Consider the equation below Cr 2 O 7 2- 6Fe 2 1... - JAMB Chemistry 2019 Question
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 option: b
Cr\(_2\)O\(_7 ^{2-}\) + 6Fe\(^{2+}\) + 14H\(^{+}\) \(\to\) 2Cr\(^{3+}\) + 6Fe\(^{3+}\) + 7H\(_2\)O
Oxidation of Cr in Cr\(_2\)O\(_7 ^{2-}\) :
Let oxidation of Cr = x
then,
2x + (-2 x 7) = -2 \(\implies\) 2x - 14 = -2
2x = 12 ; x = +6
Therefore,
the change in the oxidation of Cr = +6 to +3
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