REDOX REACTION EXERCISE # 1

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1 REDX REACTIN EXERCISE #. No. of equivalent mole n-factor ( n-) S 3 + H S + H + + e...() n-factor for R N () is () 0. n. n. n Final oidation state will be (3 ) ((3 ) ). Meq.() of K Cr 7 Meq.() of ABD n-factor of K Cr 7 in acidic medium 6. (K Cr 7 n 6) New oidation state of A n will be n (A) (A n n + 3 ) SS Na SSSS H H Na HS 7 (6, 6) Na S 6 (, 0, 0, ) + S + (B) NaS Na S 3 HS + Na (, 0) (6) (C) S 3 > S > H S > S (D) H S > S > H S > H S (6, 6). (NH ) Cr 7 N + Cr 3 + H SH Decomposition () R N. NaN 3 N 3 N /3 N H N N N + (A) N N + 6. S + I S 6 + I Re dn Redo R N () I I d. 3 S 6 S (, 0) (, 0, 0, ) 8. HI + HN 3 N + I + H (N 3C N ) Adding () and () (I I e) 3...()...() 3 6I + N 3 N + 3I 6HI + HN 3 N + 3I + H 6, Mn + C + H Mn + C + H Mn +C +H + (6) Mn + + 0C + 8H (A),, 6 0. Molarit () Normalit( ) n factor( ) (e) M H 3 P /3 M H 3 P No. of equivalent( ). M Volume of sol ( )(in L ) Meq. 0 0 M mole 0 M gm. HC.H Mass of H C.H () gm. 63 % (w/v) H C.H 00 ml contain 63 gm 63 ml 00 gm 63 Mole of H C Mole of NaH H C + NaH Na C + H mole of Acid () Mole of NaH 8 And will have ) Sol. is neutral () mole of NaH ( NaH

2 3. CaC 3 + HCl CaCl + H + C ( ml) C mole HCl M HCl N M. eq. () of H 3 P M. eq. () of Ba(H). v v 0 ml 3.. KMn n factor in Acidic medium (KMn n ) K Cr 7 n factor in acidic medium 6 (K Cr 7 n ) 6 0. V 0.3 V 6 V V V V 6. K Cr 7 have greater n factor as compaire KMn so same volume of K Cr 7 will oidise more amount of Fe +. (K Cr 7 n KMn K Cr 7 Fe + ) Mole of V Mole of V mole ~ 0. V V e I e I Mole of I Mole of V Cl + S 3 S + Cl + S 0.0 n factor () n factor S 3 n factor (i) Cl + H + Na S 3 Na S + S + HCl Balanced equation (ii) Mole of S (iii) Equivalent of oidising agent ( ) 0.00 (iv) Molarit of Na S () 0.0 M M eq.() of KMn M eq. () of C N C M mole of oalate 9 3 Wt of oalate % C % 0. M eq. of KMn M eq. of C M eq. of CaC 3 0. M eq. of Ca Mole of Ca % Ca.8 Ca %. Cr + 3H S Cr (S ) 3 + 3H + 7/ mole Cr Liberate () 7/ mole of As + H +I As + H + I molar mass () Na 3 As molar mass () 08 eq.() of As 08 0 equivalent () of Na 3 As equivalent () of I equivalent () of Na S 3.. V 0 L V 8. ml M eq. of KMn. (A) M eq. of FeS. (C) M eq. of H... (D) M eq. of SnCl. NV NV N. V V Molarit () H H + L H, Mole H give. L L H, 0.7 H V

3 Volume strength() 8. V Alternative() V S.6 N.6.8.V. M eq. of KMn. 0 0 M eq. of H. M eq. of KMn remaining (KMn 6. ) (0 ) Mole of KMn (30 ml, 0V H ) + (80 ml, NH ) (A) (B) N A N V N V.6 V V N A A B B C A B N C 7 (7/7) Ca(HC 3 ) + Ca CaC 3 + H 00 6 gm gm gm 7 VS.6 7 VS 3.6 V 7 M C Mol/L M 7 C 7 Mole / L C 7 3 gm / Ltr. Concentration ().8 gm/ltr. REDX REACTIN EXERCISE #. Fe + Fe Fe + Fe 3 0. (0.60) Mole ratio () Fe Fe FeS mole of S than mole Fe + In Fe (S ) 3 3 mole of S than mole Fe 3+ mole of S than 3 ratio () 3 Fe 3 Fe 3 3. Fe + 3 Fe 3 Let assume n mole of Iron Initial n 0 n mole Fe3+ wt. (n ) n 6..6 n n 0.33 % total Iron 3.3%. HN 3 + NH + N + N Meq of HN 3 Meq of NH + mole n-factor mole n-factor mole 6 mole of HN 3 6 H 6. Z KMn Mn Z Meq of Z Meq of KMn 0. ( ) 0.0 ( ) Z + Z + ( ) 7. Cl + KMn Mn + + Cl Meq of NaCl Meq of KMn 0 mole n-factor 8 0 mole 8 volume of Cl L 8 8. I I + I 3 I I I I 3 e I I I + 6H I 3 0e I 0I I + 6H + H I 3 + H I + H I 3 + 6H + 0e H + 6I 0I + I 3 + 6H ratio of 3 I : I 0

