Cr\(_2\)O\(_7\)\(^{2-}\) + 6Fe\(^{2+}\) + 14H\(^+\) → 2Cr\(^{3+}\) +6Fe\(^{3+}\) + 7H\(_2\)O. In the above , the oxidation number of chromium changed from?
The oxidation number of Cr in Cr\(_2\)O\(_7\)\(^{2-}\) is
Cr\(_2\)O\(_7\) = - 2
2Cr + ( - 2) x 7 = - 2
2Cr - 14 = - 2
2Cr = - 2 + 14
2Cr = 12
\(\frac{2Cr}{2}\) = \(\frac{12}{2}\)
Cr = + 6
On the RHS, the oxidation number of Cr is +3.
Thus, the oxidation number changed from +6 to +3 - option B.
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