Consider the equation below: Cr\(_2\)O\(_7 ^{2-}\) + 6Fe\(^{2+}\) + 14H\(^{+}\) \(\to\) 2Cr\(^{3+}\) + 6Fe\(^{3+}\) + 7H\(_2\)O. The oxidation number of chromium changes from
A. +5 to +3 B. +6 to +3 C. -2 to +3 D. +7 to +3
Correct Answer: B
Explanation
Cr\(_2\)O\(_7 ^{2-}\) + 6Fe\(^{2+}\) + 14H\(^{+}\) \(\to\) 2Cr\(^{3+}\) + 6Fe\(^{3+}\) + 7H\(_2\)O The oxidation of Cr in Cr\(_2\)O\(_7 ^{2-}\) : Let the oxidation of Cr = x;Â 2x + (-2 x 7) = -2 \(\implies\) 2x - 14 = -2 2x = 12 ; x = +6 Hence, the change in oxidation of Cr = +6 to +3
