For the redox reaction-IO-3aq - Rds -H2O- RdO-4aq - I-aq -H-The correct coefficient of the IO-3- H2O and H- is -

Steps for Balancing redox reactions: Identify the oxidation nbsp;and nbsp;reduction half nbsp;. Find the oxidising and reducing agent. Find the n fact ...

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

Byju's Answer

Standard XII

Chemistry

Oxidation Number Method

For the redox...

Question

For the redox reaction:
$I O_{3}^{-} (a q) + R d (s) + H_{2} O \to R d O_{4}^{-} (a q) + I^{-} (a q) + H^{+}$
The correct coefficient of the $I O_{3}^{-}$ , $H_{2} O$ and $H^{+}$ is :

1, 5, 10

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

5, 1, 10

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

7, 3, 6

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

7, 6, 4

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

Open in App

Solution

The correct option is C 7, 3, 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 simplified n-factor values.

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.

$n_{f} = (| O . S ._{P r o d u c t} - O . S ._{R e a c t a n t} | \times number of atom$
$+ 5 I O_{3}^{-} (a q) + 0 R d (s) \to + 7 R d O_{4}^{-} (a q) + {- 1 I}^{-} (a q)$

$0 R d (s) \to + 7 R d O_{4}^{-} (a q) +$ oxidation
$n_{f} = (| 7 - 0 | \times 1 = 7$
$+ 5 I O_{3}^{-} (a q) \to {- 1 I}^{-} (a q)$ reduction
$n_{f} = (| - 1 - 5 | \times 1 = 6$
$R d$ is reducing agent
$I O_{3}^{-}$ is oxidising agent.

Balance atom undergoing oxidation and reduction.
Already balanced
$I O_{3}^{-} (a q) + R d (s) \to R d O_{4}^{-} (a q) + I^{-} (a q)$

Cross mutiply the oxidising or reducing agent with n-factor.
$7 I O_{3}^{-} (a q) + 6 R d (s) \to 6 R d O_{4}^{-} (a q) + 7 I^{-} (a q)$

Balance atoms oxygen.
$7 I O_{3}^{-} (a q) + 6 R d (s) + 3 H_{2} O \to 6 R d O_{4}^{-} (a q) + 7 I^{-} (a q)$

Balance hydrogen.
$7 I O_{3}^{-} (a q) + 6 R d (s) + 3 H_{2} O \to 6 R d O_{4}^{-} (a q) + 7 I^{-} (a q) + 6 H^{+}$

balance charge
charge in reactant side = -7
charge in product side = -7
so the balanced equation is $7 I O_{3}^{-} (a q) + 6 R d (s) + 3 H_{2} O \to 6 R d O_{4}^{-} (a q) + 7 I^{-} (a q) + 6 H^{+}$

Suggest Corrections

Similar questions

Q. Consider the redox reaction giben below :

I O_{3}^{-} (a q) + R d (s) + H_{2} O \to R d O_{4}^{-} (a q) + I^{-} (a q) + H^{+}

What will be the correct coefficient of the

I O_{3}^{-}

H_{2} O

and

H^{+}

respectively.

Q. Consider the redox reaction giben below :

I O_{3}^{-} (a q) + R d (s) + H_{2} O \to R d O_{4}^{-} (a q) + I^{-} (a q) + H^{+}

What will be the correct coefficient of the

I O_{3}^{-}

H_{2} O

and

H^{+}

respectively.

Q. The coefficient of

I^{-}, I O_{3}^{-}

and

H^{+}

in the redox reaction,

I^{-} + I O_{3}^{-} + H^{+} \to I_{2} + H_{2} O

in the balanced form respectively are:

Q. For the redox reaction

C r_{2} O_{7}^{2 -} + F e^{2 +} + H^{+} \to C r^{3 +} + F e^{3 +} + H_{2} O

The correct coefficients of the following for the balanced reaction are

Q. Consider the reaction,

H^{+} + I O_{4}^{-} + I^{-} \to I_{2} + H_{2} O

. The ratio of the coefficients of

I O {_{4}}^{-}, I^{-}

and

I_{2}

, respectively, are:

Join BYJU'S Learning Program

Explore more

Oxidation Number Method

Standard XII Chemistry

Join BYJU'S Learning Program

For the redox reaction: IO−3(aq)+Rd(s)+H2O→RdO−4(aq)+I−(aq)+H+ The correct coefficient of the IO−3, H2O and H+ is :

For the redox reaction:
$I O_{3}^{-} (a q) + R d (s) + H_{2} O \to R d O_{4}^{-} (a q) + I^{-} (a q) + H^{+}$
The correct coefficient of the $I O_{3}^{-}$ , $H_{2} O$ and $H^{+}$ is :