Answer
First, Chromium(II) rapidly reduces O2(g) to water and forms chromium(III):
$2 Cr^{3+}(aq) + Zn(s) → 2 Cr^{2+}(aq) + Zn^{2+}(aq) $
Then, chromium(III) is reduced back to chromium(II) by the zinc:
$4 Cr^{2+}(aq) + 4 H^{+}(aq) + O_{2}(g) → 4 Cr_{3}+(aq) + 2 H_{2}O(l)$
Work Step by Step
First, Chromium(II) rapidly reduces O2(g) to water and forms chromium(III):
$2 Cr^{3+}(aq) + Zn(s) → 2 Cr^{2+}(aq) + Zn^{2+}(aq) $
Then, chromium(III) is reduced back to chromium(II) by the zinc:
$4 Cr^{2+}(aq) + 4 H^{+}(aq) + O_{2}(g) → 4 Cr_{3}+(aq) + 2 H_{2}O(l)$
