Every electrochemistry data sheet you will ever be handed prints its half-reactions as reductions — electrons on the left, going in. The number beside one is its standard reduction potential, $E^{\circ}$, in volts, measured against the standard hydrogen electrode, which is pinned at exactly $0.00\ \mathrm{V}$ by international agreement. The little circle is said standard, or naught: one mole per litre for every solute, one bar for every gas, 25 °C.

A cell needs two half-cells, and the table hands out the jobs. The one with the more positive $E^{\circ}$ wants the electrons more, so it runs forward as written — a reduction — and that electrode is the cathode. The other is forced to run backwards, as an oxidation: the anode. Red Cat, An Ox — reduction at the cathode, oxidation at the anode.

The pair is worth the gap between them: $E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}$. Read aloud: E-standard-cell equals E-standard-cathode minus E-standard-anode. $E^{\circ}_{\text{cell}}$ is the whole cell's standard voltage, in volts; the two on the right are the tabulated reduction potentials of the two half-cells, also in volts, subscripted for the job each one ended up doing. You are solving for the one on the left.

Now the wound this lesson exists to close. The anode's value goes into that subtraction exactly as the table prints it, sign and all. Do not flip its sign first and then subtract as well — that is one operation performed twice, and it lands you on the wrong side of zero. Which gives you a free check, every time: a galvanic cell that actually runs has $E^{\circ}_{\text{cell}} > 0$. A negative answer means you gave the jobs to the wrong electrodes.