Chemistry, asked by karishma8448, 10 months ago

In the reaction
Cr₂O₇²⁻ +14H⁺ + 6I⁻ → 2Cr³⁺ + 7H₂O + 3I₂
Which element is reduced
(a) I (b) O
(c) H (d) Cr

Answers

Answered by AnkitaBanga
1

in the above question, Chromium Oxide is reduced to Chromium

So The answer is Cr

Answered by Anonymous
1

Element which is reduced is - (d) Cr

In the given reaction, chromium oxide has inital oxidation number equivalent to +6.

After the reaction, chromium ions are released as product with oxidation number +3.

Since, chromium gains electron, hence it is the reduced element in the given reaction.

Oxidation number of chromium oxide can be calculated as follows-

Assuming X as oxidation number of chromium in chromium oxide.

2x + (-2*7) = -2

2x -14 = -2

2x = 14-2

2x = 12

X = +6

Hence, correct option is option d.

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