Standard electrode potential of copper disproportionation

Question

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 $= \pu{0.52 V}$)

and

$\ce{Cu^2+ + e- -> Cu+}$ (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 $\pu{0.36 V}$?

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) = $\pu{+0.36 V}$