Recording-2025-03-11T03:39:05

Overview of Electrochemistry

  • Focus on stoichiometry related to electrochemistry for exam preparation.

  • Emphasize on the importance of handling multiple-choice questions, especially those involving calculations.

Ampere (Amps) Calculation

  • Definition: Amps are defined as coulombs per second (C/s).

    • Example: 50 Amps = 50 coulombs per second.

    • Example: 100 Amps = 100 coulombs per second.

  • Starting Point: Never begin calculations with amps; treat it as a conversion factor.

Faraday's Constant

  • Represents how much energy is in a mole of electrons.

  • Value: 96485 coulombs (not necessary to memorize; provided in exams as 96,000 or 96,500).

  • Usage: Conversion factor in stoichiometry calculations for electrochemical reactions.

Stoichiometry Example: Copper Production

  1. Question: Calculate mass of copper metal produced when 2.5 Amps pass through a copper(II) sulfate solution for 50 minutes.

  2. Steps:

    • Ignore sulfate; focus on copper(II) reduction to copper metal.

    • Balancing reaction indicates 2 moles of electrons needed for 1 mole of copper.

    • Convert time into seconds:

      • 50 minutes x 60 seconds/minute = 3000 seconds.

    • Use amps to convert to coulombs:

      • 2.5 coulombs/second x 3000 seconds = 7500 coulombs.

    • Calculate moles of electrons:

      • Moles of electrons = 7500 C / 96485 C/mole = 0.0778 moles.

    • Find moles of copper produced:

      • 0.0778 moles electrons x (1 mole Cu / 2 moles e-) = 0.0389 moles Cu.

    • Calculate mass of copper:

      • Mass = 0.0389 moles Cu x 63.55 grams/mole = 2.47 grams Cu.

Oxygen Production Calculation

  1. Question: Determine the volume of oxygen produced at STP with similar conditions.

  2. Process:

    • Start with the same time: 50 minutes = 3000 seconds.

    • Current (amps) = 2.5 Amps implies:

      • 2.5 coulombs/second x 3000 seconds = 7500 coulombs.

    • Based on the reaction, 4 moles of electrons produce 1 mole of oxygen.

    • Moles of electrons = 7500 C / 96485 C/mole = 0.0778 moles.

    • Moles of O2 produced = 0.0778 moles x (1 mole O2 / 4 moles e-) = 0.01945 moles O2.

    • Volume of oxygen at STP:

      • Volume = 0.01945 moles O2 x 22.4 L/mole = 0.436 liters.

Amperage Calculation Example

  1. Question: Find out how many amps are needed to transform 250 grams of chromium(III) to chromium metal.

  2. Steps:

    • Chromium(III) to chromium requires 3 electrons.

    • 250 grams of chromium = 250 g / 52.00 g/mole = 4.81 moles of chromium.

    • Thus, we need:

      • Electrons needed = 4.81 moles x 3 moles e-/1 mole Cr = 14.43 moles e-.

    • Calculate coulombs required:

      • Coulombs = 14.43 moles e- x 96485 C/mole = 1393710.55 C.

    • Calculate seconds using 8 hours:

      • 8 hours = 8 x 3600 = 28800 seconds.

    • Amps calculation:

      • Amps = Total coulombs / seconds = 1393710.55 C / 28800 s = 48.32 Amps.

Time Calculation Example for Zinc

  1. Question: How much time does it take for 40 grams of zinc from Zn2+ with a current of 12 Amps?

  2. Process:

    • Reaction: Zn2+ to Zn requires 2 electrons.

    • 40 grams of zinc converts to moles:

      • 40 g Zn / 65.38 g/mole Zn = 0.611 moles Zn.

    • Moles of electrons needed = 0.611 moles Zn x (2 moles e-/1 mole Zn) = 1.222 moles e-.

    • Calculate coulombs:

      • Coulombs = 1.222 moles e- x 96485 C/mole = 117671.57 C.

    • Since current is 12 Amps:

      • Time (seconds) = Total coulombs / current = 117671.57 C / 12 A = 9805.96 seconds.

    • Convert to hours:

      • Time in hours = 9805.96 seconds / 3600 seconds/hour = 2.73 hours.

Key Tips

  • Always write balanced reactions to determine electron transfer required.

  • Start calculations with mass or moles, never with amps.

  • Use conversion factors to transition between units efficiently, especially time and charge.