We know that Nernst equation is given by:

$E=E^{\circ}+\frac{RT}{nF} \ln[\ce{M^{n+}}]$.

Now, A question asks me to derive the unit of $\frac{RT}{nF}$.

Which I think can easily be derived this way:


Now, the unit of $E-E^{\circ}$ is volts (since it is emf), and $\ln[\ce{M^{n+}}]$ is just a constant numerical value.

Hence, I think the unit of $\frac{RT}{nF}$ is Volts.

This was my approach. Can anyone tell me if it is a correct approach?


2 Answers 2


M. Farooq have given good description of how to analyse units in a equation. Therefore, your assumption of the unit of $\dfrac{RT}{nF}$ should be Volts is correct. However, your question is not to assume but to derive. What's that means is unit conversion. Let's see the units of all four variables:

  • $R$ is a constant: $\pu{8.314 J mol-1 K-1}$
  • $F$ is also a constant: $\approx \pu{96485 C mol-1}$
  • $T$ is a variable: $\pu{K}$
  • $n$ is number of total electrons in the redox reaction involved, and hence a variable but unit less.

Now lets see the unit conversion:

$$\frac{RT}{nF} = \frac{\pu{J mol-1 K-1}\times \pu{K}}{\pu{C mol-1}} = \frac{\pu{J}}{\pu{C}} $$

By definition (Ohm's law):

$$ V = \frac{\text{potential energy}}{\text{Charge}} = \frac{\pu{J}}{\pu{C}} $$

$$\therefore \:\; \frac{RT}{nF} = \frac{\pu{J}}{\pu{C}}= \pu{V} $$


You should read the article "Can One Take the Logarithm or the Sine of a Dimensioned Quantity or a Unit? Dimensional Analysis Involving Transcendental Functions" Here

The concepts mentioned there will help for life even if you don't pursue chemistry. In short, natural log factor ($\ln$) should be dimensionless. Now your equation is

$E=E^{\circ}+\frac{RT}{nF} \ln[\ce{M^{n+}}]$.

Now do a unit analysis: The left hand side has units of volts $\pu{V}$.

$[\pu{V}] = [\pu{V}] + \frac{RT}{nF}$ times (dimensionless natural log)

So you cannot add apples to oranges. It implies that $\frac{RT}{nF}$ must have units of Volts.

Now do a proper dimensional analysis of the equation in terms of SI base units of [M], [L], [n], [T], [I] etc. and see if it is all consistent.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.