Electrical Energy Calculations Chemistry Tutorial

Key Concepts

• The electrode potential of a galvanic (voltaic) cell, or a battery while it is discharging, is an EMF or voltage.
• This voltage (EMF) produced by the cell or battery can be used to calculate the electrical energy delivered by that cell or battery.
• The electrolysis of a substance requires an input of electrical energy.

  This electrical energy can be used to :

  (i) recharge a battery (eg, recharge a lead-acid battery)

  (ii) electroplate an object

  (iii) extract a metal from its ore

• Electrical energy, E, measured in joules (J) produced by a battery, or used during electrolysis, can be calculated:

  E = Q × V

  E = energy in joules (J)

  Q = quantity of electricity or charge in coulombs (C)

  V = voltage (or EMF) in volts (V)

• Rearranging this equation allows us to calculate:

  (i) quantity of electricity (charge):
  Q = E ÷ V

  (ii) voltage (emf):
  V = E ÷ Q

• The kilowatt-hour, kWH, is a unit of electrical energy.

  1 kWH = 3.6 x 10⁶ J

  To convert energy in joules to kWh, divide energy in joules by 3.6 x 10⁶.

  To convert energy in kilo-watt hours to joules, multiply kWh by 3.6 x 10⁶.

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Worked Examples: Energy of Batteries

Question 1: 96,500 C of electricity (charge) was delivered by a 1.3 volt mercury button cell.
How much energy, in joules, was produced by this cell?

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

   Calculate energy in joules
   E = ? J

2. What data (information) have you been given in the question?

   Extract the data from the question:

   Q = quantity of charge delivered = 96,500 C

   V = voltage produced (EMF) = 1.3 V

3. What is the relationship between what you know and what you need to find out?

   Write the equation for calculating electrical energy:
   E = Q × V

4. Substitute the values into the equation and solve for E:
   E = 96 500 × 1.3 = 125,450 J
5. Is your answer plausible?

   Perform a "rough" calculation by "rounding off" data to see if your answer is in the same "ball park".
   Let Q = 100,000 C
   Let V = 1 V
   Then, E = Q × V = 100,000 × 1 = 100,000 J

   Our calculated value of E = 125,450 J is in the same "ball park" as the rough calculation so we are reasonably confident our answer is plausible.

6. State your solution to the problem:

   E = 125,450 J

Question 2: A battery is made up of a series of 2.0 volt galvanic (voltaic) cells.
The battery delivers 48,250 C of electricity (charge) which provides 579 kJ of energy.
How many galvanic cells made up the battery?

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

   Calculate the number of galvanic cells in the battery.

2. What data (information) have you been given in the question?

   Extract the data from the question:

   E = energy = 579 kJ
   Convert kJ to J by multiplying by 1000 J/kJ
   E = 579 × 10³ J

   Q = quantity of charge = 48,250 C

   V_(cell) = 2 V

3. What is the relationship between what you know and what you need to find out?

   (i) Write the equation to calculate the voltage produced by the battery:
   E = Q × V_(battery)

   (ii) Rearrange this equation to calculate V:
   V_(battery) = E ÷ Q

   (iii) Substitute the values for E and Q into this equation to find the voltage for the battery, E_(battery):
   V_(battery) = E ÷ Q
   V_(battery) = 579 × 10³ ÷ 48,250 = 12 V

4. Calculate how many cells make up the battery:

   1 galvanic cell delivers 2 V
   Let x = number of cells required to produce a 12 volt battery
   x × 2 V = 12 V
   x = 12 ÷ 2 = 6 cells

5. Is your answer plausible?

   Work backwards: If you place 6 lots of 2 V galvanic cells in series to produce 48,250 C of charge, how much energy is provided?
   Voltage of battery = 2 × 6 V = 12 V
   E = Q × V = 48,250 × 12 = 579,000 J = 579 kJ

   Since this amount of energy agrees with that given in the question, we are reasonably confident that our calculated value for the number of 2 V cells in series is correct.

6. State your solution to the problem:

   There are 6 galvanic cells connected in series to make up the battery.

Question 3. A 2.0 volt cell delivers a current of 300 mA, generating 2100 J of energy in the process.
For how long did the cell operate?

