Question #a2c32

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Question #a2c32

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Answer 1 · 2017-05-14T13:12:14

$C r_{2} O_{7}^{2 -} (a q)$ $Cr_2O_7^(2-) (aq)$ + $3 S O_{2} (g)$ $3SO_2 (g)$ + $2 H^{+} (a q)$ $2H^+ (aq)$ → $2 C r^{3 +} (a q)$ $2Cr^(3+)(aq)$ + $3 S O_{4}^{2 -} (a q)$ $3SO_4^(2-) (aq)$ + $H_{2} O (l)$ $H_2O (l)$

Explanation:

$S O_{2}$ $SO_2$ gas is bubbled through a solution of $K_{2} C r_{2} O_{7}$ $K_2Cr_2O_7$

Half equations:
$C r_{2} O_{7}^{2 -} (a q)$ $Cr_2O_7^(2-) (aq)$ + $14 H^{+} (a q)$ $14H^+ (aq)$ + $6 e^{-}$ $6e^-$ → $2 C r^{3 +} (a q)$ $2Cr^(3+) (aq)$ + $7 H_{2} O (l)$ $7H_2O (l)$

$S O_{2} (g)$ $SO_2 (g)$ + $2 H_{2} O (l)$ $2H_2O (l)$ → $S O_{4}^{2 -} (a q)$ $SO_4^(2-) (aq)$ + $4 H^{+} (a q)$ $4H^+ (aq)$ + $2 e^{-}$ $2e^-$

(i'm not sure if you need this but i'll just include it)

Multiply half-eq'n 2 by 3
$3 S O_{2} (g)$ $3SO_2 (g)$ + $6 H_{2} O (l)$ $6H_2O (l)$ → $3 S O_{4}^{2 -} (a q)$ $3SO_4^(2-) (aq)$ + $12 H^{+} (a q)$ $12H^+ (aq)$ + $6 e^{-}$ $6e^-$

Remove electrons from both equations and then add them together:

$C r_{2} O_{7}^{2 -}$ $Cr_2O_7^(2-)$ + $14 H^{+}$ $14H^+$ + $3 S O_{2}$ $3SO_2$ + $6 H_{2} O$ $6H_2O$ → $2 C r^{3 +}$ $2Cr^(3+)$ + $7 H_{2} O$ $7H_2O$ + $3 S O_{4}^{2 -}$ $3SO_4^(2-)$ + $12 H^{+}$ $12H^+$

And basically just subtract the same species to get a simpler answer.

e.g. $14 H^{+}$ $14H^+$ on RHS minus $12 H^{+}$ $12H^+$ on LHS gives $2 H^{+}$ $2H^+$ on RHS.

Full equation:
$C r_{2} O_{7}^{2 -} (a q)$ $Cr_2O_7^(2-) (aq)$ + $3 S O_{2} (g)$ $3SO_2 (g)$ + $2 H^{+} (a q)$ $2H^+ (aq)$ → $2 C r^{3 +} (a q)$ $2Cr^(3+)(aq)$ + $3 S O_{4}^{2 -} (a q)$ $3SO_4^(2-) (aq)$ + $H_{2} O (l)$ $H_2O (l)$

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Question #a2c32

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