# Why oxidation potential of an electrode equals negative reduction potential of the same electrode?

My textbook states that:

$$E^\circ_\mathrm{ox} = -E^\circ_\mathrm{red}$$ (of same electrode)

I know that electrons flow from anode half cell (oxidation) to cathode half cell (reduction).

But if $$E^\circ_\mathrm{ox} = -E^\circ_\mathrm{red}$$ in each electrode, then potential in each electrode of the galvanic cell should be 0 volts (oxidation cancels reduction).

Hence potential difference in the whole circuit is 0 volts (no electricity is generated).

Do not get me wrong, I am just a high school student and I know my question might sound trivial, I am new to electrochemistry.

• Anode and cathode are not the same electrode (or I don't understand the question). I took a liberty to correct slang and notations to more standardized forms. Also, please note that citing an actual textbook instead of saying "My textbook" is a good practice. – andselisk Dec 28 '19 at 18:55
• @andselisk I am not saying they are the same electrode, you do not understand my question, let me clarify with an example if oxidation potential of zinc electrode = -0.76V then it's reduction potential = 0.76V – AmirWG Dec 28 '19 at 18:59
• Exactly, that's why only standard reduction potentials are tabulated. There must be another electrode to complete the circuit. – andselisk Dec 28 '19 at 19:02

There exist two methods for teaching electrochemistry, and of course two schools of teachers. The two methods are equivalent, but they are exactly the opposite of one another. I will take an example to explain it correctly : Let us speak of the Daniell cell Zn/Cu, which uses reduction potentials equal to -$$0.76$$ V for Zn and +$$0.34$$ V for Cu, if the ions concentrations are 1 molar. In this case, the overall voltage is $$1.10$$ V. The Daniell cell is based on the two half-reactions : $$\ce{Zn^{2+} + 2e^- -> Zn}$$ $$\ce{Cu^{2+} +2e^--> Cu}$$In the Zn/Cu cell, Zn is oxidized and acts as emitter of electron, which is in the opposite direction of the above equation, so the equation explaining what is going on in a Daniell cell contains the equation of Zn reversed. This becomes :$$\ce{Zn-> Zn^{2+} + 2e^-}$$ $$\ce{Cu^{2+} + 2e- ->Cu}$$When you want to describe the overall equation that goes in this cell, you add these two half-equations, remove the 2$$\ e^-$$ and this gives : $$\ce{Zn + Cu^{2+}-> Zn^{2+} + Cu}$$ We are here at the point where the two approaches differ. Strangely enough, one uses an addition, and the second a subtraction. Let me speak of the first method, which is used in your textbook.
First method. To obtain this overall equation, you had to add the two half-equations. To follow the same logic, you may simultaneously add the corresponding reduction potentials. But the reduction potentials of Zn and Cu are respectively $$-0.76$$V, and + $$0.34$$ V. And if you add them, you do not get $$1.10$$ V. So the chemists of this school have decided that when a half- equation is used in the opposite sense, with the electrons at the right-hand side, in an oxidation sense, like Zn here, the sign of the potential of this cell should also be opposite. $$\ E_{anode} = - E_{cathode}$$. Here : $$E(Zn)_{anode} = - E(Zn)_{cathode)}$$ = +$$0.76$$ V. With this assumption, you add $$+ 0.76$$ V plus $$+0.34$$ V and you obtain $$1.10$$ V, which is perfect.
Second method. Here you use the same half-reactions as described previously. But the values of the potential remains the same if the electrode works as a reducing agent or as an oxidant. It is always - $$0.76$$ V pour Zn and + $$0.34$$ V for Cu. You never change its sign. But remembering from the electricity course, that a measured voltage is a difference of potential, the voltage of the cell may be obtained by subtracting the redox potential of the cathode minus the potential of the anode, that is $$E_{Cu} - E_{Zn}$$. The result of this subtraction is: $$+0.34$$V -($$0.76$$ V) = $$1.10$$ V, which is perfect. This remains true even though you have obtained the overall equation with an addition of two half-reactions.