $AgN O_{3}$ solution, if you use a current of $0.1 A$ for $20 minutes$ .

A3:

Given values:
- Current ( $I$ ) = $0.1 A$
- Time ( $t$ ) = $20 min$
- Molar Mass of Ag ( $M_{A g}$ ) = $107.868 g/mol$
- Faraday's constant ( $F$ ) $\approx 96485 C/mol e^{-}$ (for $A g^{+} + e^{-} \to Ag$ ).
Convert time to seconds: $t = 20 min \times \frac{60 s}{1 min} = 1200 s$
Calculate total charge ( $Q$ ): $Q = I \times t = 0.1 A \times 1200 s = 120 C$
Calculate moles of electrons transferred: $Moles of e^{-} = \frac{Q}{F} = \frac{120 C}{96485 C/mol e ^{-}} \approx 0.0012437 mol e^{-}$
Calculate moles of silver deposited: From the reaction $A g_{(aq)}^{+} + e^{-} \to A g_{(s)}$ , $1 mol$ of $e^{-}$ deposits $1 mol$ of Ag. $Moles of Ag = Moles of e^{-} = 0.0012437 mol$
Calculate mass of silver deposited: $Mass of Ag = Moles of Ag \times M_{A g} = 0.0012437 mol \times 107.868 g/mol \approx 0.134 g$