If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Calculating the equilibrium constant from the standard cell potential edited

earn how to calculate the equilibrium constant, k, for a redox reaction using standard reduction potentials. Discover the relationship between the equilibrium constant, k, and the standard cell potentials, e zero. Understand how to balance equations and calculate the standard cell potential to find the equilibrium constant.
Visit us (http://www.khanacademy.org/science/healthcare-and-medicine) for health and medicine content or (http://www.khanacademy.org/test-prep/mcat) for MCAT related content. These videos do not provide medical advice and are for informational purposes only. The videos are not intended to be a substitute for professional medical advice, diagnosis or treatment. Always seek the advice of a qualified health provider with any questions you may have regarding a medical condition. Never disregard professional medical advice or delay in seeking it because of something you have read or seen in any Khan Academy video.
Created by Jay.

## Want to join the conversation?

• Is it true for most galvanic cells that k is very large?
• It would make sense...galvanic cells are spontaneous by definition. This means deltaG is negative and that would make K positive according to delta G= -RTlnK
• I used the equation with E= RT/n.F x lnK and I get a math error on my calculator, but I compute the numbers the exact same way as in the video. What am I doing wrong?
• This is how I solved using the natural log version:
E = 0.0257/n *lnK is the simplified equation after you plug in your constants (see the Standard cell potential and equilibrium constant video for more info on this)

We know that E is equal to 2.20V as solved in this video. I think it is easier to first simplify the constants before dividing them over. So, 0.0257/6 = 0.004283 *lnK. Divide by 0.004283 to isolate lnK. So, 2.20V / 0.004283 = 513.619. Careful with the next part! Remember that log has a base of 10 but ln does not! Natural log has a base of e. So rather than raising both sides by 10 like in the video, you should raise both sides by e. When you put e^513.619 in your calculator, you should get 1.1... * 10^223. This is equivalent to k = 10^223 since the 1.1.... number is negligible.

Hope that helped! :)