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Types of Redox Reactions

Given : XN a ...

Question

Given : $X N a_{2} H A s O_{3} + Y N a B r O_{3} + Z H C l \to N a B r + H_{3} A s O_{4} + N a C l$
The values of X, Y and Z in the above redox reaction are respectively:

2, 1, 3

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3, 1, 6

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2, 1, 2

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3, 1, 4

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Solution

The correct option is B 3, 1, 6
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:

The equation before balancing be:
$N a_{2} H A s O_{3} + N a B r O_{3} + H C l \to N a B r + H_{3} A s O_{4} + N a C l$

Oxidation state As in $N a_{2} H A s O_{3} = + 3$
Oxidation state As in $H_{3} A s O_{4} = + 5$
Oxidation state Br in $N a B r O_{3} = + 5$
Oxidation state As in $N a B r = - 1$

Cleary, $N a_{2} H A s O_{3}$ is undergoing oxidation and $N a B r O_{3}$ is undergoing reduction.

using the formula of n-factor given above,
$n_{f}$ of $N a_{2} H A s O_{3} = 2$
$n_{f}$ of $N a B r O_{3} = 6$

here 2 is common in both n-factors.

$\Rightarrow$ Ratio of n-factors=1:3

Cross multiplying these with ratio of $n_{f}$ 's of each other.
we get,
$3 N a_{2} H A s O_{3} + N a B r O_{3} + H C l \to N a B r + H_{3} A s O_{4} + N a C l$

Balancing $A s$ ,
$3 N a_{2} H A s O_{3} + N a B r O_{3} + H C l \to N a B r + 3 H_{3} A s O_{4} + N a C l$

As of now oxygen already got balanced,

Balancing $N a$ ,
$3 N a_{2} H A s O_{3} + N a B r O_{3} + 6 H C l \to N a B r + 3 H_{3} A s O_{4} + 6 N a C l$

After this when we observe every element is balanced. so no need to follow further steps.

$3 N a_{2} H A s O_{3} + N a B r O_{3} + 6 H C l \to N a B r + 3 H_{3} A s O_{4} + 6 N a C l$

This is the final balanced equation.
We can also see that charge on both sides is 0. which also indicates that the equation is balanced now.

Comparing to the equation given in question, we get,
The value of X, Y and Z are 3, 1 and 6 respectively.
Hence option (2) is the answer.

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