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8.26 Using the standard electrode potentials given in the Table 8.1, predict if the reaction between the following is feasible

(a) Fe^{3+}(aq) and I^{-}(aq)
(b) Ag^{+}(aq) and Cu(s)
(c) Fe^{3+}(aq) and Cu(s)
(d) Ag(s) and Fe^{3+} (aq)
(e) Br_{2}(aq) and  Fe^{2+}(aq).

Answer-  If  for the overall reaction is positive  feasible                                                  negativenot feasible (a)                                                                                               ---------------------------------------------------------------------------------------                                 (b)                                             ...

 8.18   Balance the following redox reactions by ion – electron method 

(d) Cr_{2}O_{7}^{2-}+SO_{2}(g) \rightarrow Cr^{3+}(aq)+SO_{4}^{2-}(aq)

in acidic medium

Answer(a)- Half-reaction oxidation half    reduction half  balancing them by multiplying oxidation half by 3 and adding the reaction  

 8.18   Balance the following redox reactions by ion – electron method 

(c) H_{2}O_{2}(aq)+Fe^{3+}(aq)\rightarrow Fe^{3+}(aq)+H_{2}O(l)

in acidic medium 

In acidic medium oxidation half reaction- reduction half reaction- Balancing the reaction multiply by 2 on oxidation half-reaction then add it with reduction half reaction

8.18   Balance the following redox reactions by ion – electron method 

(b) MnO_{4}^{-}(aq)+SO_{2}(g) \rightarrow Mn^{2+}(aq)+HSO_{4}^{-}

(In Acidic medium)

Answer- oxidation half reaction reduction half reaction Balancing the reaction multiply the oxidation half by 5 and reduction half by 2 and then add these two reactions   

8.18   Balance the following redox reactions by ion – electron method 

(a) MnO_{4}^{-}(aq)+ I^{-}(aq)\rightarrow MnO_{2}(s)+ I_{2}(s)  (In basic medium)

reduction half reaction   (+7 to +4)  Add 3 electron on LHS side and after that to balance charge add OH ions. And to balance O atom add water molecule on whichever side it needed balance it  oxidation half balance it  equalising the no. of electrons by multiplying the oxidation half by 3 and reduction half by 2 and then add it. final answer-  

8.17  Consider the reactions:

(c)C_{6}H_{5}CHO(l)+2[Ag(NH_{3})_{2}]^{+}(aq)+3OH^{-}(aq)\rightarrow C_{6}H_{5}COO^{-}(aq)+2Ag^{+}(s)+4NH_{3}(aq)+2H_{2}O(l)

(d)C_{6}H_{5}CHO(l)+2Cu^{2+}(aq)+5OH^{-}(aq)\rightarrow No change observed

Answer- Here, by looking at the reaction, we conclude that   oxidises  and in the second reaction  not able to oxidise. So we can say that is stronger oxidizing agent than .

8.17  Consider the reactions:

(a) H_{3}PO_{2}(aq)+4AgNO_{3}(aq)+2H_{2}O(l)\rightarrow H_{3}PO_{4}(aq)+4Ag(s)+4HNO_{3}(aq)

(b)H_{3}PO_{2}(aq)+2CuSO_{4}(aq)+2H_{2}O(l)\rightarrow H_{3}PO_{4}(aq)+2Cu(s)+4H_{2}SO_{4}(aq)

What inference do you draw about the behaviour of Ag^{+} and Cu^{+} from these reactions ? 

Answer- In the first reaction, we can see that   oxidizes the phosphorus from (+1  +5) also in second, we clearly see that   oxidize the phosphorus from (+1+5). Both are oxidizing agents.

8.22  Consider the elements: Cs, Ne, I and F
(a) Identify the element that exhibits only negative oxidation state.

(b) Identify the element that exhibits only positive oxidation state

(c) Identify the element that exhibits both positive and negative oxidation states

(d) Identify the element which exhibits neither the negative nor does the positive oxidation state. 

Answer- (a) (because of highly electronegative in nature) (b) (highly electropositive) (c)(has emplty d orbitals) (d)  (enert gas  

 8.24 Refer to the periodic table given in your book and now answer the following questions:

 (a) Select the possible non-metals that can show disproportionation reaction.