4 9. Eq. mass () 0. molecular mass ( ) n factor( ) As +3 As + n-factor 6 S 3 3 S total n-factor ( n-) 8 Eq. mass () m.wt. 8 6 Zn S HN Zn (N ) H S N 3 3 change in.n. of Zn (Zn ) Zn 0. From Hit and trial method S 6() 8 N 3H + CN N + C + 6H + + 7e...(i) 3CN + 9H 3N + 3C + 8H + + e e + 7N 3 + 8H + 7N + H Balance equation () 3CN + 7N 3 + 0H + 0N + 3C + H. K Cr 7 + Sn + Sn + + Cr 3+ Sn + + Fe + Fe 3+ meq. of Sn + meq of K Cr 7 meq. of Sn + meq of Fe 3+ meq. of Fe 3+ meq of K Cr N K Cr millimol () n -factor 0 millimol () 0 7. Vol. of at NTP V V ml 3 ml of H gives ml at N.T.P. (3 ml H N.T.P. ml) ml of H gives ml of at NTP hence volume strength of H '3 V' (H ) 8. Half meq of salt (Na C 3 ) in neutralize using Hph indicator ( (Na C 3 ) Hph ) meq of salt meq of HCl (0 0. ) (i) complete meq of salt (Na C 3. NaHC 3. H ) is neutralise using MeH indicator ( (Na C 3. NaHC 3. H ) MeH) Meq. of salt Meq of HCl (ii) eq (ii) eq (i) 0.0 ( ) (6 ) ( ) 0.0 ( ) 80 ml ( ) 0 9. Meq of I Meq of Hpo solution () Meq of 0ml I Meq of I Meq of 00 ml I 0. Meq of CaC 3 0. w w % purit () % I + Na S 3 I + S 6 let ml of I react with Hpo ( ml, I ) meq of I meq of Hpo N 0. N 6...(i) meq of H S used b base (H S ) meq of NaH used b I (I NaH ) (30 6) (0 ) N..(ii) from eq (i) & eq (ii) ml 30 N 6 N M M 0 0. N M n-factor

5 . Let a gm H S and (3.8 a) g H C Meq of 0 ml miture 0.3 meq of 000 ml miture meq of H S + meq of H C 30 a (3.8 a) (i) 9 In another e. meq of 00 ml miture meq of KMn 0.0 meq of 00 ml miture 0. meq of 000 ml miture meq of H C 3.8 a (ii). K Cr 7 + KI I + Cr 3+ I + Na S 3 I + S 6 Na S 3 + K Cr 7 meq of Na S 3 meq. of K Cr 7 30 N N 0 meq. of I meq. of Hpo meq. of I meq. of KI meq of KI meq. of K Cr meq. of ml K Cr 7 meq. of 00 ml K Cr w w % purit () % 0.8 REDX REACTIN EXERCISE # 3 CMPREHENSIN BASED QUESTINS Comprehensi on #. H + S 3 H S ; 8 g water combines with 80 g S 3 (8 g 80 g S 3 ). g of H combines with 0 g of S 3. g H, 0 g S 3 00 g of oleum contains 0 g of S 3 or 0% free S 3 00g 0g S 3 0% S 3. Initial moles of free S 3 present in oleum moles 8 3 S 3 moles of water that can combines with S 3 combined with water 9 mole 8 S 3 moles of free S 3 remainss 3 mole 3 6 volume of free S 3 at STP (STP S 3 ). 3.73L 6 3. Na C 3 + H S Na S + H + C moles of C formed moles of Na C 3 reacted volume of C formed at atm pressure and 300 K L. eq. of H S + eq. of S 3 eq. of NaH ( ) % of free S % Comprehensi on #. L of H (aq) provide. L of at STP moles of. 0.. n H required 0. M H n V H solution M. Strength in percentage mean how man g H present per 00 ml (00 ml g H ) M and mol. wt. of H 3 3 H present per litre of solution or 3. H present per 00 ml of solution. 3 H 3. H 00 ml 3. m.eq. of H m.eq. of KMn 0 N N volume strength ( ) of H N.6 volume strength of H.6

