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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.

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    $\begingroup$ 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. $\endgroup$ – andselisk Dec 28 '19 at 18:55
  • $\begingroup$ @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 $\endgroup$ – AmirWG Dec 28 '19 at 18:59
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    $\begingroup$ Exactly, that's why only standard reduction potentials are tabulated. There must be another electrode to complete the circuit. $\endgroup$ – andselisk Dec 28 '19 at 19:02
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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.

As you see, both approaches are equivalent. Have you understood my development ?

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  • $\begingroup$ You really did a great job explaining it to me , thanks a lot for your time. $\endgroup$ – AmirWG Dec 29 '19 at 14:33

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