Question

the following reaction represents the action of bleaching powder in water. which of the following is the oxidizing agent of this reaction?
$ca(ocl)_2 + h_2o \to ca(oh)_2 + cl_2$
a $ca(ocl)_2$
b $ca(oh)_2$
c $cl_2$
d $h_2o$

Explanation:

Step1: Determine oxidation states

In \(Ca(OCl)_2\), let oxidation state of \(Cl\) be \(x\). For \(Ca(OCl)_2\), \(Ca\) has \(+2\) oxidation state and \(O\) has \(-2\). Using formula \(+2 + 2\times(-2 + x)=0\), we get \(x = +1\). In \(Cl_2\), oxidation state of \(Cl\) is \(0\).

Step2: Analyze oxidation - reduction

\(Cl\) in \(Ca(OCl)_2\) goes from \(+1\) to \(0\) (reduction). Oxidizing agent is the substance that gets reduced.

Answer:

A. \(Ca(OCl)_2\)