The number of electrons required to reduce chromium completely in ${C r}_{2} O_{7}^{2 -}$ to ${C r}^{3 +}$ in acidic medium is

5

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3

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6

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2

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Solution

The correct option is C $6$
${C r}_{2} O_{7}^{2 -} + 14 H^{+} + 6 e^{-} ⟶ 2 {C r}^{3 +} + 7 H_{2} O$

So, 6 electrons are required to reduce.

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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 -} = x \times

x

H_{2} O_{2} (a q)

{C r}_{2} O_{7}^{2 -} (a q)

{C r}^{3 +} (a q)

H_{2} O_{2} (a q) + {C r}_{2} O_{7}^{2 -} ⟶ O_{2} (g) + {C r}^{3 +} (a q)

{F e}^{2 +}

x

M n O_{4}^{-}

{C r}_{2} O_{7}^{2 -}

{F e}^{2 +}

x = 5

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