I am having some difficulty with the below.

It says the standard potential for $\ce{2Cu+ (aq) -> Cu (s) + Cu^2+ (aq)}$ is $\pu{0.36 V}$.

The relevant half equations are:

$\ce{Cu+ + e- -> Cu}$ (potential $= \pu{0.52 V}$)


$\ce{Cu^2+ + e- -> Cu+}$ (potential $= \pu{0.16 V}$)

To get the overall equation, I flipped the second half equation and so reversed the sign of the potential also to get $\pu{-0.16 V}$.

The first half equation is therefore reduction and the second is oxidation.

If standard electrode potential is $\mathrm{RHS (reduction)} -\mathrm{LHS (oxidation)}$, would the answer not be $\pu{0.52 V} - (\pu{-0.16 V}) = \pu{0.68 V}$? Why is it $\pu{0.36 V}$?

  • $\begingroup$ It is RHS(reduction)-LHS(reduction) $\endgroup$ Aug 16, 2020 at 12:20
  • $\begingroup$ how do I know which to put on which electrode if both are reduction? as in how do I know that the 0.52V corresponds to RHS? thanks! $\endgroup$
    – Fariha
    Aug 16, 2020 at 12:29
  • 3
    $\begingroup$ Another Issue, emf of reactions are not additive; free energy is... read up on electrochemistry before you proceed.. the part undergoing oxidation is oxidation and part going reduction is reduction (seems redundant doesn't it) ;-) $\endgroup$ Aug 16, 2020 at 12:32
  • $\begingroup$ After Safdar, the standard potential of the reaction is 0.68 V/2 = 0.34 V (and not 0.36 V) $\endgroup$
    – Maurice
    Aug 16, 2020 at 16:29
  • 1
    $\begingroup$ It should be: $\pu{0.52 V} - \pu{0.16 V} = \pu{0.36 V} $. $\endgroup$ Aug 16, 2020 at 18:33

1 Answer 1


Alas, the confusions related to signs in electrochemistry will never vanish. You mention that

$\ce{2Cu+ (aq) -> Cu (s) + Cu^2+ (aq)}$ is $\pu{0.36 V}$.

Let me start with a single equation, x-y = 10; There can be indefinite solutions if you can simultaneously change the value of x and y. However, the moment you fix the value of x, the value of y is fixed.

You stated that the overall cell potential is $\pu{+0.36 V}$. Electrochemically, this means that this reaction is spontaneous.

Now you also know that,

$$E_\text{cell} = E_\text{reduction} - E_\text{anode} \tag{1}$$

You are not supposed to change any sign of the half-cell from the electrode potential tables. People should stop teaching this nonsense to relatively innocent students. Suppose, I write

\begin{align} &\ce{H2O (liquid) -> H2O (gas)} &T &= \pu{100 ^\circ C} \\ &\ce{H2O (gas) -> H2O (liquid)}, &T &=\pu{ -100 ^\circ C} ?? \end{align}

The equation (1), itself takes care of all sign flipping and all.

Your half-cell corresponding to the reduction is

$\ce{Cu+ + e- -> Cu}$ (potential $= \pu{0.52 V}$)

And your half-cell potential for the oxidation is

$\ce{Cu^2+ + e- -> Cu+}$ (potential $= \pu{0.16 V}$)

Using equation (1), what you do get (remember no sign flipping) = $\pu{+0.36 V}$


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