I'm currently working on a problem for a general chemistry class.
This is the voltaic cell given in the problem: $$\ce{Sn^4+(aq) + Cu(s) -> Sn^2+(aq) + Cu^2+(aq)}$$
$$E^\circ_{cell} = \pu{-0.21 V}$$ What is the value of the equilibrium constant at $\pu{115 ^\circ C}$?

For this I'm using this equation $$-nFE^\circ_{cell} = RT\log(K)$$ Plugging this into the equation:
$$ (\pu{2 mol e^-})(\pu{96500 C}/\pu{mol e^-})(\pu{-0.21 V}) = (\pu{8.314 J/molK})(\pu{388 K})\log(K)$$ I get a value of $$K = 3.49 \times 10^{-6}$$ However, my instructor's solution is using this equation $$- E^\circ_{cell} = (0.0592/n)\log(K)$$ and gets a value of $8.04276 \times 10^{-8}$.

I thought you could only use that equation if the temperature is $T = \pu{298 K}$ (standard state conditions). Not sure whether I'm right or wrong about this, any insight is appreciated.

  • 1
    $\begingroup$ You'd want to use the van't Hoff equation. $\endgroup$ – Buck Thorn Jul 24 '19 at 8:49
  • $\begingroup$ What @Buck Thorn said is correct and works fine: I just did the calculation quick and dirty (neglecting entropy, so enthalpy is approximated as the free energy at 298.15K) and got about $1.6 x 10^{-5}$ for the equilibrium constant at 115°C. What your instructor got for the equilibrium constant is fine at 298.15K. $\endgroup$ – Ed V Jul 24 '19 at 23:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.