Question

Which of the following relation is correct?
nFE cell
Ink -
RT
FE
cell
Ink -
nRT
nRT
Ink =
FE del
nE
Ink =
cell
RTF​

Answer 1

ELECTROCHEMISTRY:

$\boxed{ \sf{ \red{nFE_{cell} = nFE_{0} - RT ln(Q) }}}$

This is the correct Equation.

As per Nernst Equation:

$\sf{\Delta G =- n \times F \times E}$

Similarly at standard conditions:

$\sf{\Delta G_{0} = -n \times F \times E_{0}}$

According to Thermodynamics:

$\rm{\Delta G = \Delta G_{0} + RT\ln(Q)}$

$\rm{=>-nFE = -nFE_{0} + RT\ln(Q)}$

$\rm{=>-nFE = -nFE_{0} + RT\ln(Q)}$

$\rm{=>E = E_{0} - \dfrac{RT}{nF}\ln(Q)}$

This equation helps us to find the relationship between the actual EMF of a chemical cell and the reaction quotient of that chemical reaction.

So, final answer is :

$\boxed{\red{\bold{E_{cell} = E_{0} - \dfrac{RT}{nF}\ln(Q)}}}$

Answer 2

Explanation:

ELECTROCHEMISTRY:

\boxed{ \sf{ \red{nFE_{cell} = nFE_{0} - RT ln(Q) }}}

nFE

cell

=nFE

0

−RTln(Q)

This is the correct Equation.

As per Nernst Equation:

\sf{\Delta G =- n \times F \times E}ΔG=−n×F×E

Similarly at standard conditions:

\sf{\Delta G_{0} = -n \times F \times E_{0}}ΔG

0

=−n×F×E

0

According to Thermodynamics:

\rm{\Delta G = \Delta G_{0} + RT\ln(Q)}ΔG=ΔG

0

+RTln(Q)

\rm{=>-nFE = -nFE_{0} + RT\ln(Q)}=>−nFE=−nFE

0

+RTln(Q)

\rm{=>-nFE = -nFE_{0} + RT\ln(Q)}=>−nFE=−nFE

0

+RTln(Q)

\rm{=>E = E_{0} - \dfrac{RT}{nF}\ln(Q)}=>E=E

0

−

nF

RT

ln(Q)

This equation helps us to find the relationship between the actual EMF of a chemical cell and the reaction quotient of that chemical reaction.

So, final answer is :

\boxed{\red{\bold{E_{cell} = E_{0} - \dfrac{RT}{nF}\ln(Q)}}}

E

cell

=E

0

−

nF

RT

ln(Q)