Question

The reaction between \[C{{r}_{2}}O_{7}^{2-}\] and \[HN{{O}_{2}}\] in an acidic medium is:
\[C{{r}_{2}}O_{7}^{2-}+5{{H}^{+}}+3HN{{O}_{2}}\xrightarrow{{}}2C{{r}^{3+}}+3NO_{3}^{-}+4{{H}_{2}}O.\]
The rate of disappearance of \[C{{r}_{2}}O_{7}^{2-}\]is found to be \[2.4\times {{10}^{-4}}\text{ }mol/Ls~\]during a measured time interval. Find the rate of disappearance of \[HN{{O}_{2}}\] and the rate of appearance of \[C{{r}^{3+}}\] during this time interval.

Answer 1

Hint: We know that In peroxide compounds, oxidation number of oxygen atoms is taken as\[-1\]. We can calculate the overall charge of a compound by just adding respective oxidation numbers of all atoms present in the compound. From this expression, we can find the oxidation number of \[Cr.\]

Complete step-by-step answer:
In case that compound contains $-O-O-$ (peroxide) linkage, then the numbers of oxygen involved in peroxide linkage have an oxidation number equal to \[-1\text{ }.\] When oxygen atom is present in superoxide from, those atoms have an oxidation number of \[-0.5\text{ }.\] Overall charge on compound= Oxidation state of \[Cr\text{ }+\text{ }4\](Oxidation state of oxygen in peroxide form) \[+\]Oxidation state of other oxygen atom also given that $-\dfrac{\Delta \left[ C{{r}_{2}}O_{7}^{2-} \right]}{\Delta t}=2.4\times {{10}^{-4}}mol{{L}^{-1}}{{s}^{-1}}$

You need to remember that contains two $-O-O-$linkages. So, in short has four oxygen atoms that are in peroxide linkage. Now that is a rule that if oxygen atom is involved in a peroxide linkage, then both the oxygen have oxidation number \[+1\] which is different from the normal oxidation state of oxygen atom.

The equality in this case is: $-\dfrac{\Delta \left[ C{{r}_{2}}O_{7}^{2-} \right]}{\Delta t}=-\dfrac{1}{3}\dfrac{\Delta \left[ HN{{O}_{2}} \right]}{\Delta t}$ and $-\dfrac{\Delta \left[ C{{r}_{2}}O_{7}^{2-} \right]}{\Delta t}=-\dfrac{1}{2}\dfrac{\Delta \left[ C{{r}^{3+}} \right]}{\Delta t}$
On equating first equation we get $\dfrac{\Delta \left[ HN{{O}_{2}} \right]}{\Delta t}$ value as $\dfrac{\Delta \left[ HN{{O}_{2}} \right]}{\Delta t}=3\times \left[ \dfrac{\Delta \left[ C{{r}_{2}}O_{7}^{2-} \right]}{\Delta t} \right]$
Substituting the values in above equation $=3\times 2.4\times {{10}^{-4}}=7.2\times {{10}^{-4}}mol{{L}^{-1}}{{s}^{-1}}$
Similarly, we can calculate $\dfrac{\Delta \left[ C{{r}^{3+}} \right]}{\Delta t}$ and we it is given as: $\dfrac{\Delta \left[ C{{r}^{3+}} \right]}{\Delta t}=2\times \left[ \dfrac{\Delta \left[ C{{r}_{2}}O_{7}^{2-} \right]}{\Delta t} \right]$
Substituting the values in above equation $=2\times 2.4\times {{10}^{-4}}=4.8\times {{10}^{-4}}mol{{L}^{-1}}{{s}^{-1}}$

Note: Remember that do not consider the oxidation number of oxygen atoms in peroxide linkage equal to \[\left( -2 \right)\] as this mistake often occurs. In case there is overall charge on the compound, do not forget to include it in the calculation of oxidation number.