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Redox Reactions Test - 63

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Redox Reactions Test - 63
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  • Question 1
    1 / -0
    The oxidation number of all the elements in the given compound is:    Ethane ($${ C }_{ 2 }{ H }_{ 6 }$$)
  • Question 2
    1 / -0
    Of the following acids, the one that is strongest is :
    Solution
    As oxidation state of central atom increases, acidity increases.
    So, strongest acid is that which has maximum value of oxidation number (O.N) of the central atom.
    $$HBr{ O }_{ 4 }$$ ; O.N of $$Br=+7$$ (strongest acid)
    $$HOCl$$ ; O.N of $$Cl=+1$$
    $$H{NO}_{2}$$ ; O.N of $$N=+3$$
    $${H}_{3}{PO}_{3}$$ ; O.N of $$P=+3$$
  • Question 3
    1 / -0
    For a titration of $$100\: {cm}^{3}$$ of $$0.1$$ $$M$$ $${ Sn }^{ 2+ }$$ to $${ Sn }^{ 4+ }$$, $$50$$ $${cm}^{3}$$ of $$0.40$$ $$M$$ $${ Ce }^{ 4+ }$$ solution was required. The oxidation state of cerium in the reduction product is :
    Solution
    According to given conditions, we have
    $${ Sn }^{ 2+ }\longrightarrow { Sn }^{ 4+ }+2{ e }^{ - }; increase\ in\ O.N=2$$
    $${ Ce }^{ 4+ }+n{ e }^{ - }\longrightarrow { Ce }^{ (4-n) };\ decrease\ in\ O.N=n$$
    On balancing, we get
    $$n{ Sn }^{ 2+ }+2{ Ce }^{ 4+ }\longrightarrow n{ Sn }^{ 4+ }+2{ Ce }^{ (4-n) }$$
    Given,
    Millimoles of $${ Sn }^{ 2+ }=100\times 0.1=10$$
    Millimoles of $${ Ce }^{ 4+ }=50\times 0.4=20$$
    At equivalent point, number of equivalents of both reactant and product are same.
    Therefore, $$\cfrac { n }{ 2 } =\cfrac { 10 }{ 20 } $$
    or, $$n=1$$
    Thus, oxidation state of cerium in the reduced product $$({ Ce }^{ (4-1) }={ Ce }^{ 3+ })=+3$$.
  • Question 4
    1 / -0
    Oxidation number of $$Cr$$ in the following compounds are $$x,y,z$$ and $$w$$ respectively.
    $$K\left[ \underset { x }{ Cr } O({ O }_{ 2 })(OH) \right] ,\quad { K }_{ 3 }\left[ \underset { y }{ Cr } { \left( { O }_{ 2 } \right)  }_{ 4 } \right] ,\quad { \left( { NH }_{ 3 } \right)  }_{ 3 }\left[ \underset { z }{ Cr } { \left( { O }_{ 2 } \right)  }_{ 2 } \right] ,\quad \underset { w }{ Cr } { O }_{ 2 }{ Cl }_{ 2 }$$
    These values are, respectively :
    Solution
    $$\underset { \uparrow \\ +1 }{ K } \underset { \uparrow \\ +x }{ [(Cr } \underset { \uparrow \\ -2 }{ O) } \underset { \uparrow \\ -2 }{ \left( { O }_{ 2 } \right)  } \underset { \uparrow \\ -1=0 }{ OH] } $$
    $$x=4$$
    $$\underset { \uparrow \\ 3 }{ { K }_{ 3 } } \underset { \uparrow \\ +y }{ [Cr } \underset { \uparrow \\ -8=0 }{ { \left( O \right)  }_{ 4 } } ]$$
    $$y=5$$
    $$\underset { \uparrow \\ 0 }{ { \left( { NH }_{ 3 } \right)  }_{ 3 } } \left[ \underset { \uparrow \\ +z }{ Cr } \underset { \uparrow \\ -4=0 }{ { \left( { O }_{ 2 } \right)  }_{ 2 } }  \right] $$
    $$z=4$$
    $$\underset { \uparrow \\ w-4-2=0 }{ Cr } { O }_{ 2 }{ Cl }_{ 2 }$$
    $$w=6$$
    Thus, the values of $$x,y,z$$ and $$w$$ are $$4,\:5,\:4$$ and $$6$$ respectively.
  • Question 5
    1 / -0
    Oxidation number of $$Cr$$ is $$+5$$ in :
    Solution
    $$\begin{array}{l}\text { There are } 4 \text { Peroxy bonds } \\\qquad \begin{aligned}3+x &+8(-1)=0 \\\Rightarrow x &=+5\end{aligned}\end{array}$$


    $$\begin{array}{l}\left(\mathrm{NH}_{3}\right)_{3}\mathrm{CrO}_{4},\mathrm{~K}_{2}\mathrm{Cr}_{}\mathrm{O}_{4},\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{5}\mathrm{(H}_{2}\mathrm{O)}\right]\mathrm{Cl}_{3} \text { contain } \\\text { Chromium oxidation states of }+4,+6,+3 \\\text { respectively }\end{array}$$

