Can you walk through the steps for balancing ${Cr}_{2} O_{7}^{2 -} + {SO}_{3}^{2 -} + H^{+} \to {Cr}^{3 +} + {SO}_{4}^{2 -} + H_{2} O$ $"Cr"_2"O"_7^(2-) + "SO"_3^(2-) + "H"^+ → "Cr"^(3+) + "SO"_4^(2-) + "H"_2"O"$ ?

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Can you walk through the steps for balancing ${Cr}_{2} O_{7}^{2 -} + {SO}_{3}^{2 -} + H^{+} \to {Cr}^{3 +} + {SO}_{4}^{2 -} + H_{2} O$ $"Cr"_2"O"_7^(2-) + "SO"_3^(2-) + "H"^+ → "Cr"^(3+) + "SO"_4^(2-) + "H"_2"O"$ ?

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Answer 1 · 2015-10-22T18:00:25

Here are the detailed steps.

Explanation:

Your equation is

${Cr}_{2} O_{7}^{2 -} + {SO}_{3}^{2 -} + H^{+} \to {Cr}^{3 +} + {SO}_{4}^{2 -} + H_{2} O$ $"Cr"_2"O"_7^(2-) + "SO"_3^(2-) + "H"^+ → "Cr"^(3+) + "SO"_4^(2-) + "H"_2"O"$

1. Identify the oxidation number of every atom.

Left hand side: $Cr = +6$ $"Cr = +6"$ ; $O = -2$ $"O = -2"$ ; $S = +4$ $"S = +4"$ ; $H = +1$ $"H = +1"$
Right hand side: $Cr = +3$ $"Cr = +3"$ ; $S = +6$ $"S = +6"$ ; $O = -2$ $"O = -2"$ ; $H = +1$ $"H = +1"$

2. Determine the change in oxidation number for each atom that changes.

$Cr: +6 \to +3$ $"Cr: +6 → +3"$ ; Change = -3
$S: +4 \to +6$ $"S: +4 → +6"$ ; Change = +2

3. Make the total increase in oxidation number equal to the total decrease in oxidation number.

We need 2 atoms of $Cr$ $"Cr"$ for every 3 atoms of $S$ $"S"$ . This gives us total changes of -6 and +6.

4. Put the appropriate coefficients in front of the formulas containing those atoms.

$1 {Cr}_{2} O_{7}^{2 -} + 3 {SO}_{3}^{2 -} + H^{+} \to 2 {Cr}^{3 +} + 3 {SO}_{4}^{2 -} + H_{2} O$ $color(red)(1)"Cr"_2"O"_7^(2-) + color(red)(3)"SO"_3^(2-) + "H"^+ → color(red)(2)"Cr"^(3+) + color(red)(3)"SO"_4^(2-) + "H"_2"O"$

5. Balance all remaining atoms other than $H$ $"H"$ and $O$ $"O"$ .

Done.

6. Balance O.

We have fixed 16 atoms of $O$ $"O"$ on the left, so we need 16 atoms of $O$ $"O"$ on the right.

And we have fixed 12 atoms of $O$ $"O"$ on the right, so we need 4 more.

Put a $4$ $color(blue)(4)$ in front of the $H_{2} O$ $"H"_2"O"$ .

$1 {Cr}_{2} O_{7}^{2 -} + 3 {SO}_{3}^{2 -} + H^{+} \to 2 {Cr}^{3 +} + 3 {SO}_{4}^{2 -} + 4 H_{2} O$ $color(red)(1)"Cr"_2"O"_7^(2-) + color(red)(3)"SO"_3^(2-) + "H"^+ → color(red)(2)"Cr"^(3+) + color(red)(3)"SO"_4^(2-) + color(blue)(4)"H"_2"O"$

7. Balance H.

We have fixed 8 atoms of $H$ $"H"$ on the right, so we need 8 on the left.

Put an $8$ $color(green)(8)$ in front of the $H^{+}$ $"H"^+$ .

$1 {Cr}_{2} O_{7}^{2 -} + 3 {SO}_{3}^{2 -} + 8 H^{+} \to 2 {Cr}^{3 +} + 3 {SO}_{4}^{2 -} + 4 H_{2} O$ $color(red)(1)"Cr"_2"O"_7^(2-) + color(red)(3)"SO"_3^(2-) + color(green)(8)"H"^+ → color(red)(2)"Cr"^(3+) + color(red)(3)"SO"_4^(2-) + color(blue)(4)"H"_2"O"$

8. Check that everything is balanced.

(a) Atoms

On the left: $2Cr; 16 O; 3 S; 8 H$ $"2Cr; 16 O; 3 S; 8 H"$
On the right: $2Cr; 3 S; 16 O; 8 H$ $"2Cr; 3 S; 16 O; 8 H"$

(b) Charge

On the left: 0
On the right: 0

Everything checks!

The final balanced equation is

${Cr}_{2} O_{7}^{2 -} + 3 {SO}_{3}^{2 -} + 8 H^{+} \to 2 {Cr}^{3 +} + 3 {SO}_{4}^{2 -} + 4 H_{2} O$ $"Cr"_2"O"_7^(2-) + 3"SO"_3^(2-) + 8"H"^+ → 2"Cr"^(3+) + 3"SO"_4^(2-) + 4"H"_2"O"$

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Can you walk through the steps for balancing ${Cr}_{2} O_{7}^{2 -} + {SO}_{3}^{2 -} + H^{+} \to {Cr}^{3 +} + {SO}_{4}^{2 -} + H_{2} O$ $"Cr"_2"O"_7^(2-) + "SO"_3^(2-) + "H"^+ → "Cr"^(3+) + "SO"_4^(2-) + "H"_2"O"$ ?

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Explanation:

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Can you walk through the steps for balancing Cr2O2−7+SO2−3+H+→Cr3++SO2−4+H2O"Cr"_2"O"_7^(2-) + "SO"_3^(2-) + "H"^+ → "Cr"^(3+) + "SO"_4^(2-) + "H"_2"O"?

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Can you walk through the steps for balancing ${Cr}_{2} O_{7}^{2 -} + {SO}_{3}^{2 -} + H^{+} \to {Cr}^{3 +} + {SO}_{4}^{2 -} + H_{2} O$ $"Cr"_2"O"_7^(2-) + "SO"_3^(2-) + "H"^+ → "Cr"^(3+) + "SO"_4^(2-) + "H"_2"O"$ ?