$$
2 H _{2} O \rightarrow O _{2}+4 H +4 e - E ^{\circ}=-1.23 V \ldots \text { (i) }
$$
\(2 H _{2} O _{2} \rightarrow 2 O _{2}+4 H ++4 e ^{-}, E ^{0}=-0.68 V \ldots\) (ii)
We are to get: \(2 H _{2} O _{2} \rightarrow 2 H _{2} O + O _{2}\)
To have this, equation (i) is reversed and then added to equation (ii)
Reversing equation (i), we have:
\(O _{2}+4 H ^{+}+4 e ^{-}=2 H _{2} O , E ^{\circ}=+1.23\)
Note that the sign is reversed as well.dding the equation so obtained to cquation (ii), we have:
\(2 H _{2} O + O _{2}+4 H ^{+}+4 e ^{-}=2 O _{2}+4 H ^{+}+4 e ^{-}+2 H _{2} O\)
\begin{aligned} ^{\circ} &=(-0.68+1.23) V \\
&=+0.55 V
\end{aligned}verything cancels out as already shown. Also, \(1 mol\) of \(O _{2}\) cancels \(1 mol\) of \(O _{2}\) on both sides.
Then, we have:
\(H _{2} O _{2}= O _{2}+2 H _{2} O , E ^{\circ}=+0.55 V\)