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 ?
