Cr\(_2\)O\(_7\)\(^{2-}\)(aq) + 14H\(^+\)(aq) + 6l\(^-\)(aq) → 2Cr\(^{3+}\)(aq) + 3l(g) + 7H\(_2\)O(I).
The change in the oxidation number of oxygen in the equation above is?
0
1
2
7
Explanation
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Discussions (16)
I will try my best to explain the little i know Cr2O7^2- (-2*2+7O=-2)
-4+7O=-2
7O=-2+4
7O=+2
O=2/7*7
(in order to eliminate the denominator)
O=+2
equii
H2O
+2+O=0
Therefore O=-2
Then the change will be +2-2=0
I hope it helps
In the given redox reaction, the oxidation number of oxygen does not change. Oxygen has an oxidation number of -2 in both the reactants (Cr2O7^2-) and the products (H2O). So, the correct answer is A. 0.
Am I the only one seeing an incomplete question with no instructions whatsoever
Be water is a polar solvent and any polar solvent has one positive end and one negative end
