In the reaction
Cr₂O₇²⁻ +14H⁺ + 6I⁻ → 2Cr³⁺ + 7H₂O + 3I₂
Which element is reduced
(a) I (b) O
(c) H (d) Cr
Answers
Answered by
1
in the above question, Chromium Oxide is reduced to Chromium
So The answer is Cr
Answered by
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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