6 . m-eq. of Ba(Mn ) m. eq. of H REDX REACTIN. KMn + X +n X mole mole Eq. () of KMn Eq. of X +n ( n) + Mn + n 6 M 3. M 6. CuS + Cu S + KMn Mn + + Cu + + S N 33.6 M 3. w 37 Eq. wt. () of CuS M /6 Eq. wt. () of Cu S M /8 Eq. wt. () of KMn M 3 / M n-factor 9. H S + NH 3 (NH ) S (30 ) Meq. Meq. (30 0.) Meq. V NH ; w 8. % purit () w ml 0. n 6 + n 7....() n + n 0....() n 6 % of KH 00 3%. Ca(H) 6%. n 06 + n 8...() n + n 0. V () V 7.89 ml. Eq. of H S Eq. of NaH n n [ ] mole of H S total. [N 98 0.] mass of H added mole of H mole of S 3 % of S % Comprehensi on # 3. n-factor () 0. H 3 P is a monobasic acid () n-factor 3. n-factor M eq. wt n-factor of V 3; Fe 3 ; and are and 3 EXERCISE # [A] 3. CaC 3 + HCl CaCl + C + H 0. mole HCl + KH KCl + H 0.0 V V 0.0 L ml. NaH + NaH P Na 3 P + H V 0 0. Mole V 0. V 0. lit 00 ml.. CaC 3 + HCl CaCl + H + C mole MgC 3 + HCl MgCl + H + C mole ( ) () () % CaC %.6 % MC 3.% 6. M Eq.() of CaC 3 M Eq. of HCl M Eq. of NaH w (00 / ) w. gm % CaC % 7. Na C 3 + NaHC 3 g milli mole...() + 0.., Na C 3 06 mg 0. mg NaHC 3 0. gm

7 8. Na C 3 NaH m mole () ().77 m mole Na C NaH.8 gm/lit. 3. gm/lit. 9. NaH + Na C 3 m mole () NaH Na C () , 0. m mole gm 0.06 gm 0.06 gm 0. Na C 3 NaHC 3 Meq () () Same as 9.. Ce + + Sn + Sn + + Ce ml M M Meq. of Ce + Meq. Sn ( n) 0 ( n) n 3 3. Se + CrS Cr +3 + Se + Meq. of Se Meq. of CrS ( n).. n ~ n 0. K C.3H C.H +Mn Mn + + C V ml, 0. M V 0. V 3.68 ml. H + KMn Mn + + / 00 (3 / ) N N 6. S n + K Cr 7 SnCl + Cr N V ml M / V V 337 ml 7. Meq. of Cu w (63./) [ ] w.076 gm % Cu % 8. Meq. of Fe Meq. of K Cr / N 6 N H C.H + KHC.H +NaH product mole mole 8.9ml, 0. N H C.H +KHC.H +KMn Mn + +C mol mol.ml, 0.N () () % H C. H.36% % KH C.H 8.7% 3 0. Meq. of Ca(H) Meq of HCl w 7 / 000 ( ) w % Ca(H) Meq. of Na C 3 Meq of HCl w / % purit w % 3. gm substance () 0.6 gm NaCl, 0.37 gm KCl gm N N N N. M M 0.7 gm/lit Volume strength of final solution ( ).6. 8.

