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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.

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    $\begingroup$ $$\ce{Cr^{3+} + 3e^- -> Cr}\quad\quad\quad\text{E = -0.74}$$ $$\ce{Cr -> Cr^{3+} + 3e^- }\quad\quad\quad\text{E = +0.74}$$ $\endgroup$ – MaxW Dec 16 '18 at 8:59
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    $\begingroup$ 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
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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.

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Good question.

The short answer is just throw the numbers into the equation.

In all other thermodynamic equations, everything is "added," and if there are subtractions, it comes from the fact that values themselves are negative. In this case, technically, the potential for the reaction of chromium is +0.74 V (because it is the reverse reaction of the tabulated value). However, there exists a convention to keep the value and thus the sign of the standard reduction potential. The sign change is already accounted for by the minus sign in the equation: \begin{equation} E_{cell}=E_{cathode}-E_{anode}. \end{equation} So the $E_{cell}$ from your problem is thus \begin{equation} E_{cell}=-0.4 V -(-0.74 V)=0.34 V. \end{equation}

In the end, whether or not you change the sign partly depends on who looks at your work. As long as you get the right answer, some people will not care if you stuck to convention or not, but others might. Some do not even know the convention.

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Note that the formal potential of a redox reaction $E=- nF \Delta G$ and the measurable electrode potential are 2 different things.

You change the sign of the former when reverting the equation, but you cannot change the sign of the latter. The electrode keeps just one sign and the reaction occurs there in both directions.

E.g. the potential of $\ce{Zn/Zn^2+}$ electrode ( not reaction! ) is always negative, with ongoing bidirectional $\ce{Zn <-> Zn^2+ + 2e-}$

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