How many electrons are involved in the following redox reaction?
$C r_{2} O_{7}^{2 -} + F e^{2 +} + C_{2} O_{4}^{2 -} \to C r^{3 +} + F e^{3 +} + C O_{2}$ (Unbalanced)

3

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4

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6

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5

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Solution

The correct option is B $6$
$C r_{2} O_{7}^{2 -} + 2 F e^{2 +} + 2 C_{2} O_{4}^{2 -} \to 2 C r^{3 +} + 2 F e^{3 +} + 4 C O_{2}$

The oxidation number of chromium in $C r_{2} O_{7}^{2 -}$ is $+ 6$ and it reduces to $+ 3$ in $C r^{3 +}$ . On balancing the equation, we will see $2$ moles of chromium ion goes from $+ 6$ to $+ 3$ .

Hence, there are $6$ electrons involved in the above redox reaction.

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