Redox Reactions
Unit 8
Redox Reactions
I. Multiple Choice Questions (Type-I)
1. Which of the following is not an example of redox reaction?
(i) $CuO+H_2 \longrightarrow Cu+H_2 O$
(ii) $Fe_2 O_3+3 CO \longrightarrow 2 Fe+3 CO_2$
(iii) $2 K+F_2 \longrightarrow 2 KF$
(iv) $BaCl_2+H_2 SO_4 \longrightarrow BaSO_4+2 HCl$
Show Answer
Answer: (iv)2. The more positive the value of $E^{\ominus}$, the greater is the tendency of the species to get reduced. Using the standard electrode potential of redox couples given below find out which of the following is the strongest oxidising agent.
$\mathbf{E}^{\ominus}$ values: $Fe^{3+} / Fe^{2+}=+0.77 ; I_2(s) / I^{-}=+0.54$;
$ Cu^{2+} / Cu=+0.34 ; Ag^{+} / Ag=+0.80 V $
(i) $Fe^{3+}$
(ii) $I_2(s)$
(iii) $Cu^{2+}$
(iv) $Ag^{+}$
Show Answer
Answer: (iv)3. $E^{\ominus}$ values of some redox couples are given below. On the basis of these values choose the correct option.
$\mathbf{E}^{\ominus}$ values : $Br_2 / Br^{-}=+1.90 ; Ag^{+} / Ag(s)=+0.80$
$ Cu^{2+} / Cu(s)=+0.34 ; I_2(s) / I^{-}=+0.54 $
(i) $Cu$ will reduce $Br^{-}$
(ii) $Cu$ will reduce $Ag$
(iii) $Cu$ will reduce I
(iv) $Cu$ will reduce $Br_2$
Show Answer
Answer: (iv)4. Using the standard electrode potential, find out the pair between which redox reaction is not feasible.
$\mathbf{E}^{\ominus}$ values : $Fe^{3+} / Fe^{2+}=+0.77 ; I_2 / I^{-}=+0.54 ;$
$ Cu^{2+} / Cu=+0.34 ; Ag^{+} / Ag=+0.80 V $
(i) $Fe^{3+}$ and $I^{-}$
(ii) $Ag^{+}$and $Cu$
(iii) $Fe^{3+}$ and $Cu$
(iv) $Ag$ and $Fe^{3+}$
Show Answer
Answer: (iv)5. Thiosulphate reacts differently with iodine and bromine in the reactions given below:
$ \begin{aligned} & 2 S_2 O_3^{2-}+I_2 \to S_4 O_6^{2-}+2 I^{-} \\ & S_2 O_3^{2-}+2 Br_2+5 H_2 O \to 2 SO_4^{2-}+2 Br^{-}+10 H^{+} \end{aligned} $
Which of the following statements justifies the above dual behaviour of thiosulphate?
(i) Bromine is a stronger oxidant than iodine.
(ii) Bromine is a weaker oxidant than iodine.
(iii) Thiosulphate undergoes oxidation by bromine and reduction by iodine in these reactions.
(iv) Bromine undergoes oxidation and iodine undergoes reduction in these reactions.
Show Answer
Answer: (i)6. The oxidation number of an element in a compound is evaluated on the basis of certain rules. Which of the following rules is not correct in this respect?
(i) The oxidation number of hydrogen is always +1 .
(ii) The algebraic sum of all the oxidation numbers in a compound is zero.
(iii) An element in the free or the uncombined state bears oxidation number zero.
(iv) In all its compounds, the oxidation number of fluorine is -1 .
Show Answer
Answer: (i)7. In which of the following compounds, an element exhibits two different oxidation states.
(i) $NH_2 OH$
(ii) $NH_4 NO_3$
(iii) $N_2 H_4$
(iv) $N_3 H$
Show Answer
Answer: (ii)8. Which of the following arrangements represent increasing oxidation number of the central atom?
