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Question: In the equation \(NO_2^\Theta + {H_2}O \to NO_3^\Theta + 2{H^ \oplus } + n{e^ - }\) n stands for...

In the equation NO2Θ+H2O→NO3Θ+2H⊕+ne−NO_2^\Theta + {H_2}O \to NO_3^\Theta + 2{H^ \oplus } + n{e^ - } n stands for

A

1

B

2

C

3

D

4

Answer

2

Explanation

Solution

To find the value of 'n', we need to balance the equation in terms of both atoms and charge.

1. Balancing Atoms:

The atoms (N, O, H) are already balanced on both sides.

2. Balancing Charge:

  • Reactant Side Total Charge:

    • Charge of NO2ΘNO_2^\Theta: -1
    • Charge of H2OH_2O: 0
    • Total Reactant Charge = -1 + 0 = -1

  • Product Side Total Charge:

    • Charge of NO3ΘNO_3^\Theta: -1
    • Charge of 2H⊕2H^\oplus: 2 * (+1) = +2
    • Charge of ne−n e^-: n * (-1) = -n
    • Total Product Charge = -1 + 2 - n = 1 - n

For the equation to be balanced, the total charge on the reactant side must be equal to the total charge on the product side.

So, we set up the equation:

Total Reactant Charge = Total Product Charge

−1=1−n-1 = 1 - n

Now, solve for n:

n=1−(−1)n = 1 - (-1)

n=1+1n = 1 + 1

n=2n = 2

Alternative Method: Oxidation State Change

We can also determine 'n' by calculating the change in the oxidation state of Nitrogen.

  • Oxidation state of N in NO2ΘNO_2^\Theta: Let the oxidation state of N be x. x+2(−2)=−1x + 2(-2) = -1 x−4=−1x - 4 = -1 x=+3x = +3

  • Oxidation state of N in NO3ΘNO_3^\Theta: Let the oxidation state of N be y. y+3(−2)=−1y + 3(-2) = -1 y−6=−1y - 6 = -1 y=+5y = +5

The oxidation state of Nitrogen changes from +3 to +5. This is an increase of 2, meaning Nitrogen loses 2 electrons.

Therefore, n = 2.