Correct Answer: Option C

Explanation

Cr\(_2\)O\(_7\)\(^{2-}\)(aq) + 14H\(^+\)(aq) + 6e\(^-\) → 2Cr\(^{3+}\)(ag) + 7H\(_2\)O(1).

All we need to do is to calculate the oxidation number of Cr on the reactant side (LHS) and compare with the one on the RHS.

                    Cr\(_2\)O\(_7\)\(^{2-}\)     →    Cr\(^{3+}\)

                   Cr\(_2\)O\(_7\) = - 2         →    Cr  = + 3

                    2Cr + 7(O) = - 2 

                    2Cr + 7(- 2) = - 2

                    2Cr - 14  = - 2

                    2Cr = - 2 + 14

                    2Cr = + 12

                     Cr = \(\frac{12}{2}\)

                     Cr = + 6    →      Cr  = + 3

Therefore, the change in oxidation number of Cr changed from +6 to +3 - Option C

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