(i) $CrO_2^{-}, ClO_3^{-}, CrO_4^{2-}, MnO_4^{-}$
(ii) $ClO_3^{-}, CrO_4^{2-}, MnO_4^{-}, CrO_2^{-}$
(iii) $CrO_2^{-}, ClO_3^{-}, MnO_4^{-}, CrO_4^{2-}$
(iv) $CrO_4^{2-}, MnO_4^{-}, CrO_2^{-}, ClO_3^{-}$
Show Answer
Answer: (i)9. The largest oxidation number exhibited by an element depends on its outer electronic configuration. With which of the following outer electronic configurations the element will exhibit largest oxidation number?
(i) $3 d^{1} 4 s^{2}$
(ii) $3 d^{3} 4 s^{2}$
(iii) $3 d^{5} 4 s^{1}$
(iv) $3 d^{5} 4 s^{2}$
Show Answer
Answer: (iv)10. Identify disproportionation reaction
(i) $CH_4+2 O_2 \longrightarrow CO_2+2 H_2 O$
(ii) $CH_4+4 Cl_2 \longrightarrow CCl_4+4 HCl$
(iii) $2 F_2+2 OH^{-} \longrightarrow 2 F^{-}+OF_2+H_2 O$
(iv) $2 NO_2+2 OH^{-} \longrightarrow NO_2^{-}+NO_3^{-}+H_2 O$
Show Answer
Answer: (iv)11. Which of the following elements does not show disproportionation tendency?
(i) $Cl$
(ii) $Br$
(iii) $F$
(iv) I
II. Multiple Choice Questions (Type-II)
In the following questions two or more options may be correct.
Show Answer
Answer: (iii)
II. Multiple Choice Questions (Type-II)
12. Which of the following statement(s) is/are not true about the following decomposition reaction.
$ 2 KClO_3 \to 2 KCl+3 O_2 $
(i) Potassium is undergoing oxidation
(ii) Chlorine is undergoing oxidation
(iii) Oxygen is reduced
(iv) None of the species are undergoing oxidation or reduction
Show Answer
Answer: (i), (iv)13. Identify the correct statement (s) in relation to the following reaction:
$ Zn+2 HCl \to ZnCl_2+H_2 $
(i) Zinc is acting as an oxidant
(ii) Chlorine is acting as a reductant
(iii) Hydrogen ion is acting as an oxidant
(iv) Zinc is acting as a reductant
Show Answer
Answer: (iii), (iv)14. The exhibition of various oxidation states by an element is also related to the outer orbital electronic configuration of its atom. Atom(s) having which of the following outermost electronic configurations will exhibit more than one oxidation state in its compounds.
(i) $3 s^{1}$
(ii) $3 d^{1} 4 s^{2}$
(iii) $3 d^{2} 4 s^{2}$
(iv) $3 s^{2} 3 p^{3}$
Show Answer
Answer: (iii), (iv)15. Identify the correct statements with reference to the given reaction
$ P_4+3 OH^{-}+3 H_2 O \to PH_3+3 H_2 PO_2^{-} $
(i) Phosphorus is undergoing reduction only.
(ii) Phosphorus is undergoing oxidation only.
(iii) Phosphorus is undergoing oxidation as well as reduction.
(iv) Hydrogen is undergoing neither oxidation nor reduction.
Show Answer
Answer: (iii), (iv)16. Which of the following electrodes will act as anodes, when connected to Standard Hydrogen Electrode?
(i) $Al / Al^{3+}$ $\quad$ $E^{\ominus}=-1.66$
(ii) $Fe / Fe^{2+}$ $\quad$ $E^{\ominus}=-0.44$
(iii) $Cu / Cu^{2+}$ $\quad$ $E^{\ominus}=+0.34$
(iv) $F_2$ (g) $/ 2 F^{-}$(aq) $\quad$ $E^{\ominus}=+2.87$
III. Short Answer Type
Show Answer
Answer: (i), (ii)
III. Short Answer Type
17. The reaction
$ Cl_2(g)+2 OH^{-}(aq) \longrightarrow ClO^{-}(aq)+Cl^{-}(aq)+H_2 O(l) $
represents the process of bleaching. Identify and name the species that bleaches the substances due to its oxidising action.
