Give reasons for the following: (1) `HNO_3` acts only as an oxidising agent while `HNO_2` can act both as a reducing agent and an oxidising agent
(2) Chlorine liberates iodine from Kl solution.

Explanation

Step 1: The given equation is
`Cu+HNO_(3) to Cu(NO_(3))_(2)+NO+H_(2)O`
Step 2: Writing the oxidation number of atoms we have
`Cu+overset(+1)(H)overset(+5)(N) overset(-2)(O_(3)) to Cu(overset(+5)(N)overset(-2)(O_(3))_(2)+overset(+2) overset(-2)(O)+overset(+1)(H_(2)) overset(-2)(O)`
Step 3: Identify the redox couples in the reaction and split the given reaction into two half-reactions.
`O : Cu to Cu(overset(+5)(N) overset(-2)O_(3))_2+2e^(-)`
`R: overset(+1)(H) overset(+5)(N) overset(-2)(O_(3))+3e^(-) to overset(+2)(N) overset(-2)O`
(O=oxidation half reaction, R=reduction half reaction).
Step 5: Balancing half reaction, Balance all other atoms except hydrogen and oxygen.
`O : overset(0)(Cu)+2HNO_(3) to overset(+2)(Cu) (overset(+5)(N) overset(-2)(O_(3)))_(2)+2e^(-)`
`R : overset(+1)(H) overset(+5)(N) overset(-2)(O_(3)) +3e^(-) to overset(+2)(N) overset(-2)(O)`
(b) Balance the charge : For reactions in an acidic solution, balance the charge so that both sides have the same total charge by adding a `H^+` ion to the side deficient in positive charge.
`O: overset(0)(Cu)+2HNO_(3) to overset(+2)(Cu) (overset(+5)(N) overset(-2)(O_(3)))_(2)+2e^(-)+2H^(+)`
`R : overset(+1) overset)(+5)(N) overset(-2)(O_(3))+3e^(-)+3H^(-) to overset(+2)(N)overset(-2)(O)`
(c ) Balance the oxygen atomms :
`O : overset(0)(Cu)+2HNO_(3) to overset(+2)(Cu) (overset(+5)(N) overset(-2)(NO_(3)))_(2)+2e^(-)+2H^(+)`
`R: overset(+1) overset(+5) overset(-2)O_(3)+3e^(-)+3H^(+) to overset(+2)(N) overset(-2)O +2H_(2)O`
Step 6 : Make electron gain equivalent to electron lost. The electrons lost in the oxidation half-reaction must be equal to the electrons gained in the reduction half- reaction. To make the two equals, multiply the coefficients of all species by integers producing the lowest common multiple between the half-reactions.
`O : overset(0)(Cu) +2HNO_(3) to overset(+2) Cu (overset(+5)(N) overset(-2)(O_(3))_(2)+2e^(-)+2H^(+) |xx3`
`R : overset(+1)(H) overset(+5)(N) overset(-2)(O_(3))+3e^(-)+3H^(+) to overset(+2)(N)overset(-2)(O)+2H_(2)O`
Step 7 : Add the half -reactions together.
`overset(+1)(K_(2)) overset(+5)(Cr_(2)) overset(-2)(O_(7)) +14HCl +6e^(-) +6H^(+) to 3 overset(0)(Cl_(2)) +2 overset(+3)(Cr) overset(-1)(Cl_(3))+6e^(-)+2KCl+6H^(+) +7H_(2)O`
Simplify the equation by crossing out common parameters on opposite sides of the arrow.
`overset(+1)(K_(2)) overset(+5)(Cr_(2)) overset(-2)O_(7) +14HCl Leftrightarrow 3 overset(0)(Cl_(2))+2 overset(+3)(Cr) overset(-1)(Cl_(3))+2KCl+7H_(2)O`
Step 8: At last see the equation is balanced and verify that the equation contains the same type and number of atoms on both sides of the equation. Also check for sum of charges both the sides of equation are equal. Thus, the balanced equation would be:
`K_(2)Cr_(2)O_(7) +14HCl to 2KCl+2CrCl_(3)+7H_(2)O+3Cl_2`
(ii) 1 Concider `HNO_3` Oxidation number of H=+1, O=-2, N =-x, then substituting oxidation number values in neutral molecules `HNO_(3)` we get
(+1) +(x)+(-2) `xx 3=0`
x=+6-1
x=+5
The oxidation number of N in `HNO_(3)` is +5.
Maximum oxidation number of Nitrogen is +5 as it has 5 valence electrons.
Minimum oxidation number of Nitrogen is -3 as it can accept 3 more of Nitrogen is -3 as it can accept 3 more electrons to attain noble gas configuration.
As oxidation number of N in `HNO_(3)` is +5 which is maximum, can act as only oxidising agent.
While in `HNO_(2)`, oxidation number of H=+1, O=-2, N=x then substituting oxidation no. values in neutral molecules `HNO_2` we get
`(+1) +(x)+(-2) xx 2=0`
x=+4-1
x=+3
The oxidation number of N in `HNO_(2)` is +3.
As oxidation number of N in `HNO_2` is +3, it can either increase by losing or decrease by accepting electrons.
Thus `HNO_(2)` can act as both oxidising as well as reducing agent.
(2) The oxidising power of halogens on going down the group (17) decreases.
Fluorine having higest oxidising power and iodine having the least.
Thus, chlorine placed above iodine can replace `I^(-)` ions in solution.
Chlorine undergoes metal displacement reaction and displaces iodine from KI solution.
`C_(2)(g)+2KI^(-) (aq) to 2KCl (aq) +overset(0)(I_(2)(s)`