In a reaction , $C r_{2} O_{7}^{2 -}$ is reduced to $C r^{3 +}$ . What is concentration of 0.1 M $K_{2} C r_{2} O_{7}$ in equivalent per litre?
$C r_{2} O_{7}^{2 -} + 14 H^{+} + 6 e \to 2 C r^{3 +} + 7 H_{2} O$

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Solution

$N = M \times$ valence factor
$∴ N = 0.1 \times 6 (C r_{2}^{6 +} + 6 e \to 2 C r^{3 +})$

or $N_{K_{2} C r_{2} O_{7}} = 0.6 N$

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Similar questions

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

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

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

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

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

K_{2} C r_{2} O_{7}

= M

25^{\circ} C

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

p O H = 11

0.01 M

C r^{3 +}

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

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

E^{\circ} = 1.33 V

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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