How many electrons should be added to the reactant to balance the atom whose oxidation number is changing?

$C r_{2} O_{7}^{2 -}$ $⟶$ $C r^{+ 3}$

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Solution

The correct option is D
6

First multiply $C r^{+ 3}$ $\times$ 2 to balance no of Cr atoms

$C r_{2} O_{7}^{- 2}$ $⟶$ $2 C r^{3 +}$

First let us calculate the oxidation number of the Cr atom on the LHS.

2n-14 = -2 2n =12

$⟹$ n =6

Clearly, the change in oxidation state is $2 + 6 C r ⟶ 2 + 3 C r$

As you can see, each Cr atom undergoes a change in oxidation of +3 - (+6) = -3. So For each $C r$ atom 3 electrons must be added to the LHS. So a total of 6 electrons shall be added

Therefore, $C r_{2} O_{7}^{2 -} + 6 e^{-} ⟶$ $2 C r^{3 +}$ . Do note that we are not balancing the whole reaction here. This is more like balancing a half reaction.

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Similar questions

How many electrons should be added to the reactant to balance the atom whose oxidation number is changing?

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How many electrons should be added to the reactant to balance the atom whose oxidation number is changing? Cr2O2−7⟶Cr+3

How many electrons should be added to the reactant to balance the atom whose oxidation number is changing?

$C r_{2} O_{7}^{2 -}$ $⟶$ $C r^{+ 3}$