Potential energy given to each coulomb of charge

In summary, the potential energy given to each coulomb of charge in a chemical cell is about 300 kJ per mole of metal attacked, with each zinc atom losing 2 electrons to become a Zn2+ ion. This means that a mole of zinc would lose 2 x (6.0 ´ 10^23) = 1.2 ´ 10^24 electrons. Additionally, 1 coulomb of charge is equivalent to the inverse of the charge of an electron, and a volt is defined as a joule per coulomb. The molar mass of zinc is not relevant to this calculation.
  • #1
benhorris
21
0
This is the info I am given...

e = 1.6 ´ 10–19 C
NA = 6.0 ´ 1023 particles / mole
Molar mass of zinc = 65.4 grams
In a chemical cell, the energy released by chemical attack on a metal, say zinc, as in many chemical reactions, is about 300 kJ per mole of metal attacked.

What is the potential energy given to each coulomb of charge (i.e. the potential difference)?
Remember that each zinc atom loses 2 electrons to become a Zn2+ ion. :confused:
 
Physics news on Phys.org
  • #2
how many elecrons are lost in a mole of zinc if each zinc atom loses two?

How many electrons do you need to have a coulomb? (this number is the inverse of the charge of the electron).

A volt is a joule per coulomb.

molar mass is a red herring.
 
  • #3


To calculate the potential energy given to each coulomb of charge, we can use the following formula:

Potential energy per coulomb = (Energy released per mole of metal / Avogadro's number) * (Molar mass of metal / Number of electrons transferred)

In this case, the energy released per mole of zinc is 300 kJ, the Avogadro's number is 6.0 x 10^23 particles/mole, the molar mass of zinc is 65.4 grams, and each zinc atom loses 2 electrons to become a Zn2+ ion.

Substituting these values into the formula, we get:

Potential energy per coulomb = (300 kJ/mol / 6.0 x 10^23 particles/mol) * (65.4 g/mol / 2 electrons)

= (5 x 10^-22 kJ/particle) * (32.7 g/electron)

= 1.64 x 10^-20 kJ/electron

Therefore, the potential energy given to each coulomb of charge, or the potential difference, is 1.64 x 10^-20 kJ/electron. This means that for every coulomb of charge transferred in the chemical reaction, 1.64 x 10^-20 kJ of energy is released.
 

1. What is potential energy given to each coulomb of charge?

Potential energy given to each coulomb of charge refers to the amount of energy a single unit of charge (one coulomb) has in a particular electric field.

2. How is potential energy given to each coulomb of charge calculated?

The potential energy given to each coulomb of charge is calculated by multiplying the electric potential (measured in volts) by the amount of charge (measured in coulombs).

3. What is the unit of measurement for potential energy given to each coulomb of charge?

The unit of measurement for potential energy given to each coulomb of charge is joules per coulomb (J/C).

4. How does the distance between charges affect potential energy given to each coulomb of charge?

The potential energy given to each coulomb of charge is directly proportional to the distance between the charges. As the distance increases, the potential energy decreases and vice versa.

5. What is the significance of potential energy given to each coulomb of charge in the study of electricity?

Potential energy given to each coulomb of charge is an important concept in the study of electricity as it helps us understand the behavior of electric charges and their interactions in electric fields. It also plays a crucial role in determining the amount of work needed to move a charge in an electric field.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
658
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Biology and Chemistry Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
7
Views
4K
  • Electromagnetism
Replies
18
Views
2K
  • Materials and Chemical Engineering
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
5K
  • Introductory Physics Homework Help
Replies
7
Views
6K
  • Introductory Physics Homework Help
Replies
1
Views
3K
Back
Top