Question

In the reaction
Cr₂O²⁻₇ +14H⁺ + 6I⁻ ⟶ 2Cr³⁺ + 7H₂O + 3I₂
Which element is reduced
(a) I (b) O
(c) H (d) Cr

Answer 1

Answer:

$I^{-}$ is oxidised form $I^{-}$ to I₂ because ON changes from -1 to 0

Cr₂O₇$^{2-}$ is reduced into $Cr^{3+}$ because ON changes from +6 to +3

Explanation:

Answer 2

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Option.B

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