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 $\frac{96485C}{mol}$ .

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

$E=-\frac{\Delta G}{nF}$

Let's plug in the values in it:

$E=-\frac{55kJ}{3mol*96485C.mol^-}$

$E=-\frac{0.00019kJ}{C}$

since, $\frac{1kJ}{C}=1000V$

Where C stands for coulombs and V stands for volts.

So, $E=-\frac{0.00019kJ}{C}(\frac{1000V}{\frac{1kJ}{C}})$

E = -0.19 V

So, the cell potential is -0.19 volts.