Consider the following redox reaction -S 4 O 62- aq - Al s - H 2 S aq - Al 3- aq What will be the net balanced equation in an acidic medium by oxidation number method-A- 2 S 4 O 62- aq -6 Al s -15 H - 2 H 2 S aq -6 Al 3- aq -6 H 2 O l B- S 4 O 62- aq -6 Al s -20 H - 4 H 2 S aq -6 Al 3- aq -6 H 2 O l - 3 S 4 O 62- aq -6 Al s -2 H - 3 H 2 S aq -6 Al 3- aq -6 H 2 O l D- S 4 O 62- aq -5 Al s -20 H - 4 H 2 S aq -5 Al 3- aq -4 H 2 O l

+2.5S_4O^2 _6(aq) + 0Al(s) → H_2 2S(aq) + +3Al^3+(aq) S_4O^2 _6 is oxidising agent Al is reducing agent. n_f=(|O.S._Product O.S._Reactant|×number of atom +2. ...

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Standard XII

Chemistry

Oxidation Number Method

Consider the ...

Question

Consider the following redox reaction :

$S_{4} O_{6}^{2 -} (a q) + A l (s) \to H_{2} S (a q) + A l^{3 +} (a q)$

What will be the net balanced equation in an acidic medium (by oxidation number method).

2 S_{4} O_{6}^{2 -} (a q) + 6 A l (s) + 15 H^{+} \to 2 H_{2} S (a q) + 6 A l^{3 +} (a q) + 6 H_{2} O (l)

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S_{4} O_{6}^{2 -} (a q) + 6 A l (s) + 20 H^{+} \to 4 H_{2} S (a q) + 6 A l^{3 +} (a q) + 6 H_{2} O (l)

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3 S_{4} O_{6}^{2 -} (a q) + 6 A l (s) + 2 H^{+} \to 3 H_{2} S (a q) + 6 A l^{3 +} (a q) + 6 H_{2} O (l)

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S_{4} O_{6}^{2 -} (a q) + 5 A l (s) + 20 H^{+} \to 4 H_{2} S (a q) + 5 A l^{3 +} (a q) + 4 H_{2} O (l)

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Solution

The correct option is B $S_{4} O_{6}^{2 -} (a q) + 6 A l (s) + 20 H^{+} \to 4 H_{2} S (a q) + 6 A l^{3 +} (a q) + 6 H_{2} O (l)$
${+ 2.5 S}_{4} O_{6}^{2 -} (a q) + 0 A l (s) \to H_{2} - 2 S (a q) + + 3 A l^{3 +} (a q)$

$S_{4} O_{6}^{2 -}$ is oxidising agent
$A l$ is reducing agent.

$n_{f} = (| O . S ._{P r o d u c t} - O . S ._{R e a c t a n t} | \times number of atom$

${+ 2.5 S}_{4} O_{6}^{2 -} (a q) \to H_{2} - 2 S (a q)$ Reduction
$n_{f} = (| - 2 - 2.5 | \times 4 = 18$

$0 A l (a q) \to + 3 A l^{3 +} (a q)$ Oxidation

$n_{f} = (| 3 - 0 | \times 1 = 3$
Ratio of $n_{f}$ for oxidation to reduction is $6 : 1$

Cross mutiply the oxidising and reducing agents with ratio of n-factors.

$S_{4} O_{6}^{2 -} (a q) + 6 A l (s) \to H_{2} S (a q) + A l^{3 +} (a q)$
Balancing elements except $O$ and $H$

$S_{4} O_{6}^{2 -} (a q) + 6 A l (s) \to 4 H_{2} S (a q) + 6 A l^{3 +} (a q)$

Balance atoms oxygen by adding $H_{2} O$ .
$S_{4} O_{6}^{2 -} (a q) + 6 A l (s) \to 4 H_{2} S (a q) + 6 A l^{3 +} (a q) + 6 H_{2} O (l)$

Balance hydrogen by adding $H^{+}$
$S_{4} O_{6}^{2 -} (a q) + 6 A l (s) + 20 H^{+} \to 4 H_{2} S (a q) + 6 A l^{3 +} (a q) + 6 H_{2} O (l)$

Balancing charge:

charge in reactant side = -2+20=+18
charge in product side = +18
so the balanced equation is

$S_{4} O_{6}^{2 -} (a q) + 6 A l (s) + 20 H^{+} \to 4 H_{2} S (a q) + 6 A l^{3 +} (a q) + 6 H_{2} O (l)$

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Similar questions

Q. Write balanced net ionic equation for the following reaction in acidic solution.

S_{4} O_{6}^{2 -} (a q) + A l (s) \to H_{2} S (a q) + A l^{3 +} (a q)

Q. Consider the following redox reaction :

S_{4} O_{6}^{2 -} (a q) + A l (s) \to H_{2} S (a q) + A l^{3 +} (a q)

What will be the net balanced equation in an acidic medium (by oxidation number method).

Write balanced net ionic equation for the following reactions in acidic solution.

S_{2} O {_{6}}^{2 -} (a q) + C r_{2} O {_{7}}^{2 -} (a q) \to S_{4} O {_{6}}^{2 -} (a q) + C r^{3 +} (a q)

Balance the following redox equations by oxidation number method.

occur in acidic medium

S n_{(a q)}^{2 +} + I O_{4 (a q)}^{-} \to S n_{(a q)}^{4 +} + I_{(a q)}^{-}

Q. For the equation in acidic medium:

S_{2} O_{3}^{2 -} (a q) + C r_{2} O_{7}^{2 -} (a q) \to S_{4} O_{6}^{2 -} (a q) + C r^{3 +} (a q)

The coefficient of

H^{+}

in the balanced reaction by oxidation number method will be :

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Oxidation Number Method

Standard XII Chemistry

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Consider the following redox reaction : S4O2−6(aq)+Al(s)→H2S(aq)+Al3+(aq) What will be the net balanced equation in an acidic medium (by oxidation number method).

Consider the following redox reaction :

$S_{4} O_{6}^{2 -} (a q) + A l (s) \to H_{2} S (a q) + A l^{3 +} (a q)$

What will be the net balanced equation in an acidic medium (by oxidation number method).