How many electrons are lost or gained by Cr in the given chemical reaction below:
$C r_{2} O_{7}^{2 -} \to C r^{3 +}$

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Solution

The correct option is D 6
Let’s go step by step here.
First multiply $C r^{3 +} \times 2$ to balance the number of Cr atoms.
$C r_{2} O_{7}^{2 -} \to 2 C r^{3 +}$
Oxidation state of $C r$ on the LHS:
$2 \times C r + 7 \times (- 2) = - 2, C r = + 6$
Oxidation state of $C r$ on the RHS:
$C r = + 3$
Since there are two $C r$ atoms, the total change is $- 6$
So, we need to add 6 electrons to the LHS to balance this reaction.
$6 e^{-} + C r_{2} O_{7}^{2 -} \to 2 C r^{3 +}$

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