0
$\begingroup$

I have a cell equation as:

$$\ce{2Cr^0 + 3Cd^2+ -> 2Cr^3+ + 3Cd^0}$$

I am provided the standard reduction potentials as:

$$\begin{align}\ce{Cd^2+ + 2e- &-> Cd} && E= \pu{-0.4V}\\ \ce{Cr^3+ + 3e- &-> Cr} && E= \pu{-0.74V}\end{align}$$

Now do I use chromium’s reduction potential as $\pu{-0.74V}$ or do I flip the equation and use $\pu{+0.74V}$ (as it is getting oxidised and I have the equation for reduction)?

This is using the formula $E(\text{cell}) = E(\text{cathode}) -E(\text{anode})$, all reduction potentials.

$\endgroup$
  • 1
    $\begingroup$ $$\ce{Cr^{3+} + 3e^- -> Cr}\quad\quad\quad\text{E = -0.74}$$ $$\ce{Cr -> Cr^{3+} + 3e^- }\quad\quad\quad\text{E = +0.74}$$ $\endgroup$ – MaxW Dec 16 '18 at 8:59
  • 1
    $\begingroup$ A few notes: I (the pronoun) is capitalised in English, colons are typeset directly attached to the preceeding words (no space), dashes after colons are discouraged, physical values need a unit and there is a space before the opening bracket. $\endgroup$ – Jan Dec 16 '18 at 16:03
0
$\begingroup$

The reduction potentials are a special case of standard reaction enthalpy, because you can transform $\Delta E$ into $\Delta H$ just by calculating with a few constants.

Standard reaction enthalpy and all other thermodynamic values related to reactions change sign when the direction of reaction is reversed.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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