Show Answer
Answer: Hypochlorite ion18. $MnO_4^{2-}$ undergoes disproportionation reaction in acidic medium but $MnO_4^{-}$ does not. Give reason.
Show Answer
Answer: In $MnO_4^{-}, Mn$ is in the highest oxidation state i.e. +7. Therefore, it does not undergo disproportionation. $MnO_4^{2-}$ undergoes disproportionation as follows :
$3 MnO_4^{2-}+4 H^{+} \longrightarrow 2 MnO_4^{-}+MnO_2+2 H_2 O$
19. $PbO$ and $PbO_2$ react with $HCl$ according to following chemical equations :
$ \begin{aligned} & 2 PbO+4 HCl \longrightarrow 2 PbCl_2+2 H_2 O \\ & PbO_2+4 HCl \longrightarrow PbCl_2+Cl_2+2 H_2 O \end{aligned} $
Why do these compounds differ in their reactivity?
Show Answer
Answer: $2 PbO+4 HCl \longrightarrow 2 PbCl_2+2 H_2 O$ (Acid base reaction)
$PbO_2+4 HCl \longrightarrow PbCl_2+Cl_2+2 H_2 O$ (Redox reaction)
(Hint : Note the oxidation number of lead in the oxides)
20. Nitric acid is an oxidising agent and reacts with $PbO$ but it does not react with $PbO_2$. Explain why?
Show Answer
Answer: $PbO$ is a basic oxide and simple acid base reaction takes place between $PbO$ and $HNO_3$. On the other hand in $PbO_2$ lead is in +4 oxidation state and cannot be oxidised further. Therefore no reaction takes place. Thus, $PbO_2$ is passive, only $PbO$ reacts with $HNO_3$.
$2 PbO+4 HNO_3 \longrightarrow 2 Pb(NO_3)_2+2 H_2 O$ (Acid base reaction)
21. Write balanced chemical equation for the following reactions:
(i) Permanganate ion $(MnO_4^{-})$reacts with sulphur dioxide gas in acidic medium to produce $Mn^{2+}$ and hydrogensulphate ion.
(Balance by ion electron method)
(ii) Reaction of liquid hydrazine $(N_2 H_4)$ with chlorate ion $(ClO_3^{-})$in basic medium produces nitric oxide gas and chloride ion in gaseous state.
(Balance by oxidation number method)
(iii) Dichlorine heptaoxide $(Cl_2 O_7)$ in gaseous state combines with an aqueous solution of hydrogen peroxide in acidic medium to give chlorite ion $(ClO_2^{-})$and oxygen gas.
(Balance by ion electron method)
22. Calculate the oxidation number of phosphorus in the following species.
(a) $HPO_3^{2-}$ and $\quad$ (b) $PO_4^{3-}$
Show Answer
Answer: (a) +3 , $\quad$ (b) +523. Calculate the oxidation number of each sulphur atom in the following compounds:
(a) $Na_2 S_2 O_3$ $\quad$ (b) $Na_2 S_4 O_6$ $\quad$ (c) $Na_2 SO_3$ $\quad$ (d) $Na_2 SO_4$
Show Answer
Answer: (a) +2 $\quad$ (b) $+5,0,0,+5$ $\quad$ (c) +4 $\quad$ (d) +6
Justification :
Write Lewis structure of each ion then assign electron pair shared between atoms of different electronegativity to more electronegative atom and distribute the electron pair shared between atoms of same element equally. Now count the number of electrons possessed by each atom. Find out the difference in number of electrons possessed by neutral atom and that possessed by atom in the compound. This difference is the oxidation number. If atom present in the compound possesses more electrons than the neutral atom, the oxidation number is negative. If it possesses less electrons then oxidation number is positive.
