Consider the following unblanced reaction in an acidic medium-S 4 O 62- aq - Al s - H 2 S aq - Al 3- aq Choose the correct options for a balanced chemical reaction-A- The coefficient of H -is 20B- The coefficient of Al is 5C- The coefficient of H 2 O is 6D- The coefficient of Al 3- is 8

The correct options are A The coefficient of H^+ is 20 C The coefficient of H_2O is 6+2.5S_4O^2 _6(aq) + 0Al(s) → H_2 2S(aq) + +3Al^3+(aq) S_4O^2 _6 is oxidisi ...

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Standard XII

Chemistry

Oxidation Number Method

Consider the ...

Question

Consider the following unblanced reaction in an acidic medium:
$S_{4} O_{6}^{2 -} (a q) + A l (s) \to H_{2} S (a q) + A l^{3 +} (a q)$
Choose the correct options for a balanced chemical reaction.

The coefficient of

H^{+}

20

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The coefficient of

A l

5

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The coefficient of

H_{2} O

6

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The coefficient of

A l^{3 +}

8

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Solution

The correct options are
A The coefficient of $H^{+}$ is $20$
C The coefficient of $H_{2} O$ is $6$
${+ 2.5 S}_{4} O_{6}^{2 -} (a q) + 0 A l (s) \to H_{2} - 2 S (a q) + + 3 A l^{3 +} (a q)$

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

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

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

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

$n_{f} = (| 3 - 0 | \times 1 = 3$
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:
$S_{4} O_{6}^{2 -} (a q) + 6 A l (s) + 20 H^{+} \to 4 H_{2} S (a q) + 6 A l^{3 +} (a q) + 6 H_{2} O (l)$

The coefficient of $H^{+}$ is $20$ and coefficient of $H_{2} O$ is 6.
Option A and C are correct.

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Similar questions

Q. Consider the following unblanced reaction in an acidic medium:

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

Choose the correct options for a balanced chemical reaction.

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

On balncing the above unblanced reaction (in acidic medium), the ratio of coefficient of

H^{+}

to that of

H_{2} O

is:

Q. What would be the coefficient of

H^{+}

and

H_{2} O

respectively in the balanced net ionic equations for the given redox reactions 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. What would be the coefficient of

H^{+}

and

H_{2} O

respectively in the balanced net ionic equations for the given redox reactions 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. 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)

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Oxidation Number Method

Standard XII Chemistry

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Consider the following unblanced reaction in an acidic medium: S4O2−6(aq)+Al(s)→H2S(aq)+Al3+(aq) Choose the correct options for a balanced chemical reaction.

Consider the following unblanced reaction in an acidic medium:
$S_{4} O_{6}^{2 -} (a q) + A l (s) \to H_{2} S (a q) + A l^{3 +} (a q)$
Choose the correct options for a balanced chemical reaction.