Question

 For the following problems write the balanced chemical equation, indicate which element is oxidized and which-is reduced, and indicate the oxidation numbers for all species involved.

 (a) N_2(g) + 3 H_2(g) → 2 NH_3(g)

 (b) Iron(II) Nitrate reacts with solid Aluminum to form Iron and Aluminum Nitrate.

 (c) Lead sulphate and water are formed by the reaction of Lead Sulfide and hydrogen peroxide.

 (d) Chlorine gas and Sodium Iodide react together to form Iodine and Sodium Chloride.

Answer 1

9. (a)

This is a balanced chemical reaction as all atoms are balanced on both sides.

Now we know that oxidation is gain of oxygen and loss of hydrogen whereas reduction is gain of hydrogen and loss of oxygen. Here N₂ is gaining hydrogen (H₂) so N₂ is getting reduced and H₂ is getting oxidized (as it is the reducing agent). Also for an element in free state, the oxidation state is 0. Thus N₂ and H₂ being in their free state have 0 oxidation state. As N is more electronegative than H so in NH₃, H has +1 oxidation state and there are three such H. So to balance the 3x+1=+3 on three H, N has -3 oxidation state.

(b) Iron (II) nitrate is Fe(NO₃)₂ as nitrate ion NO₃^- has a -1 charge, so to balance +2 oxidation state of Fe there are two nitrate ions in this compound

Aluminium nitrate is Al(NO₃)₃as nitrate ion NO₃^- has a -1 charge, so to balance +3 oxidation state of Al (Al is in group 13 and has valency 3 so aluminium ion carries +3 charge) there are three nitrate ions in this compound. So the reaction is as shown

Now in given compound there are two nitrate groups on left side and three on right so to balance it out write 2 as coefficient of Al(NO₃)₃ and 3 as coefficient of Fe(NO₃)₂

There are 3 Fe and 1 Al on left and 1 Fe and 2 Al on right. Now to balance Al and Fe write 2 as coefficient of Al (s) and 3 as coefficient of Fe (s)

This is a balanced chemical reaction as all atoms are balanced now on both sides.

Here Al(s) and Fe(s) are in their free state, so these have 0 oxidation state.

In Fe(NO₃)₂ and Al(NO₃)₃ as discussed above Fe and Al have +2 and +3 oxidation state. Also each O has -2 oxidation state and each N has +5 oxidation state because NO₃^- carries -1 charge and O being more electronegative than N has -2 oxidation state (as it is in group 16, has 6 valence electrons so it gains 2 electrons to attain stable configuration) and to balance -2x3=-6 on three oxygen atoms, N has +5 oxidation state so the overall charge on nitrate ion is -1.

Now oxidation is loss of electrons (increase in oxidation state) and reduction is gain of electrons (decrease in oxidation state).

Thus Fe in Fe(NO₃)₂ is getting reduced (going from +2 oxidation state to 0 oxidation state) and Al is getting oxidized (going from 0 to +3 oxidation state)

(c) We know that in lead sulphide, Lead (Pb) has +2 charge and sulphur has -2 charge (commonly lead forms Pb²⁺ and S being in group 16 has 6 valence electrons and it gains 2 electrons to form S^2-). Similarly lead sulphate is PbSO₄ as sulphate is SO₄^2- and lead is Pb²⁺ and thus they balance each other. We know that hydrogen peroxide has formula H₂O₂ and each O (peroxide) means O^- and Hydrogen in this is H⁺.

So the reaction is as shown.

Now there are 2 O on left side and (4 in PbSO₄+1 in H₂O)=5 on right. So to balance it out write 4 as coefficient of H₂O₂ and H₂O then all atoms get balanced and the chemical reaction is balanced.

Now lead remains Pb⁺² on both sides. S is S^2- on left side but in SO₄^2- it is in +6 oxidation state as each O (being more electronegative than S) carries has -2 oxidation state and 4 O will have -8 to balance which S is +6 in oxidation state to make the overall charge -2 on SO₄^2-. Similarly in H₂O each H is +1 oxidation state and O is in -2 oxidation state to balance +2 oxidation state of 2 H. So O is getting reduced from -1 oxidation state in H₂O₂ to -2 oxidation state in H₂O. S is getting oxidized from -2 in PbS to +6 in PbSO_{​​​​​​4}.

(d)

Sodium iodide is NaI as Sodium (Na) is in group 1 and iodine (I) is in group 17. Thus Na has 1 valence electron and it loses this to form Na⁺ also I has 7 valence electrons and it gains 1 electron to form iodide I^-. Thus sodium iodide is NaI.

Similarly sodium chloride is NaCl as sodium form Na⁺ ion and Chloride is Cl^- (Cl is in group 17 similar to I) and they balance each other's charge to form NaCl. So reaction is as given above.

Now in the given equation, there are two Cl atoms on left and 1 on right. Similarly there are two I atoms on right and 1 on left. So to balance these out, write 2 as coefficient of NaI and NaCl.

This is the balanced chemical reaction as atoms are balanced in both sides of the equation.

Now Cl₂ and I₂ are in their free elemental state so these have 0 oxidation state. In both NaCl and NaI, Na is more electronegative than Cl and I respectively and hence it carries +1 oxidation state (as it belongs to group 1 and has valency 1. Similarity Cl and I in NaCl and NaI have -1 oxidation state, as these belong to group 17 and have 7 valence electrons and thus these gain 1 electron each to form Cl^- and I^-.

So Cl₂ is getting reduced (going from 0 to -1). NaI is getting oxidized (as I goes from -1 oxidation state in NaI to 0 oxidation state in I₂).