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?