The value of n in the equation- Cr2O72- - 14H- - ne- 2Cr3- - 7H2O is-

The correct option is D 6 Using hit and trial method: Cr_2O_7^2 + 14H^+ + 6e^ → 2Cr^3+ + 7H_2O #160;In the above equation, n=6 electrons are ...

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Oxidation Number Method

The value of ...

Question

The value of $n$ in the equation, $C r_{2} O_{7}^{2 -} + 14 H^{+} + n e^{-} \to 2 C r^{3 +} + 7 H_{2} O$ is:

2

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3

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4

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6

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^{'} n^{'}

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

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

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

M

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

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

1 m o l

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

14 H^{+} + C r_{2} O_{7}^{2 -} + 3 N i \to 2 C r^{3 +} + 7 H_{2} O + 3 N i^{2 +}

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