(i) Lewis structure of $S_2 O_4^{2-}$ can be written as follows :
Electron pair shared between sulphur and oxygen atom is assigned to oxygen atom because of more electronegativity of oxygen. Thus each sulphur atom is deficient of two electrons with respect to neutral sulphur atom. Hence, each sulphur atom is in +2 oxidation state. Each oxygen atom gets two excess electrons hence it is in -2 oxidation state. Lewis structure of $S_4 O_6^{2-}$ can be written as follows :
To find out oxidation state of each atom we distribute electrons of electron pair shared between two sulphur atoms equally (i.e. one electron is assigned to each sulphur atom). Both electrons of electron pair shared between sulphur and oxygen atom are assigned to oxygen as oxygen is more electronegative. Thus we find that each of the central sulphur atoms obtains six electrons. This number is same as that in the outer shell of neutral sulphur atom. Hence, oxidation state of each central sulphur atom is zero. Each of the sulphur atoms attached to oxygen atoms finally obtains only one electron as its share. This number is less by five electrons in comparison to the neutral sulphur atom. So, outer sulphur atoms are in +5 oxidation state. Therefore average oxidation state of sulphur atoms is :
$ \frac{5+0+0+5}{4}=\frac{10}{4}=2.5 $
By using the formula we obtain average oxidation state of the particular type of atoms. Real oxidation state can be obtained only by writing the complete structural formula. Similarly we can see that each oxygen atom is in -2 oxidation state.
In the same way one can find out the oxidation state of each atom in $SO_3^{2-}$ and $SO_4^{2-}$ ions. Oxidation state of metal atoms will be +1 as these will lose one electron in each case.
IV. Matching Type
24. Balance the following equations by the oxidation number method.
(i) $Fe^{2+}+H^{+}+Cr_2 O_7^{2-} \longrightarrow Cr^{3+}+Fe^{3+}+H_2 O$
(ii) $I_2+NO_3^{-} \longrightarrow NO_2+IO_3^{-}$
(iii) $I_2+S_2 O_3^{2-} \longrightarrow I^{-}+S_4 O_6^{2-}$
(iv) $MnO_2+C_2 O_4^{2-} \longrightarrow Mn^{2+}+CO_2$
25. Identify the redox reactions out of the following reactions and identify the oxidising and reducing agents in them.
(i) $3 HCl(aq)+HNO_3$ (aq) $\longrightarrow Cl_2$ (g) $+NOCl$ (g) $+2 H_2 O($ l)
(ii) $HgCl_2$ (aq) $+2 KI$ (aq) $\longrightarrow HgI_2$ (s) $+2 KCl$ (aq)
(iii) $Fe_2 O_3(s)+3 CO(g) \xrightarrow{\Delta} 2 Fe(s)+3 CO_2(g)$
(iv) $PCl_3$ (l) $+3 H_2 O$ (l) $\longrightarrow 3 HCl$ (aq) $+H_3 PO_3$ (aq)
(v) $4 NH_3+3 O_2(g) \longrightarrow 2 N_2(g)+6 H_2 O(g)$
26. Balance the following ionic equations
(i) $Cr_2 O_7^{2-}+H^{+}+I^{-} \longrightarrow Cr^{3+}+I_2+H_2 O$
(ii) $Cr_2 O_7^{2-}+Fe^{2+}+H^{+} \longrightarrow Cr^{3+}+Fe^{3+}+H_2 O$
(iii) $MnO_4^{-}+SO_3^{2-}+H^{+} \longrightarrow Mn^{2+}+SO_4^{2-}+H_2 O$
(iv) $MnO_4^{-}+H^{+}+Br^{-} \longrightarrow Mn^{2+}+Br_2+H_2 O$
IV. Matching Type
27. Match Column I with Column II for the oxidation states of the central atoms.
Column I | Column II |
---|---|
(i) $Cr_2 O_7^{2-}$ | (a) +3 |
(ii) $MnO_4^{-}$ | (b) +4 |
(iii) $VO_3^{-}$ | (c) +5 |
(iv) $FeF_6^{3-}$ | (d) +6 |
(e) +7 |
Show Answer
Answer: (i) $\to$ (d) $\quad$ (ii) $\to$ (e) $\quad$ (iii) $\to$ (c) $\quad$ (iv) $\to$ (a)28. Match the items in Column I with relevant items in Column II.