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

   Calculate time
   t = ? seconds

2. What data (information) have you been given in the question?

   Extract the data from the question:

   V = voltage = 2.0 V

   I = current = 300 mA
   Convert milliamps to amps by dividing by 1000 mA/A
   I = 300 mA ÷ 1000 mA/A = 0.300 A

   E = energy = 2100 J

3. What is the relationship between what you know and what you need to find out?

   (i) Calculate quantity of charge, Q
   E = Q × V
   Rearrange this equation to find Q:
   Q = E ÷ V
   Substitute in the values and solve for Q:
   Q = 2100 ÷ 2.0 = 1050 C

   (ii) Calculate time taken, t
   Q = I × t
   rearrange this to find t
   t = Q ÷ I

4. Substitute the values into the equation and solve for t:

   t = Q ÷ I
   t = 1050 ÷ 0.300
       = 3500 seconds

5. Is your answer plausible?

   Work backwards: if you run a 2 V cell with a current of 300 mA for 3500 seconds, how much energy is delivered?
   (i) E = Q × V,
   (ii) Q = I × t,
   so E = (I × t) × V = (300/100 × 3500) × 2 = 2100 J

   
   Since this is the same amount of energy given in the question, we are reasonably confident that our calculated value for time is correct.
6. State your solution to the problem:

   time = 3500 seconds

Worked Example: Electrolysis and Energy Calculations

Question: An EMF of 4.5 V produces 1 kg of sodium metal by the electrolysis of Na⁺.
Calculate the minimum number of kilowatt-hours of electricity needed to produce the sodium metal.

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

   Calculate energy in kilowatt-hours
   E = ? kWh

2. What data (information) have you been given in the question?

   Extract the data from the question:

   V = EMF (voltage) = 4.5 V

   m(Na) = mass of sodium metal produced = 1 kg
   Convert mass in kg to g by multiplying by 1000
   m(Na) = 1 kg × 1000 g/kg = 1000 g

3. What is the relationship between what you know and what you need to find out?

   (i) Calculate moles of sodium metal produced:
   moles(Na) = mass(Na) ÷ molar mass(Na)
   n(Na) = 1000 g ÷ 22.99 g mol^-1 = 43.5 mol

   (ii) Calculate moles of electrons required to produce 43.5 moles of sodium metal from sodium ions:
   Na⁺ + e^- → Na_(s)
   mole (stoichiometric) ratio e^- : Na_(s) is 1:1
   Therefore 43.5 moles electrons needed to produce 43.5 moles of sodium metal.

   (iii) Calculate the quantity of charge, Q, required:
   Q = n(e^-)F
       n(e^-) = 43.5 mol
       F = 96,500 C mol^-1 (from data sheet)
   Q = 43.5 × 96,500 = 4,197,750 C

   (iv) Calculate the required energy, E, in joules
   E = Q × V
       Q = 4,197,750 C
       V = 4.5 V
   E = 4,197,750 × 4.5 = 18,889,875 J

4. Convert energy in joules to energy in kilowatt-hours

   1 kilowatt-hour = 3.6 × 10⁶ J
   1 J = 1 ÷ (3.6 × 10⁶) = 2.78 × 10^-7 kWh
   18,889,875 J = 18,889,875 × 2.78 × 10^-7 = 5.25 kWh

5. Is your answer plausible?

   Work backwards: Calculate the EMF needed to produce 1 kg of Na_(s) using 5.25 kWh of energy.
   E = Q × V
   So, V = E ÷ Q

   E = 5.25 kWk = 5.25 kWh × 3.6 × 10⁶ J kWh^-1 = 1.89 × 10⁷ J
   so, V = 1.89 × 10⁷ ÷ Q

   Since Q = n(e)F and F = 96,500, Q = 96,500n(e)
   so, V = 1.89 × 10⁷ ÷ (96,500n(e))

   1 mole e produced 1 mole Na⁺,
   n(e) = n(Na⁺) = mass/molar mass = 1000/22.99 = 43.5
   so, V = 1.89 × 10⁷ ÷ (96,500 × 43.5) = 4.5 V

   
   Since this voltage agrees with that given in the question, we are reasonably confident that our calculated value for energy is correct.
6. State your solution to the problem:

   E = 5.25 kWh