JAMB - Chemistry (2019 - No. 40)
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
+5 to +3
+6 to +3
-2 to +3
+7 to +3
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;
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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