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
jamb 2019chemistry

The correct answer is B. +6 to +3

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; 

2x + (-2 x 7) = -2 \(\implies\) 2x - 14 = -2

2x = 12 ; x = +6

Hence, the change in oxidation of Cr = +6 to +3

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