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: Option 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;
2x + (-2 x 7) = -2 (implies) 2x – 14 = -2
2x = 12 ; x = +6
Hence, the change in oxidation of Cr = +6 to +3
There is an explanation video available below.