8 3. 0 N 0 0. N 0.0 M M 0.0 gm/ gm/ ppm REDX REACTIN ppm Meq. of H Meq. of KMn 3 / / % Purit () % N 8..6 N N ppm. meq. of Hpo meq. of I ( I ) moles of I. m moles (I. m moles) CuS + KI Cu I + K S + I from reaction moles of CuS ( CuS ). m moles M w of hdrated CuS ( CuS M w ) so () meq. of Hpo meq. of I () (I ) meq of Cl 3 (Cl 3 ) 0 m moles of Cl 3 (Cl 3 m moles) 6H + 6Cl 0Cl + Cl 3 + H + so m moles moles of Cl 6 m moles ( Cl 6 m moles) 6e + H + + Cr 7 Cr H (Cl Cl + e ) 3 H + + Cr 7 + 6Cl 3Cl + Cr H 6 m moles m moles of Cr 7 m moles wt. of Cr g (Cr 7 ) % purit () 8.8 % ml.6 mg Ca(HC 3 ) ml mg Ca(H) + Ca(HC 3 ) CaC 3 + H w w gm. Bleaching powder + Mohr salt ecess product. (+ ) Mohr salt () + KMn product (.. Meq. of Se Meq. of Br 3 3 used 3. w 000 M 0 60 w 0.08 gm 8 mg 0. M n n No. of electron taken up b oidant. () EXERCISE # [B] 3. Let H C. H g in 00 ml ( H C. H 00 ml g ) n reaction with NaH with phenolphthalein ( NaH ) g Eq. of acid in 0 ml (0 ml ) 6 g Eq. of NaH (NaH ) so g so mass of Na C.. g Na C Now, in 0.g of same miture(, 0.g) H C. H 0.3 g Na C 0. g so g Eq. of H C.H (H C.H ) g Eq. of Na C (Na C ) 0. g Eq. of KMn (KMn ) V V V 77.6 ml

9 . First HCl will react with KI 3 to from I & Cl then this Cl produced will again react with KI to form I. ( HCl, KI 3 I Cl Cl KI I ) Let initiall moles of KI 3 were mied with moles of HCl then(ki 3 HCl ) I 3 + 0Cl I + Cl 0 Cl + KI I + KCl Total moles of I formed ( I ) so mole so concentration of HCl ( HCl ) M N 0.0 moles of KI 3 consumed ( KI 3 ) volume of KI 3 consumed ( KI 3 ) L 0. ml. As 3 + 6HCl AsCl 3 + 3H AsCl 3 + H HAs + 3H + + 3Cl gram equivalent of I gram Eq. of HAs I HAs gram Eq. of AsCl 3 (AsCl 3 ) gram Eq. of As 3 (As 3 ) gram equivalent of As (As 3 ) gram equivalent of KMn (KMn ) Let amount of KMn used w g then (KMn w g ) w w 0.06 g 6. H + H + S S + H + e H + S S + H + e...() H + H + e H + H H + e H...() Eq. () () + () H + H + S S + H...(3) NaH + HCl NaCl + H Meq ~ From equation (3) ( (3) ) H + H + S S + H moles of S (S ) wt. () % of S sample(s %) % 7. (Mn + Mn 3+ + e )... () e + 8H + + Mn Mn 3+ + H... () equation() () + equation() () 8H + + Mn + + Mn Mn 3+ + Mn 3+ + H or 8H + + Mn + + Mn Mn 3+ + H...() e + 8H + + Mn 3 3Mn + + H...() from equation () milli equivalent of Mn N V (() Mn ) M V.f. V milli equivalent of Mn + milli equivalent of Mn (Mn + Mn ) from equation (3) milli eq. of Mn 3 3 milli equivalent of Mn + ((3) Mn 3 3 Mn+ ) equivalent of Mn (Mn 3 ) W W % of Mn 3 in the sample ( Mn 3 %) % 0.

10 8. H S + S S S (n -factor 6) ((n- 6)) for H S (H S ) ( ) S S for S (S ) (n -factor ) (n- ) 6 ( ) g concentration of H S(H S ) mg concentration of S (S ) mg S /L 9. Let mass of KCl 3 (KCl 3 ) g Let mass of KCl (KCl ) g KCl 3 /. AgCl KCl / e + 6H + + Cl 3 Cl + 3H (i) for complete oidation of an oidizing agent reacted FeS solution unreacted FeS ( FeS FeS ) N V N V N 3 milli eq.() put above value in eq. (i) ((i) ) (ii) 7 moisture()(+7) g 0. BaCr 0.09 g Cr g 3 % of Cr (Cr %) % 0 Cr mole g Eq. of Mn (Mn ) wt. () wt. of Mn (Mn ) g % of Mn (Mn %) % 0. CH 3 (CH ) n CH + (n + )C + H a (n + )a C + NaH Na C 3 (n + )a b b (n + )a (n + )a solution has ( ) NaH b (n + )a Na C 3 (n + )a n dividing in equal part moles get halfed. () Part - I (-I) : b (n )a (n )a + Part - II (-II) : (i) b (n )a (n )a (ii) (ii) (i) (n )a 0.03 (n + )a (iii) and.6 60 n a... (iv) from equation (iii) & (iv) ( (iii) (iv) ) n (n ) 9.33n n.33n.33 n from equation (iii) ((iii) ) 6a 0.06 a 0.0 from equation (i) ( (i) ) b b b moles of NaH (NaH 0.08 ) mass () g

