If Standard Reduction Potential and Standard Oxidation potential are equal in magnitude and opposite in sign?

If Standard Reduction Potential(SRP) and Standard Oxidation potential(SOP) are equal in magnitude and opposite in sign ,


Then as per this
As we know SRP of H2O is -0.83 V as per this SOP of H2O should be + 0.83 V but this is not the case the value is SOP of H2O is 1.23 V.


Why ? And how ?




As $$\Delta\_\text{r} G = -nF E^{\circ}$$


and as


$$\Delta\_\mathrm{r,forward} G = - \Delta\_\mathrm{r,backward} G,$$


it should not be surprising that


$$E^{\circ}\_\text{ox} = - E^{\circ}\_\text{red}$$


Be aware that


* The formal redox potential of a half reaction flips the sign with the reaction direction.
* The above potentials are conventionally tabelizec for oxidation direction.
* The physical value of measured or predicted electrode potential has just one sign, measured wrt SHE.


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There is another way of taking care of the potentials in a galvanic cell, without changing the sign of the anodic half-cell. It is coming from the fact that in physics, all tensions are measured in volts. And they all can be considered as differences of potentials. I repeat : a DIFFERENCE of potential.


So when coupling two half-equations like in the Daniel cell, the two half-equations are : $$\ce{Zn -> Zn^{2+} + 2 e^-}$$ $$\ce{Cu -> Cu^{2+} + 2 e^-}$$The redox potentials of these half-equations are $\pu{E°\_{Zn} = - 0.76}$ V for the $\ce{Zn}$ electrode, and $\pu{E°\_{Cu} = +0.34}$ V for the $\ce{Cu}$ electrode. It is not necessary to change the sign of the $\ce{Zn}$ electrode (which gets oxidized) when considering the Daniel cell produced by the two half-equations.


The tension $\epsilon\_{cell} $ of the cell formed by the two half-equations can be obtained by the DIFFERENCE of the two redox potentials of $\ce{Zn}$ and $\ce{Cu}$ : $$\epsilon\_{cell} = \pu{E\_o(red)- E\_o(ox) = E°\_{Cu} - E°\_{Zn} = 0.34 \mathrm{V} - ( -0.76 \mathrm{V}) = 1.10 \mathrm{V}} $$ As the result must be positive, it is necessary to subtract the smallest potential (or the most negative - here $\pu{E°\_{Zn}}$ ) from the most positive value ($\pu{E°\_{Cu}}$), and not the contrary.


With this approach, it is not necessary to change the sign of redox potentials when a half-equation is reversed. Redox potentials are maintained always at the same numerical values, whatever their behavior (oxidation or reduction).
$$\pu{\epsilon\_{cell} = E\_o(red)- E\_o(ox)}$$


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