Question

In the following reaction

$Cr_{2}O_{7}^{-} + 14H^{+} + 6I^{-} \rightarrow 2Cr^{3 +} + 3H_{2}O + 3I_{2}$

Which element is reduced.

• A. $Cr$
• B. $H$
• C. $O$
• D. $I$

Answer 1

Answer: (A)

$Cr$

Answer 2