2H2O (l) + 2F2(g) → 4HF(aq) + O2(g).
In the reaction above, the substance that is being reduced is

A.

O2(g)

B.

H2O(l)

C.

F2(g)

D.

HF(aq)

View answer & explanation

Correct answer is C

F(2) + 2e \(\rightarrow\) 2F- x 2

2F2 + 4e \(\rightarrow\) 4F-

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