Question

Find the cell potential for a system whose ∆G = +55 kJ and 3 moles of electrons are exchanged.

Answer 1

Answer:- cell potential = -0.19 volts

Solution:- The equation that shows the connection between
`\Delta G` and cell potential, E is written as:

`\Delta G=-nFE`

in this equation, n stands for moles of electrons, E stands for cell potential and F stands for faraday constant and it's value is
`(96485C)/(mol)` .

It asks to calculate the value of E, so let's rearrange the equation:

`E=-(\Delta G)/(nF)`

Let's plug in the values in it:

`E=-(55kJ)/(3mol*96485C.mol^-)`

`E=-(0.00019kJ)/(C)`

since,
`(1kJ)/(C)=1000V`

Where C stands for coulombs and V stands for volts.

So,
`E=-(0.00019kJ)/(C)((1000V)/((1kJ)/(C)))`

E = -0.19 V

So, the cell potential is -0.19 volts.