The number of electrons required to reduce chromium completely in Cr $_{2}$ O $_{7}$ $^{2 -}$ to Cr $^{3 +}$ in acidic medium, is

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Solution

The correct option is C
6

$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} \to C r^{3 +} + 7 H_{2} O$

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Similar questions

The number of electrons required to reduce chromium completely in Cr $_{2}$ O $_{7}$ $^{2 -}$ to Cr $^{3 +}$ in acidic medium, is

C r_{2} O_{7}^{2 -}

C r^{3 +}

C r_{2} O_{7}^{2 -}

C r^{3 +}

The number of electrons involved in the reduction of $C r_{2} O_{7}^{2 -}$ in acidic solution to $C r^{3 +}$ is

C r_{2} O_{7}^{2 -} t o C r^{3 +}

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