Answer:- cell potential = -0.19 volts
Solution:- The equation that shows the connection between [tex]\Delta G[/tex] and cell potential, E is written as:
[tex]\Delta G=-nFE[/tex]
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 [tex]\frac{96485C}{mol}[/tex] .
It asks to calculate the value of E, so let's rearrange the equation:
[tex]E=-\frac{\Delta G}{nF}[/tex]
Let's plug in the values in it:
[tex]E=-\frac{55kJ}{3mol*96485C.mol^-}[/tex]
[tex]E=-\frac{0.00019kJ}{C}[/tex]
since, [tex]\frac{1kJ}{C}=1000V[/tex]
Where C stands for coulombs and V stands for volts.
So, [tex]E=-\frac{0.00019kJ}{C}(\frac{1000V}{\frac{1kJ}{C}})[/tex]
E = -0.19 V
So, the cell potential is -0.19 volts.
