## King 3 years ago Balance the charges and masses in foll. equation: I shall post equations in reply Pls help wid explanantion !!

1. King

\[CuS + NO _{3}^{-} --> Cu ^{2+}+ \S _{8}+NO\]

2. King

sry thats S8 in the product side

3. heena

re u sure equation i correct as if u see in the reaction O is not balanced first try to balance it

4. King

no my sir said that fr O we should balance using H20 and then add H+ ion to reactant side to balance H...oh yeah this is a acid base reaction

5. heena

hey is thiss redox reaction? right in which medium u have to do balancing?

6. heena

i mean acidic medium or basic medium u are provided this info?

7. King

acidic medium

8. heena

ok so first write down half oxidation and half reduction reaction if u are having problem in this step say ok i ll help u

9. King

i dont understand hw to do this prob.if u could explain i could try the remaining..as fr the prob i know that Copper loses 2 e- and i am not sure if Nitrogen loses 1 e- or gains 5 e- bcuz N oxidation state is +5 in reactant side

10. heena

u know NO3- here it charge is +5 and in product ide NO its charge is -2 means from being +5 to -2 is reduction right means its half reduction reaction

11. King

in NO charge is 0

12. King

haan mujhe hindi pata he

13. King

u der?

14. heena

rry actauly O have -2 charge always wid it ok its not evrytime gonna shown u

15. King

oh ok!

16. King

wat abt remaining can u do this 1 fr me wid steps so i can apply and do fr my other 2

17. King

and in NO N charge is +2 not -2 so its oxidation

18. heena

ok now see half reduction reaction half oxidation reaction NO3- -------------> NO CuS----------------S8 +5 ---------------> +2 -2--------->0 [charge]

19. heena

get this much ?

20. King

ok i got it!thnk!

21. heena

now we will balance NO3- -------------> NO H+ + NO3 -----------> NO + H2O ( by adding water balance this equation)

22. King

ok luk at this prob i noted down oxidation and reduction reactions but still its coming wrong

23. King

i got the 1st one

24. King

\[Cr _{2}O _{7^{2-}}+I ^{-} -->Cr ^{3+}+I _{2}\]

25. King

i multiply I by 2 in reactant side,and Cr by 2 in product side first

26. heena

first u say den i ll solve

27. King

then i find that Cr undergoes reduction by gaining 3 e- as it is +6->+3 and I undergoes oxidation by losing 2 e- +2->0

28. King

so i cross multiply and get 28H+ + 2Cr2O7- + 6I- --->4Cr3+ +3I2 + 14H20

29. King

but charges are not balanced so what to do?

30. King

@heena u der?

31. heena

give me a sec

32. King

kk

33. heena

srry wait

34. heena

i m gettin 14H+ + cr2o7-2----------------> 2Cr3+ + 7/2H2O + 15e-

35. King

if u put 7/2 H20 then masses are not balanced

36. King

and in ure equation charges are not balanced

37. heena

wen u do blcing in equtaion masses dont make any effect ok only mole volume dey do so

38. King

no but my sir wants masses and chrges both balanced

39. heena

iabove eqn is wrong 7H+ + cr2o7-2----------------> 2Cr3+ + 7/2H2O + 8e- +5 -5 the only prob is sigm is different in charge :(

40. Callisto

Hello :)

41. heena

ok u done this with ion electron method

42. Callisto

Eh... It's based on change in oxidation number..

43. heena

but the way d way i m doing here wats troubling here @callistion can u guide?

44. Callisto

+6, -1, +3, 0 are oxidation numbers.. and one marked with pencil is the change

45. heena

yea i get that thing wat u done but i m thig wat i have done wrong here (:

46. Callisto

gimme some time to write , please?

47. heena

anyway i ll do dis qn later hope callisto will guide u in above problem @king bye

48. King

@heena bye@callisto thnx a lot and bye!

49. Callisto

People are leaving.. :(

50. Callisto

Do you understand...?

51. heena

y u upsad ? dont worry i m still here :)

52. Callisto

Good :) Do you understand?

53. heena

yes mam :)

54. Callisto

Great! and my mission's completed?

55. heena

:) are u good in bio i need help?

56. Callisto

i don't study Biology, sorry :(

57. heena

no prob :) u knw hydrocarbon?