Question

consider the following reaction, and identify the substance being reduced.
ncl₃(l) + 3 h₂o(l) → nh₃(aq) + 3 hocl(aq)
n in nh₃
n in ncl₃
cl in ncl₃
cl in hocl

Explanation:

Step1: Calculate oxidation state of \(N\) in \(NCl_3\)

Let oxidation state of \(N\) be \(x\). Oxidation state of \(Cl=- 1\).
Using formula \(x + 3\times(-1)=0\) (since \(NCl_3\) is neutral).
So, \(x = + 3\).

Step2: Calculate oxidation state of \(N\) in \(NH_3\)

Let oxidation state of \(N\) be \(y\). Oxidation state of \(H = + 1\).
Using formula \(y+3\times(+1)=0\) (since \(NH_3\) is neutral).
So, \(y=-3\).

Step3: Calculate oxidation state of \(Cl\) in \(NCl_3\)

Oxidation state of \(Cl=-1\) (as calculated in Step1).

Step4: Calculate oxidation state of \(Cl\) in \(HOCl\)

Let oxidation state of \(Cl\) be \(z\). Oxidation state of \(H = + 1\), oxidation state of \(O=-2\).
Using formula \((+1)+(-2)+z = 0\) (since \(HOCl\) is neutral).
So, \(z = + 1\).

Step5: Determine the reduced substance

Reduction is decrease in oxidation state.
Oxidation state of \(N\) changes from \(+3\) (\(NCl_3\)) to \(-3\) (\(NH_3\)).

Answer:

N in \(NCl_3\)