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

20, 6

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

6, 20

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10, 4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4, 10

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 20, 6
Steps for Balancing redox reactions:

Identify the oxidation and reduction half .

Find the oxidising and reducing agent.

Find the n-factor of oxidising and reducing agent.

Balance atom undergoing oxidation and reduction.

Cross multiply the oxidising or reducing agent with n-factor.

Balance atoms other than oxygen and hydrogen.

Balancing oxygen atoms

Balancing hydrogen atoms

Balance charge

For acidic medium:
As soon as we add x $H_{2} O$ units, we add 2x $H^{+}$ ions on the opposite side.

$n_{f} = (| O . S ._{P r o d u c t} - O . S ._{R e a c t a n t} | \times number of atom$
${+ 2.5 S}_{4} O_{6}^{2 -} (a q) + 0 A l (s) \to H_{2} - 2 S (a q) + + 3 A l^{3 +} (a q)$

${+ 2.5 S}_{4} O_{6}^{2 -} (a q) \to H_{2} - 2 S (a q)$ ; Reduction
$n_{f} = (| - 2 - 2.5 | \times 4 = 18$
$0 A l (a q) \to + 3 A l^{3 +} (a q)$ ; Oxidation
$n_{f} = (| 3 - 0 | \times 1 = 3$
$S_{4} O_{6}^{2 -}$ is oxidising agent
$A l$ is reducing agent.

Balance atom undergoing oxidation and reduction.
$S_{4} O_{6}^{2 -} (a q) + A l (s) \to 4 H_{2} S (a q) + A l^{3 +} (a q)$

Ratio of n- factor's =18:3
dividing the n-factors by common value in them. which is 3 in this case.
$\Rightarrow$ Ratio of n- factor's =6:1

Cross mutiply the oxidising and reducing agents with ratio of n-factors. and balancing the main elements other than oxygen and hydrogen.

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

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

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

balance charge
charge in reactant side = -2+20=+18
charge in product side = +18
so the balanced equation is $S_{4} O_{6}^{2 -} (a q) + 6 A l (s) + 20 H^{+} \to 4 H_{2} S (a q) + 6 A l^{3 +} (a q) + 6 H_{2} O (a q)$
the coefficient of $H^{+}$ and $H_{2} O$ respectively are 20 and 6.

Suggest Corrections

Similar questions

H^{+}

H_{2} O

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

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

M g

V O_{4}^{3 -}

M g (s) + V O_{4}^{3 -} (a q) \to M g^{2 +} (a q) + V^{2 +} (a q)

H^{+}

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)

M g

V O_{4}^{3 -}

M g (s) + V O_{4}^{3 -} (a q) \to M g^{2 +} (a q) + V^{2 +} (a q)

Join BYJU'S Learning Program