Redox Reactions Test - 61

$$K_2Cr_2O_7+ SO_2 + H_2SO_4 \rightarrow  K_2SO_4 + Cr_2 (SO4)3+ H_2O$$
Oxidation number of Cr in $$K_2Cr_2O_7$$ is +6 and in $$Cr_2 (SO_4)_3$$ is +3.
Oxidation number of S in $$SO_2$$ is +4 and in $$(SO_4)^{2-}$$ ion is +6.

Correct option: $$D$$

Explanation:

Step: 1 calculate the oxidation state of central atom$$ Mn$$ In $$MnO_4^{-}$$

Let us assume the $$oxidation$$ $$state$$ of $$Mn$$ be x.
Overall charge on the complex: $$−1$$

Thus, $$Mn + 4 \times O = - 1$$
          $$x+4(−2) = −1$$ ($$O$$ has $$-2$$ charge )
    So, $$x=+7$$
Hence, the $$oxidation$$ $$state$$ is $$+7$$.

Step: 2 calculate the oxidation state of central atom $$Cr$$ In $$\left [ Cr\left ( CN \right )_6 \right ]^{3-}$$

Let us assume the $$oxidation$$ $$state$$ of $$Cr$$ be x.
Overall charge on the complex: $$−3$$

Thus, $$Cr + 6 \times CN = -1 $$
         $$x+6(−1) = −3$$ (Where $$CN$$ has $$-1$$ overall charge )
So, $$x=+3$$
Hence, the $$oxidation$$ $$state$$ is $$+3$$.
Step: 3 calculate the oxidation state of central atom $$Cr$$ In $$Cr_2O_3$$

Let us assume the $$oxidation$$ $$state$$ of $$Cr$$ be x.
Overall charge on the complex: $$0$$

Thus, $$2 \times Cr + 3 \times O = 0$$
        $$2x+3(−2) = 0$$ ($$O$$ has $$-2$$ charge )
So, $$x=+3$$
Hence, the $$oxidation$$ $$state$$ is $$+3$$.
Step: 4 calculate the oxidation state of central atom $$Cr$$ In $$CrO_2Cl_2$$

Let us assume the $$oxidation$$ $$state$$ of $$Cr$$ be x.
Overall charge on the complex: $$0$$

Thus, $$1 \times Cr + 2 \times O + 2 \times Cl$$
         $$x+2(−2)-2 = 0$$   ($$O$$ has $$-2$$ charge while $$Cl$$ has $$-1$$ charge)
So,$$x=+6$$
Hence, the $$oxidation$$ $$state$$ is $$+6$$.

Hence, the species having oxidation state of $$+6$$  is $$CrO_2​Cl_2$$​.

 

 

Moles of acidic $$KMnO_4$$ needed for complete oxidation of $$1$$ mole equimolar mixture of $$FeSO_4$$ and $$FeC_2O_4$$ is $$\dfrac {2}{5}$$.

$$1$$ mole equimolar mixture of $$FeSO_4$$ and $$FeC_2O_4$$  will contain $$0.5$$ moles of $$FeSO_4$$ and $$0.5$$ moles of $$FeC_2O_4$$.

Thus, it contains $$1$$ mol of $$Fe^{2+}$$ ions which will be oxidized to $$Fe^{3+}$$ ions and $$0.5$$ moles of $$C_2O_4^{2-}$$ ions which will be oxidized to $$CO_2$$.
The reactions are as follows:
$$2MnO_4^-  + 5C_2O_4^{2-}+16H^+ \rightarrow 2Mn^{2+} + 10CO_2 +8H_2O$$
$$MnO_4^- + 5Fe^{2+}+8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O$$

Thus, $$2$$ moles $$KMnO_4 =$$ $$5$$ moles $$C_2O_4^{2-}$$
$$1$$ mole $$KMnO_4 =$$ $$5$$ moles $$Fe^{2+}$$

Hence, moles of acidic $$KMnO_4$$ needed for complete oxidation of $$1$$ mole equimolar mixture of $$FeSO_4$$ and $$FeC_2O_4$$ is $$\dfrac {0.5}{5}+\dfrac {0.5}{5}+\dfrac {1}{5}=\dfrac {2}{5}$$.

$$CrO_4^{2-} \rightarrow Cr_2O_7^{2-}$$. In both the anions, oxidation number of chromium is $$+6$$.
Hence, no change has occurred.
 Acidic medium, due to $$CO_2$$, converts $$CrO_4^{2-}$$ into $$Cr_2O_7^{2-}$$.
$$CO_2 + H_2O \rightleftharpoons H_2CO_3  \rightleftharpoons 2H^+ + CO_3^{2-}$$ &  $$2CrO_4^{2-} + 2H^+ \rightarrow Cr_2O_7^{2-} + H_2O$$

Let $$x$$ be the oxidation number of sulphur in $$Na_2S_2O_3$$.
For a neutral compound, sum of oxidation states of all elements must be zero.
$$2+2x+3(-2)=0$$
$$2x=6-2$$
$$2x=4$$
$$x=+2$$
Suppose $$y$$ be the oxidation number of sulphur in $$Na_2S_4O_6$$.
$$2+4y+3(-2)=0$$
$$4y=12-2$$
$$4y=10$$
$$y=+2.5$$
The average oxidation states of sulphur in $$Na_2S_2O_3$$ and $$Na_2S_4O_6$$ are respectively $$+2$$ and $$+2.5$$.

Change in oxidation of Mn as the reactions are given below:
$$\overset{+7}{Mn}O_{4}^{-} + e \rightarrow  \overset{+6}{Mn}O_{4}^{2-}$$ ; Change in oxidation is 1
$$ \overset{+7}{Mn}O_{4}^{-}+ 3e \rightarrow \overset{+4}{Mn}O_{2}$$ ; Change in oxidation is 3
$$\overset{+7}{Mn}O_{4}^{-} + 4e \rightarrow  \overset{+3}{Mn}_2O_{3}$$ ; Change in oxidation is 4
$$\overset{+7}{Mn}O_{4}^{-}+ 5e \rightarrow \overset{+2}{Mn}$$ ; Change in oxidation is 5
Hence, the change is 1, 3, 4, 5 respectively.

$$(NH_4)_2Cr_2O_7 \overset{\Delta }{\rightarrow} N_3 +Cr_2O_3 +4H_2O$$. 

Seven substances have underlined atoms of different nature. These include
$$Na_2\underline{S_4}O_6,Na_2\underline{S_2}O_3,CaO\underline{Cl_2},\underline{N_2}H_4O_3,\underline{Fe_3}O_4,K\underline{I_3},Cr\underline{O_5}$$. The underlined atoms in these substances have different oxidation numbers.

In Sodium tetrathionate, sulphur is in $$+5, 0, 0, +5$$ oxidation state.

For the following examples, anhydride of an acid is formed by dehydration and oxidation number of central atom remains unchanged.
(A) $$H_2N_2O_2 \rightarrow N_2O+H_2O$$;  The oxidation number of N is $$+1$$.
(B) $$2HNO_2 \rightarrow 2N_2O_3+H_2O$$; The oxidation number of N is $$+3$$.
(C) $$2HNO_3 \rightarrow 2N_2O_5+H_2O$$; The oxidation number of N is $$+5$$.