Balance the following redox reactions:

$A l (s) + H_{2} {S O}_{4} (a q) ⟶ {A l}_{2} {({S O}_{4})}_{3} (a q) + H_{2} (g)$

2 A l (s) + {3 H}_{2} {S O}_{4} (a q) ⟶ {A l}_{2} {({S O}_{4})}_{3} (a q) + 3 H_{2} (g)

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5 A l (s) + {4 H}_{2} {S O}_{4} (a q) ⟶ {A l}_{2} {({S O}_{4})}_{3} (a q) + 2 H_{2} (g)

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4 A l (s) + H_{2} {S O}_{4} (a q) ⟶ {A l}_{2} {({S O}_{4})}_{3} (a q) + 3 H_{2} (g)

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5 A l (s) + {4 H}_{2} {S O}_{4} (a q) ⟶ {A l}_{2} {({S O}_{4})}_{3} (a q) + 5 H_{2} (g)

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Solution

The correct option is A $2 A l (s) + {3 H}_{2} {S O}_{4} (a q) ⟶ {A l}_{2} {({S O}_{4})}_{3} (a q) + 3 H_{2} (g)$
The balanced reaction is as follows:
$2 A l (s) + {3 H}_{2} {S O}_{4} (a q) ⟶ {A l}_{2} {({S O}_{4})}_{3} (a q) + 3 H_{2} (g)$
Aluminium is getting oxidized from $0$ to $+ 3$ .
So, there is a factor of $3$ before sulphuric acid and hydrogen.
Hydrogen is getting reduced from $1$ to $0$ .
Here, there are $2 \times 1$ electrons involved.
Hence, there will be a factor of $2$ before Al and aluminium sulphate.
Hence, the correct equation is in option A.

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Similar questions

H_{2} {S O}_{4} (a q) + {B a (O H)}_{2} \to

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H C l (g) + A q ⟶ H C l (a q); Δ H = - 17.32 k c a l

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9 g

2 A l + \frac{3}{2} O_{2} \to {A l}_{2} O_{3}

K_{2} {S O}_{4} . {A l}_{2} {({S O}_{4})}_{3} .24 H_{2} O

{({N H}_{4})}_{2} {S O}_{4} . {A l}_{2} {({S O}_{4})}_{3} .24 H_{2} O

\frac{1}{2}

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2 A l + \frac{3}{2} O_{2} ⟶ {A l}_{2} O_{3}

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