 (b) Select three metals that can show disproportionation reaction

8.30 Depict the galvanic cell in which the reaction 

Zn(s)+2Ag^{+}\rightarrow Zn^{2+}(aq)+2Ag(s)
takes place, Further show:

(i) which of the electrode is negatively charged,

(ii) the carriers of the current in the cell, and

(iii) individual reaction at each electrode.

Answers- (i)  electrode is negatively charged because it loses electrons (act as an anode) (ii) electron flow from negatively charged electrode to a positively charged electrode (anode to cathode) and the flow of current is just reversed. So current flow through silver cathode to the zinc anode. (iii) At Anode-      At Cathode 

8.29 Given the standard electrode potentials

K^{+}/K = -2.93 V   Ag^{+}/Ag = 0.80V

Hg^{2+}/Hg= 0.79V

Mg^{2+}/Mg = -2.37V  Cr^{3+}/Cr = -0.74V

arrange these metals in their increasing order of reducing power. 


Answer- A negative electrode potential means redox couple is a stronger reducing agent. So as per data the increasing order of the following is-

8.27 Predict the products of electrolysis in each of the following:

(iv) An aqueous solution of CuCl2 with platinum electrodes

Answer- (iv) In aqueous solution  ionise into  and  At the cathode, the copper ion will be deposited because it has a higher reduction potential than the water molecule At the anode, the lower electrode potential value will be preferred but due to overpotential of oxygen, chloride ion gets oxidized at the anode. @ Anode-   @ cathode- 

8.27 Predict the products of electrolysis in each of the following:

(iii) A dilute solution of H_{2}SO_{4} with platinum electrodes 

Answer- given sulphuric acid is dilute.  ionize into     At cathode  At anode, There will be -(liberation of oxygen gas)  

8.27 Predict the products of electrolysis in each of the following:

(ii) An aqueous solution AgNO_{3} with platinum electrodes 

Answer- (ii) since platinum electrode cannot easily oxidize. So at the anode    will oxidize and liberate oxygen and at cathode   will be deposited. At cathode-  At anode-  

8.27 Predict the products of electrolysis in each of the following:

(i) An aqueous solution of AgNO_{3} with silver electrodes 

Answer-  (i)  dissociate into  and  @ Cathode -  (reduction potential of silver is higher than )  @Anode -  (oxidation potential of silver is higher than water molecule.So silver electrode oxidized )

 8.28 Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn

Answer- In order to displace a metal from its metal salt is done only when the other metal has higher electrode potential.

8.25 In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g. of ammonia and 20.00 g of oxygen?

Answer- we have,  number of moles(n) = given mass/ molecular mass   ------------------------------(eq.1) No. of moles of ammonia  = 10/17 = 0.588 No. of moles of oxygen  = 20/32= 0.625 Balanced Reaction    Here we see that 4 moles of ammonia required 5 moles of oxygen. So 0.588 moles of ammonia =  moles of  . But we have only 0.625 moles of . It means oxygen is a limiting reagent and the...

8.23  Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water

Answer- Base equation-  --------------(have to remember) Now we have to balance the oxidation half and reduction half. oxidation half =  balancing-  Oxygen is balanced by adding water molecule, Hydrogen is balanced by  ion and for charge add (electron) Reduction hallf=  Balancing-  to balance charge add an electron    Now add both balanced oxidation half and reduction half,  we get 

8.21  The Mn^{3+}  ion is unstable in solution and undergoes disproportionation to give Mn^{2+} , MnO_{2} and H^{+} ion. Write a balanced ionic equation for the reaction

Answer- The base equation  write oxidation half with their oxidation state Balance the charge on  by adding 1  on RHS side. To balance charge add  ions on RHS side and then for oxygen balance add  molecule on LHS side. reduction half  balancing the reduction half by adding 1 on LHS side   Add both balanced reduction half and oxidation half

8.20  What sorts of information can you draw from the following reaction?

(CN)_{2}(g)+2OH^{-}(aq)\rightarrow CN^{-}(aq) +CNO^{-}(aq)+H_{2}O(l)

Answer- Carbon shows different oxidation state according to the compound formula.  here we can clearly say that Carbon is in its +3 oxidation state. The oxidation state of carbon is increased(oxidized) and decreased(reduced) as well in the product side. So it is a redox reaction and more specifically we can say it disproportion redox reaction.