What would be the coefficient of $H^{+}$ ion in the balanced net ionic equations for the following reactions in acidic solution: $S_{2} O_{3}^{2 -} (a q) + C r_{2} O_{7}^{2 -} (a q) \to S_{4} O_{6}^{2 -} (a q) + C r^{3 +} (a q)$

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Solution

The correct option is C 14
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.

$n_{f} = (| O . S ._{P r o d u c t} - O . S ._{R e a c t a n t} | \times number of atom$
${+ 2 S}_{2} O_{3}^{2 -} (a q) + + 6 C r_{2} O_{7}^{2 -} (a q) \to {+ 2.5 S}_{4} O_{6}^{2 -} (a q) + + 3 C r^{3 +} (a q)$

${+ 2 S}_{2} O_{3}^{2 -} (s) \to {+ 2.5 S}_{4} O_{6}^{2 -} (a q)$ oxidation
$n_{f} = (| + 2.5 - 2 | \times 2 = 1$
$+ 6 C r_{2} O_{7}^{2 -} (a q) \to + 3 C r^{3 +} (a q)$ reduction
$n_{f} = (| 3 - 6 | \times 2 = 6$
$C r_{2} O_{7}^{2 -}$ is oxidising agent
$S_{2} O_{3}^{2 -}$ is reducing agent.

Balance atom undergoing oxidation and reduction.
$2 S_{2} O_{3}^{2 -} (a q) + C r_{2} O_{7}^{2 -} (a q) \to S_{4} O_{6}^{2 -} (a q) + 2 C r^{3 +} (a q)$

Cross mutiply the oxidising or reducing agent with n-factor.
$12 S_{2} O_{3}^{2 -} (a q) + C r_{2} O_{7}^{2 -} (a q) \to 6 S_{4} O_{6}^{2 -} (a q) + 2 C r^{3 +} (a q)$

Balance atoms oxygen.
$12 S_{2} O_{3}^{2 -} (a q) + C r_{2} O_{7}^{2 -} (a q) \to 6 S_{4} O_{6}^{2 -} (a q) + 2 C r^{3 +} (a q) + 7 H_{2} O$

Balance hydrogen.
$12 S_{2} O_{3}^{2 -} (a q) + C r_{2} O_{7}^{2 -} (a q) + 14 H^{+} \to 6 S_{4} O_{6}^{2 -} (a q) + 2 C r^{3 +} (a q) + 7 H_{2} O$
Balance charge
charge in reactant side = -12
charge in product side = -6
so the balanced equation is $12 S_{2} O_{3}^{2 -} (a q) + C r_{2} O_{7}^{2 -} (a q) + 14 H^{+} \to 6 S_{4} O_{6}^{2 -} (a q) + 2 C r^{3 +} (a q) + 7 H_{2} O + 6 e^{-}$
the coefficient of $H^{+}$ is 14.

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Similar questions

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)

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)

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)

H^{+}

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

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)

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