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