What will be the ratio nCr2O2-7nSO2-3 when the following redox reaction Cr2O2-7 aq - SO2-3 aq -Cr3- aq - SO2-4 aq is balanced in an acidic medium- Where n represents the stoichiometric coefficient of the respective species-

The redox reaction is given below : +6Cr_2O_7^2 +3 +4SO_3^2 →+3Cr^3+ + +6SO^2 _4 So , oxidizing agent : Cr_2O_7^2 Reducing agent : SO_3^2 n factor : for Cr ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XIII

Chemistry

Oxidation Number Method

What will be ...

Question

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.

2 : 1

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

1 : 3

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

1 : 2

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

2 : 3

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

Open in App

Solution

The correct option is B $1 : 3$
The redox reaction is given below :
${+ 6 C r}_{2} O_{7}^{2 -} + 3 + 4 S O_{3}^{2 -} \to + 3 C r^{3 +} + + 6 S O_{4}^{2 -}$

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

n-factor :
for $C r_{2} O_{7}^{2 -} : 6$ ,
for $S O_{3}^{2 -} : 2$
$n_{f}$ ratio is $1 : 3$
Cross multiply the oxidising and reducing agent with $n_{f}$ ratio,

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

Balancing all atoms except $O$ and $H$ :

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

Balance oxygen atoms by adding $H_{2} O$

$C r_{2} O_{7}^{2 -} + 3 S O_{3}^{2 -} \to 2 C r^{3 +} + 3 S O_{4}^{2 -} + 4 H_{2} O$
Balance hydrogen atoms by adding $H^{+}$

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

Balance charge
charge in reactant side = 0
charge in product side = 0

So the balanced equation is :

$C r_{2} O_{7}^{2 -} + 3 S O_{3}^{2 -} + 8 H^{+} \to 2 C r^{3 +} + 3 S O_{4}^{2 -} + 4 H_{2} O$
So the ratio $\frac{n_{C r_{2} O_{7}^{2 -}}}{n_{S O_{3}^{2 -}}} = \frac{1}{3}$

Suggest Corrections

Similar questions

Q. 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)

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.

Q. Balance the following redox reactions by ion electron method.

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

(in acidic solution)

Q. What would be the value of stoichiometric coefficients of

M g

and

V O_{4}^{3 -}

respectively, in the balanced net ionic equation for the given redox reaction in acidic medium:

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

Q. Balance the following redox reaction.

(C r_{2} O_{7})_{(a q)}^{2 -} + F e_{(a q)}^{2 +} \to C r_{(a q)}^{3 +} + F e_{(a q)}^{3 +}

(acidic medium)

Join BYJU'S Learning Program

Explore more

Oxidation Number Method

Standard XIII Chemistry

Join BYJU'S Learning Program

What will be the ratio nCr2O2−7nSO2−3 when the following redox reaction Cr2O2−7 (aq)+SO2−3 (aq)→Cr3+ (aq)+SO2−4 (aq) is balanced in an acidic medium. Where n represents the stoichiometric coefficient of the respective species.