No sign flipping while figuring out the emf of voltaic cell?

I learnt that for a voltaic cell, the value for the $E\_\text{cell}^\circ$ when the reaction is *spontaneous* is given by 


$$E\_\text{cell}^\circ = E\_\text{cathode}^\circ - E\_\text{anode}^\circ, \label{eqn:1}\tag{1}$$


so that the difference in the right gives us a positive value for $E\_\text{cell}^\circ$. 


But suppose we are given two half-reactions:


$$
\begin{align}
\ce{Ag+(aq) + e- &→ Ag(s)} &\qquad E^\circ &= \pu{0.80 V} \\
\ce{Sn^2+(aq) + 2 e- &→ Sn(s)} &\qquad E^\circ &= \pu{-0.14 V}
\end{align}
$$


When finding the overall spontaneous reaction, we must flip the second reaction, multiply it by $2$, and then add it with the first to get our desired equation. 




---


But when determining the $E\_\text{cell}^\circ$, why don't we negate the minus sign of the second half-reaction and make positive, before we put it in $\eqref{eqn:1}$ to figure out the $E\_\text{cell}^\circ$? Shouldn't we do that because we reversed the second equation?


My book tells me to keep the $E\_\text{half-cells}^\circ$ as they are written in the tables and simply put them in $\eqref{eqn:1}$. But why?




Take a look at the two half reactions:


$$
\begin{align}
\ce{Ag+(aq) + e- &→ Ag(s)} &\qquad E^\circ &= \pu{0.80 V} \\
\ce{Sn^2+(aq) + 2 e- &→ Sn(s)} &\qquad E^\circ &= \pu{-0.14 V}
\end{align}
$$


If there is an electron for grabs (like the ones in the wire of a voltaic cell), $\ce{Ag+(aq)}$ and $\ce{Sn^2+(aq)}$ are competing for it. Whichever half reaction has the higher (more positive) reduction potential will win. If the reduction potentials are equal, it is a draw and the reaction is at equilibrium. So we are taking the difference of the reduction potentials to see in which direction the reaction will go.



> 
> No sign flipping while figuring out the emf of voltaic cell?
> 
> 
> 


Take a look at the equation you are using to figure out the emf. You are already treating the oxidation half reaction differently than the reduction half reaction because there is a negative sign in front of the anode reduction potential.


$$E\_\text{cell}^\circ = E\_\text{cathode}^\circ - E\_\text{anode}^\circ$$


If you switch the anode and cathode half reaction, you would get the opposite sign for the emf. (Not that the reaction would go in that direction.)


 3


> 
> My book tells me to keep the E∘half-cells as they are written in the
>  tables and simply put them in
> 
> 
> 


Your book is then one of the few books which teaches electrochemistry properly. The sign of the electrode reduction potential is *invariant*. If reflects the *sign of the electrostatic charge* of the electrode with respect to hydrogen electrode. I don;t know if you like history or not, long time ago in the 1950s-60s showing the electrostatic sign of a cell by means of a specially designed electroscope was a standard experiment in physics.


Imagine if I say H2O (l) --> H2O (g) at 100 oC


Does this mean reversing the reaction


H2O(g) --> H2O (l) will be at -100 oC?


You can see logical fallacy in reversing the sign of electrode potentials. 


Yes, there was a lot of confusion in electrode signs for more than 100 years but now it has been sorted out. 


 3

The Nernst equation and electrochemical potentials relate to redox systems, not to reagents and products. The forward and reversed reactions are the same redox system.


Imagine you would want to make a galvanical cell with the same electrodes. Flipping the sign would grant you a Nobel price for inventing a perpetuum mobile.


 1