MyQuestionIcon

1

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

Oxidation Number Method

Balance the f...

Question

Balance the following chemical equation in basic medium: $M n O_{4}^{-} (a q) + B r^{-} (a q) \to M n O_{2} (s) + B r O_{3}^{-} (a q)$

A

2 M n O_{4}^{-} + B r^{-} + H_{2} O \to 2 M n O_{2} + B r O_{3}^{-} + 2 O H^{-}

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

B

M n O_{4}^{-} + B r^{-} + H_{2} O \to 2 M n O_{2} + B r O_{3}^{-} + 2 O H^{-}

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

C

2 M n O_{4}^{-} + B r^{-} + H_{2} O \to M n O_{2} + B r O_{3}^{-} + 2 O H^{-}

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

D

M n O_{4}^{-} + 2 B r^{-} + H_{2} O \to M n O_{2} + 2 B r O_{3}^{-} + 2 O H^{-}

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

Open in App

Solution

The correct option is A $2 M n O_{4}^{-} + B r^{-} + H_{2} O \to 2 M n O_{2} + B r O_{3}^{-} + 2 O H^{-}$
Steps for Balancing redox reactions:

Identify the oxidation and reduction halves.

Find the oxidising and reducing agent.

Find the n-factor of oxidising and reducing agents.

Cross multiply the oxidising and reducing agent with the n-factor of each other.

Balance the atoms other than oxygen and hydrogen.

Balance oxygen atoms.

Balance hydrogen atoms.

For basic medium:
If x oxygens are less on one side than the other side add x $H_{2} O$ units to that side. As soon as we add x $H_{2} O$ units for balancing oxygen, we add generally 2x (or whatever suitable) $H^{+}$ ions on the opposite side to balance hydrogen.Then after that we add equal (equal to number of $H^{+}$ added) number of $O H^{-}$ ions to the both sides and combine $H^{+}$ and $O H^{-}$ ions to form $H_{2} O$ . Then we cancel $H_{2} O$ which is common to both sides and which can be eliminated thus finally $O H^{-}$ is left on one side and the equation is balanced. Then we can verify whether the equation is balanced or not by checking the balance of charge on both sides.
formula used for the n-factor calculation,
$n_{f} = (| O . S ._{P r o d u c t} - O . S ._{R e a c t a n t} | \times number of atoms$

Taking the given equation and following the above mentioned steps:
$M n O_{4}^{-} + B r^{-} \to M n O_{2} + B r O_{3}^{-}$

$M n O_{4}^{-} + B r^{-} \to M n O_{2} + B r O_{3}^{-}$
Oxidation state of Mn in $M n O_{4}^{-} + 7$
Oxidation state of Br in $B r O_{3}^{-} = + 5$
Oxidation state of Mn in $M n O_{2} = + 4$
Oxidation state of Br in $B r^{-} = - 1$

Cleary, $B r$ is undergoing oxidation and $M n O_{4}^{-}$ is undergoing reduction.
using the formula of n-factor given above,
$n_{f}$ of $M n O_{4}^{-} = 3$
$n_{f}$ of $B r^{-} = 6$
ratio of n factors =1:2
Cross multiplying these with the ratio of $n_{f}$ 's of each other.
we get,
$2 M n O_{4}^{-} + B r^{-} \to M n O_{2} + B r O_{3}^{-}$

Balancing the elements other than oxygen and hydrogen on both sides,
$2 M n O_{4}^{-} + B r^{-} \to 2 M n O_{2} + B r O_{3}^{-}$

Adding the $H_{2} O$ to balance the oxygen,
LHS of the equation contains 8 oxygens while RHS contains only 7 oxygens, so adding 1 $H_{2} O$ to RHS.
$2 M n O_{4}^{-} + B r^{-} \to 2 M n O_{2} + B r O_{3}^{-} + 1 H_{2} O$
adding $H^{+}$ to balance hydrogen,
As LHS of the equation contains 0 hydrogens while RHS contains 2 hydrogens, so adding 2 $H^{+}$ to LHS.
$2 M n O_{4}^{-} + B r^{-} + 2 H^{+} \to 2 M n O_{2} + B r O_{3}^{-} + 1 H_{2} O$
Now adding $O H^{-}$ to both sides to combine with $H^{+}$ and make it $H_{2} O$ ,
$2 M n O_{4}^{-} + B r^{-} + 2 H^{+} + 2 O H^{-} \to 2 M n O_{2} + B r O_{3}^{-} + 1 H_{2} O + 2 O H^{-}$

$2 M n O_{4}^{-} + B r^{-} + 2 H_{2} O \to 2 M n O_{2} + B r O_{3}^{-} + 1 H_{2} O + 2 O H^{-}$
Eliminating the unrequired common $H_{2} O$ which is present on the both sides,
$2 M n O_{4}^{-} + B r^{-} + H_{2} O \to 2 M n O_{2} + B r O_{3}^{-} + 2 O H^{-}$
This is the final balanced equation.
We can also see that charge on both sides is -3. which also indicates that the equation is balanced now.

Suggest Corrections

thumbs-up

130

Similar questions

Q. The equation

M n O_{4}^{-} (a q) + B r^{-} (a q) \to M n O_{2} (s) + B r O_{3}^{-} (a q)

when balanced in a basic medium by oxidation number method will be :

Q. In the redox ionic equation, the equivalent weight of potassium permaganate of molar mass

M

g m o l^{- 1}

2 M n O_{4}^{-} + B r^{-} + H_{2} O \to 2 M n O_{2} + B r O_{3}^{-} + 2 O H^{-}

Q. How to balance the following equation by oxidation number method.

B r^{-} + M n O_{4}^{-} \to B r O_{3}^{-} + M n O_{2}

Q. Write balanced net ionic equation for the following reaction in basic solution:

M n (O H)_{2} (s) + M n O_{4}^{-} (a q) \to M n O_{2} (s)

. . . B r O_{3}^{-} (a q) + . . . B r^{-} (a q) + . . . H^{+} (a q) \to . . . B r_{2} (l) + . . . H_{2} O

After balancing the above equation. What is the ratio of coefficient

B r O_{3}^{-}

B r^{-}

View More

Join BYJU'S Learning Program

explore_more_icon

Explore more

Oxidation Number Method

Standard XII Chemistry

Join BYJU'S Learning Program

CrossIcon