The definitions of electromotive force of a cell that i have read include:

1.When no current is drawn from a cell,i.e., when the cell is in open circuit, then potential difference between the terminals of the cell is its electromotive force.

2.The electromotive force of a cell is defined as the energy spent or the work done per unit charge in taking a positive charge around the complete circuit of the cell i.e., in the circuit outside the cell as well as in the electrolyte inside the cell.

I do not understand what definition 1 means. I think it is probably refers to the work done in moving the charge inside the cell since in the definition it is said that the circuit is open but then its meaning does not match with definition 2. My question is, what is the actual meaning of definition 1 and how is it even related with definition 2.

 

It's the amount of Voltage that is available across the positive and negative terminals.

 

Voltage is "potential" energy per unit charge when that charge is moved between the same two points between which the voltage is measured.

The reason that definition 1 specifies open circuit is that, in general, the voltage measured across a cell drops as current is drawn from it due to internal mechanisms in the cell that we usually model as some kind of "internal resistance"; so they just try to make this uncertainty go away by stipulating that the voltage of a cell is the open circuit voltage.

 

@WBahn How does the first definition match with the second definition?

 

@WBahn How does the first definition match with the second definition?

Again, voltage is the "potential" for applying energy to a unit charge. It is similar to a height difference. Imagine a 50 story building that is 10 feet per story. The difference in height between the 20th story and the 30th story is 100 feet. That is a measure of the potential energy (in a given gravitational field) that could be converted to some other form (kinetic, heat, a combination of the two, etc.) per unit mass of a mass that traveled between those two floors. The same could be said of something like the high end a teeter-totter (see-saw) with one end held down to the ground with a spring. But in this case as you actually start to move a mass up to the top it causes the high end to move downward and so, when it is dropped off the end, it is not as high as it was when the height was measured with no mass up there. When no mass was there, the high end still has the larger potential because it is just that -- the "potential" to apply work to a mass. You may not be able to realize that entire potential because, in attempting to do so, some of the energy gets diverted to other uses.

 

@WBahn I still don't get it and by the way can we measure the emf of a cell when it is in an open circuit(1st definition)?

 

@WBahn I still don't get it and by the way can we measure the emf of a cell when it is in an open circuit(1st definition)?

Sure. Put a voltmeter across it. In case you want to point out that a voltmeter has a finite resistance and thus this is not truly open circuit, then use a null-voltage technique using a reference circuit and a galvanometer.

 

So in the 1st definition the circuit is technically not open?

 

So in the 1st definition the circuit is technically not open?

It is if you use a null measurement technique.