Electrolysis is the process of using an electric current to drive a non-spontaneous chemical reaction. A common application is the deposition of a metal onto an electrode. To determine the amount of substance deposited, we use Faraday’s laws of electrolysis. Here’s a breakdown:

Key Concepts and Formulas

  1. Charge (Q): The total amount of electric charge passed through the electrolytic cell.
    • Formula: Q = I * t
      • Where:
        • Q = Charge (Coulombs, C)
        • I = Current (Amperes, A)
        • t = Time (seconds, s)
  2. Electrochemical Equivalent (Z): The mass of a substance deposited per unit of charge.
    • Formula: Z = M / (n * F)
      • Where:
        • Z = Electrochemical equivalent (grams per Coulomb, g/C)
        • M = Molar mass of the substance (grams per mole, g/mol)
        • n = Number of electrons transferred per ion (dimensionless)
        • F = Faraday’s constant (approximately 96,485 Coulombs per mole, C/mol)
  3. Mass Deposited (m): The actual mass of the substance deposited at the electrode.
    • Formula: m = Q * Z
      • Where:
        • m = Mass deposited (grams, g)
  4. Combined Formula: By substituting the formulas for Q and Z into the equation for m, we get a combined formula that directly relates the mass deposited to the current, time, molar mass, and number of electrons transferred:
    • m = (M * I * t) / (n * F)

Steps for Calculation

There are two ways to approach the calculation:

Method 1: Step-by-Step (using Q and Z)

  1. Calculate the Charge (Q): Use Q = I * t.
  2. Calculate the Electrochemical Equivalent (Z): Use Z = M / (n * F).
  3. Calculate the Mass Deposited (m): Use m = Q * Z.

Example Problem

Let’s say we want to calculate the mass of copper deposited during the electrolysis of copper sulfate (CuSO4) solution. A current of 2 amperes is passed through the solution for 1 hour.

  1. Calculate Charge (Q):
    • I = 2 A
    • t = 1 hour = 3600 seconds
    • Q = 2 A * 3600 s = 7200 C
  2. Calculate Electrochemical Equivalent (Z):
    • M (copper) = 63.55 g/mol
    • n (for Cu2+ + 2e- → Cu) = 2
    • F = 96485 C/mol
    • Z = 63.55 g/mol / (2 * 96485 C/mol) ≈ 0.000329 g/C
  3. Calculate Mass Deposited (m):
    • Q = 7200 C
    • Z ≈ 0.000329 g/C
    • m = 7200 C * 0.000329 g/C ≈ 2.37 g

Therefore, approximately 2.37 grams of copper would be deposited during the electrolysis.

Method 2: Direct Calculation (using the combined formula)

  1. Identify all the variables: Determine the current (I), time (t), molar mass (M), and the number of electrons transferred (n).
  2. Calculate the Mass Deposited (m): Use the combined formula: m = (M * I * t) / (n * F).

Example Problem (using the combined formula)

Let’s calculate the mass of copper deposited during the electrolysis of copper sulfate (CuSO4) solution. A current of 2 amperes is passed through the solution for 1 hour.

  1. Identify Variables:
    • I = 2 A
    • t = 1 hour = 3600 s
    • M (copper) = 63.55 g/mol
    • n (for Cu2+ + 2e- → Cu) = 2
    • F = 96485 C/mol
  2. Calculate Mass (m):
    • m = (63.55 g/mol * 2 A * 3600 s) / (2 * 96485 C/mol)
    • m ≈ 2.37 g

Therefore, approximately 2.37 grams of copper would be deposited during the electrolysis.

Advantages of the Combined Formula

  • Efficiency: It allows for directly calculating the mass deposited without the intermediate steps of calculating Q and Z.
  • Simplicity: It reduces the number of calculations required, minimizing potential rounding errors.

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