Column I | Column II |
---|---|
(i) Ions having positive charge | (a) +7 |
(ii) The sum of oxidation number of all atoms in a neutral molecule | (b) -1 |
of all atoms in a netural molecule | (c) +1 |
(iii) Oxidation number of hydrogen ion $(H^{+})$ | (d) 0 |
(iv) Oxidation number of fluorine in $NaF$ | (e)Cation |
(v) Ions having negative charge | (f) Anion |
V. Assertion and Reason Type
In the following questions a statement of assertion (A) followed by a statement of reason (R) is given. Choose the correct option out of the choices given below each question.
Show Answer
Answer: (i) $\to$ (e) $\quad$ (ii) $\to$ (d) $\quad$ (iii) $\to$ (c) $\quad$ (iv) $\to$ (b) $\quad$ (v) $\to$ (f)
V. Assertion and Reason Type
29. Assertion (A) : Among halogens fluorine is the best oxidant.
Reason (R) : Fluorine is the most electronegative atom.
(i) Both $A$ and $R$ are true and $R$ is the correct explanation of $A$.
(ii) Both $A$ and $R$ are true but $R$ is not the correct explanation of $A$.
(iii) $A$ is true but $R$ is false.
(iv) Both $A$ and $R$ are false.
30. Assertion (A): In the reaction between potassium permanganate and potassium iodide, permanganate ions act as oxidising agent.
Reason (R) : Oxidation state of manganese changes from +2 to +7 during the reaction.
(i) Both $A$ and $R$ are true and $R$ is the correct explanation of $A$.
(ii) Both $A$ and $R$ are true but $R$ is not the correct explanation of $A$.
(iii) $A$ is true but $R$ is false.
(iv) Both $A$ and $R$ are false.
31. Assertion (A) : The decomposition of hydrogen peroxide to form water and oxygen is an example of disproportionation reaction.
Reason (R) : The oxygen of peroxide is in -1 oxidation state and it is converted to zero oxidation state in $O_2$ and -2 oxidation state in $H_2 O$.
(i) Both $A$ and $R$ are true and $R$ is the correct explanation of $A$.
(ii) Both $A$ and $R$ are true but $R$ is not the correct explanation of $A$.
(iii) $A$ is true but $R$ is false.
(iv) Both $A$ and $R$ are false.
32. Assertion (A) : Redox couple is the combination of oxidised and reduced form of a substance involved in an oxidation or reduction half cell.
Reason (R) : In the representation $E _{Fe^{3+} / Fe^{2+}}^{\ominus}$ and $E _{Cu^{2+} / Cu}^{\ominus}, Fe^{3+} / Fe^{2+}$ and $Cu^{2+} / Cu$ are redox couples.
(i) Both $A$ and $R$ are true and $R$ is the correct explanation of $A$.
(ii) Both $A$ and $R$ are true but $R$ is not the correct explanation of $A$.
(iii) $A$ is true but $R$ is false.
(iv) Both $A$ and $R$ are false.
VI. Long Answer Type
33. Explain redox reactions on the basis of electron transfer. Give suitable examples.
34. On the basis of standard electrode potential values, suggest which of the following reactions would take place? (Consult the book for $E^{\ominus}$ value).
(i) $Cu+Zn^{2+} \longrightarrow Cu^{2+}+Zn$
(ii) $Mg+Fe^{2+} \longrightarrow Mg^{2+}+Fe$
(iii) $Br_2+2 Cl^{-} \longrightarrow Cl_2+2 Br^{-}$
(iv) $Fe+Cd^{2+} \longrightarrow Cd+Fe^{2+}$
35. Why does fluorine not show disporportionation reaction?
36. Write redox couples involved in the reactions (i) to (iv) given in question 34.
37. Find out the oxidation number of chlorine in the following compounds and arrange them in increasing order of oxidation number of chlorine.
$NaClO_4, NaClO_3, NaClO, KClO_2, Cl_2 O_7, ClO_3, Cl_2 O, NaCl, Cl_2, ClO_2$.
Which oxidation state is not present in any of the above compounds?
38. Which method can be used to find out strength of reductant/oxidant in a solution? Explain with an example.