  • Question 6
    1 / -0
    Consider the following unbalanced reactions:
    $$I: Zn+dil.\:{ H }_{ 2 }S{ O }_{ 4 }\longrightarrow ZnS{ O }_{ 4 }+{ H }_{ 2 }$$
    $$II: Zn+conc.\:{ H }_{ 2 }S{ O }_{ 4 }\longrightarrow ZnS{ O }_{ 4 }+S{ O }_{ 2 }+{ H }_{ 2 }O\quad $$
    $$III: Zn+HN{ O }_{ 3 }\longrightarrow Zn{ \left( { NO }_{ 3 } \right)  }_{ 2 }+{ NH }_{ 4 }{ NO }_{ 3 }+{ H }_{ 2 }O$$
    Oxidation number of hydrogen changes in :
    Solution
    Change in oxidation state of H is shown below:
    $$I: Zn+\underset { +1 }{ 2{ H }^{ + } } \longrightarrow \underset { 0 }{ { H }_{ 2 } } $$ oxidation state changes (reduction)
    $$II: Zn+\underset { +1 }{ { H }^{ + } } +S{ O }_{ 4 }^{ 2- }\longrightarrow { Zn }^{ 2+ }+S{ O }_{ 2 }+\underset { +1 }{ { H }_{ 2 } } O$$
    no change in oxidation state of H
    $$III: Zn+\underset { +1 }{ { H }^{ + } } +N{ O }_{ 3 }^{ - }\longrightarrow { Zn }^{ 2+ }+\underset { \uparrow \\ +1 }{ { NH }_{ 4 } } \underset { \uparrow \\ +1 }{ { H }_{ 2 } } O$$ no change
    Therefore, from the above reactions, we can see oxidation number of $$H$$ changes in $$I$$.
  • Question 7
    1 / -0
    Out of $$\underset { (I) }{ { H }_{ 2 }S{ O }_{ 4 } } ,\ \underset { (II) }{ { H }_{ 2 }{ S }_{ 2 }{ O }_{ 7 } } ,\ \underset { (III) }{ { H }_{ 2 }S{ O }_{ 5 } }$$ and $$\underset { (IV) }{ { H }_{ 2 }{ S }_{ 2 }{ O }_{ 3 } } $$ :
    Solution
    In the structures I, II and III shown below, S atom is attached to two oxygen atoms vial double bonds and to other two oxygen atoms via single bonds. Hence, the oxidation number of S in these compounds is +6. In the compound IV, the oxidation number of two S atoms is +4 and -2 respectively. Thus, the statement A is correct.

    Peroxy linkage is present in III. Thus, options B and C are incorrect.

    The oxidation number of S in I is +6 and in IV is +4 and -2. Thus, the option D is incorrect.

  • Question 8
    1 / -0
    In the presence of $$SO_3^{\displaystyle2-}$$ and $$S^{\displaystyle2-}$$ in a mixture, on adding $$dil.\:H_2SO_4$$, one notices :
    Solution
    The reaction is given below:
    $$SO_3^{-2}+S^{-2} \xrightarrow {H^+} SO_2+H_2S \rightarrow H_2O+3S$$
    $$SO_2$$ and $$H_2S$$ formed during change undergoes a redox change forming colloidal sulphur and thus, no smell is observed.
  • Question 9
    1 / -0
    $$Mn{O}_{4}^{2-}$$ (1 mole) in neutral aqueous medium disproportionate to :
    Solution
    Option (A) is correct.
    $$Mn{O}_{4}^{2-}$$ (1 mole) in neutral aqueous medium disproportionate to $$2/3$$ mole $${Mn}{O}_{4}^{-}$$ and $$1/3$$ mole $$Mn{O}_{2}$$.
    $$3K_2MnO_4 + 2H_2O \longrightarrow 2KMnO_4 + MnO_2 +4KOH$$
  • Question 10
    1 / -0
    Compound(s) of $$Cu$$ and $$Ag$$ showing $$+3$$ oxidation state is\are :
    Solution
    Compound of $$Cu$$ and $$Ag$$ showing $$+3$$ oxidation state are $$K_3 [CuF_6]$$ and $$Na[AgF_4]$$.
    Let $$x$$ be the oxidation number of $$Cu$$ in $$K_3 [CuF_6]$$.
    Since the overall charge on the compound is $$0$$, the sum of oxidation states of all elements in it should be equal to $$0$$.
    Therefore, $$3(+1)+x+6(-1)=0$$
    or, $$x=-3$$
    Let $$y$$ be the oxidation number of Ag in $$Na[AgF_4]$$.
    Since the overall charge on the compound is $$0$$, the sum of oxidation states of all elements in it should be equal to $$0$$.
    Therefore, $$+1+y+4(-1)=0$$
    or, $$y=+3$$
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