Still taught to reverse oxidation half cells in electrochemistry?

In a question, Oxidation of metals/halogens by oxygen gas in acidic aqueous solution, there was a point made that reduction half-cells should not be reversed.

I was taught ($\approx 1970$) in the following manner. Given the reduction half cells:

$$ \begin{align} \ce{O2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l)} &\quad E^\circ &= \pu{+1.23 V} \\ \ce{Ag+ + e- &→ Ag(s)} &\quad E^\circ &= \pu{+0.799 V} \\ \ce{Br2(l) + 2 e- &→ 2 Br-(aq)} &\quad E^\circ &= \pu{+1.065 V} \end{align} $$

To get the the standard oxidation potentials you reverse the products and reactants and you must also flip the sign of the reaction. So:

$$ \begin{align} \ce{Ag(s) &→ Ag+ + e- } &\quad E^\circ &= \pu{-0.799 V} \\ \ce{2 Br-(aq) &→ Br2(l) + 2 e- } &\quad E^\circ &= \pu{-1.065 V} \end{align} $$

Now you can write a balanced chemical reaction as such:

$$ \begin{align} \ce{O2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l)} &\quad E^\circ &= \pu{+1.23 V} \\ \ce{4Ag(s) &→ 4Ag+ + 4e- } &\quad E^\circ &= \pu{-0.799 V} \\ \ce{O2(g) + 4 H+(aq) + 4 Ag(s)&→ 4Ag+ + 2 H2O(l)} &\quad E_\text{Total}^\circ &= \pu{+1.23 - 0.799 = +0.43 V} \end{align} $$

and to get the cell potential the two half cells are added, one being positive and the other negative. If the EMF is positive the forward reaction is spontaneous, if negative the reverse reaction is spontaneous.

Is flipping the reduction reaction not the standard method taught these days to balance redox reactions?