Balance the following chemical equation- Cr 2 O 72- aa SO 32- aq Cr 3- aq SO 42- Acidic medium

The correct option is C (aq)Cr_2O_7^2 + (aq)3SO_3^2 +(aq)8H^+→(aq)2Cr^3+ + (aq)3SO^2 _4 + (aq)4H_2O So , oxidizing agent : Cr_2O_7 Reducing agent : SO_3 n ...

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Standard XII

Chemistry

Oxidation Number Method

Balance the f...

Question

Balance the following chemical equation: $C r_{2} O_{7}^{2 -} (a q) + S O_{3}^{2 -} (a q) \to C r^{3 +} (a q) + S O_{4}^{2 -} (a q)$ (Acidic medium)

C r_{2} O_{7}^{2 -} (a q) + 3 S O_{3}^{2 -} (a q) + H^{+} (a q) \to 2 C r^{3 +} (a q) + 3 S O_{4}^{2 -} (a q) + 4 H_{2} O (a q)

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C r_{2} O_{7}^{2 -} (a q) + S O_{3}^{2 -} (a q) + 8 H^{+} (a q) \to 2 C r^{3 +} (a q) + 3 S O_{4}^{2 -} (a q) + 4 H_{2} O (a q)

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C r_{2} O_{7}^{2 -} (a q) + 3 S O_{3}^{2 -} (a q) + 8 H^{+} (a q) \to 2 C r^{3 +} (a q) + 3 S O_{4}^{2 -} (a q) + 4 H_{2} O (a q)

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C r_{2} O_{7}^{2 -} (a q) + 3 S O_{3}^{2 -} (a q) + 8 H^{+} (a q) \to C r^{3 +} (a q) + 3 S O_{4}^{2 -} (a q) + 4 H_{2} O (a q)

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Solution

The correct option is C $C r_{2} O_{7}^{2 -} (a q) + 3 S O_{3}^{2 -} (a q) + 8 H^{+} (a q) \to 2 C r^{3 +} (a q) + 3 S O_{4}^{2 -} (a q) + 4 H_{2} O (a q)$

So , oxidizing agent : $C r_{2} O_{7}$
Reducing agent : $S O_{3}$

n-factor :
for $C r_{2} O_{7}^{2 -} : 6$ , for $S O_{3} : 2$
Equalising the decrease/increase in oxidation number :
$C r_{2} O_{7}^{2 -} + S O_{3}^{2 -} \to 2 C r^{+ 3} + S O_{4}^{2 -}$
Cross multiply the oxidising or reducing agent with simplified n-factor number,
$C r_{2} O_{7}^{2 -} + 3 S O_{3}^{2 -} \to 2 C r^{+ 3} + 3 S O_{4}^{2 -}$
Balance oxygen atoms.
$C r_{2} O_{7}^{2 -} + 3 S O_{3}^{2 -} \to 2 C r^{+ 3} + 3 S O_{4}^{2 -} + + 4 H_{2} O$
Balance Hydrogen atoms.
$C r_{2} O_{7}^{- 2} + 3 S O_{3}^{2 -} + 8 H^{+} \to 2 C r^{3 +} + 3 S O_{4}^{2 -} + + 4 H_{2} O$
balance charge
charge in reactant side = 0
charge in product side = 0
so the balanced equation is
$C r_{2} O_{7}^{2 -} + 3 S O_{3}^{- 2} + 8 H^{+} \to 2 C r^{3 +} + 3 S O_{4}^{2 -} + 4 H_{2} O$

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Similar questions

Q. Balance this equation.

C r_{2} O_{7}^{2 -} + S O_{3}^{2 -} \to S O_{4}^{2 -} + C r^{3 +}

Q. What will be the ratio

\frac{n_{C r_{2} O_{7}^{2 -}}}{n_{S O_{3}^{2 -}}}

when the following redox reaction

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

is balanced in an acidic medium. Where

n

represents the stoichiometric coefficient of the respective species.

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 reactions by ion-electron method:

(a) (aq) + I^– (aq) → MnO₂ (s) + I₂(s) (in basic medium)

(b) (aq) + SO₂ (g) → Mn²⁺ (aq) + (aq) (in acidic solution)

(d) + SO₂(g) → Cr³⁺ (aq) + (aq) (in acidic solution)

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

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Balance the following chemical equation: $C r_{2} O_{7}^{2 -} (a q) + S O_{3}^{2 -} (a q) \to C r^{3 +} (a q) + S O_{4}^{2 -} (a q)$ (Acidic medium)