11 . Total m mol of AgCl from 0 ml solution (0 ml AgClm mol) m moles of AgCl from HCl 0.8 m moles of AgCl from CaCl. mass % of NaH (original) (NaH 000 %) Now, let us assume that in 0 ml, m mol of NaH has got converted to Na C 3 (HCl AgCl m mol 0.8 CaCl AgCl m.). m mole of CaCl was consumed for precipitation of oalate from 0 ml solution. (0 ml. m CaCl ) Hence, total m mol of oalic acid in 0 ml solution (, 0 ml m ) m % of oalic acid ( %) In presence of methl orange, the whole NaH and Na C 3 are neutralized (NaH Na C 3 ) meq of HCl 6 0. meq of (NaH + Na C 3 ) meq. of NaH original (HCl 6 0. (NaH + Na C 3 ) NaH ) Total meq of NaH in original.0 g sample 0 (.0 g NaH ) (, 0 ml, m mol NaH, Na C 3 ) In 0 ml, m mol of NaH m mol of Na C 3 In nd titration, HCl used in titration of NaH + Na C (nd, NaH + Na C 3 HCl) upto phenolphthalein end point, m mol of HCl required + 3. (, HCl m mol).6 Total Na C 3 formed ( Na C 3 ) m mol of NaH left unreacted (NaH m mol) 0 weight of.0 g of eposed sample (06 8) g (.0 g ) weight % of Na C 3 in eposed sample % (Na C 3 %)

12 REDX REACTIN. (i) 7. Ca (ii) 7 6 (iii) 7 3 (iv) 7 3 Cl + Cl REDX REACTIN. + ( + ) 0 [ Ba(H P ) is neutral molecule] or 0 +. TIPS/ Formulae : (i) Write balance chemical equation for given change. oidation state. (ii) Identif most electronegative element in the reaction and has the oidation states of (in H ) and (in BaS ). In H, peroide ion is present.. TIPS/ formulae : Balanced the reaction b ion electron method idation reaction : C C +e ] Reduct ion react ion : Mn + 8H + + e Mn + + H ] Net react ion : Mn + 6H + + C Mn + + 0C + 8H. TIPS/ Formulae : (i) In an ion sum of oidation states of all atoms is equal to charge on ion and in a compound sum of oidation states of all atoms is alwas zero. idation state of Min in Mn +7 idation state of Cr in Cr(CN) idation state of Ni in NiF 6 + idation state of Cr in Cr Cl +6. TIPS/ Formulae : (i) Mass of one electron (ii) mole of electron electrons weight of mole of electron Mass of one electron Avogadra number kg No. of moles of electrons in kg EXERCISE # [A] 3. In this oidation number of N is changing (0) Final product will be Cr 3 in this oidation state of Cr is TIPS/ Formulae : EXERCISE # [B] The highest.s. of an element is equal to the number of its valence electrons (i) [Fe(CN) 6 ] 3,.N. of Fe +3 [Co(CN) 6 ] 3,.N. f Co +3 (ii) Cr Cl,.N. of Cr +6, (Highest.S. of Cr) [Mn ].N. of Mn +7, (Highest.S. of Mn) (iii) Ti 3,.N. of Ti +6, Mn.N. of Mn + (iv) [Co(CN) 6 ] 3,.N. of Co +3. TIPS/ formulae : Mn3,.N. of Mn +6 Use molarit equation to find volume of H S solutions. CuC H S g 3. g CuS + H + C For 3. gms of Cu(II) carbonate 98 g of H S are required. For 0. gms of Cu(II) carbonate weight of H S reqd g g H S Weight of required H S g Weight of solution in grams mol.wt. Molarit Volume in ml V 000 or V ml Volume of H S solution ml

in acid medium. In the experiment number of A after oxidation is :- Cr 2 were used for moles of ABD.

in acid medium. In the experiment number of A after oxidation is :- Cr 2 were used for moles of ABD. EXERCISE-01 CHECK YUR GRASP SELECT THE CRRECT ALTERNATIVE (NLY NE CRRECT ANSWER) 1. 50 ml of 0.1 M solution of a salt reacted with 25 ml of 0.1 M solution of sodium sulphite. The half reaction for the