Given the following standard potentials calculate the overall $E^\circ_\mathrm{cell}$:

$$ \begin{align} \ce{Pb(SO4)(s) + 2 e- &→ Pb(s) +SO4^2-(aq)} &\quad E^\circ &= \pu{-0.356 V} \\ \ce{Pb^2+(aq) + 2 e- &→ Pb(s)} &\quad E^\circ &= \pu{–0.125V} \end{align} $$

I know the answer should be $\pu{0.231 V}$. I was told that $E^\circ$ cell always has to be positive. I tried doing this:

$$E^\circ_\mathrm{cell} = E^\circ_\mathrm{cathode} - E^\circ_\mathrm{anode} = \pu{0.231 V}$$

The only way I get the correct answer is if I do not switch the negative sign of the standard reduction potential of the first equation to a positive sign. Is that correct? I thought you always had to switch signs if you reversed the equation?

  • 3
    $\begingroup$ Since you flipped the equation from the "reduction form" you must also flip the sign of the potential. Thus: $$\ce{Pb(SO4)_{(s)}+ 2e- → Pb_{(s)} +SO4^{2–}_{(aq)}} ~~~~ –0.356~\mathrm{V}$$ becomes: $$\ce{Pb_{(s)} + SO4^{2–}_{(aq)} → Pb(SO4)_{(s)} + 2e-}~~~~ +0.356~\mathrm{V}$$ since the reaction is reversible. $\endgroup$
    – MaxW
    Commented Dec 19, 2015 at 1:39

1 Answer 1


Now, lets first clear this up. $E_\mathrm{cell}^\circ$ can be negative, it just means that the reaction won't spontaneously occur. This means that an external current is required to be applied to push the electrons in that direction for the reaction to occur. When $E_\mathrm{cell}^\circ > 0$, that just means that reaction occurs spontaneously. So in voltaic or galvanic cells, $E_\mathrm{cell}^\circ > 0$.

Now to calculate $E_\mathrm{cell}^\circ$, the formula is: $$E_\mathrm{cell}^\circ = E_\mathrm{red}^\circ \mathrm{(cathode)} + E_\mathrm{ox}^\circ \mathrm{(anode)}$$ As you correctly stated, if you switch the direction of the reaction (so it goes from an oxidising to reducing half equation), the sign of $E^\circ$ changes. Hence the equation for $E_\mathrm{cell}^\circ$ can also be written as: $$E_\mathrm{cell}^\circ = E_\mathrm{red}^\circ \mathrm{(cathode)} - E_\mathrm{red}^\circ \mathrm{(anode)}$$ since standard potentials are given as reduction values. This is the formula that you have used and hence this is why you get the correct answer when you do this.


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