Question

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

Answer 1

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

Answer 2

sry thats S8 in the product side

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

acidic medium

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

in NO charge is 0

Answer 12

haan mujhe hindi pata he

Answer 13

u der?

Answer 14

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

Answer 15

oh ok!

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

get this much ?

Answer 20

ok i got it!thnk!

Answer 21

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

Answer 22

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

Answer 23

i got the 1st one

Answer 24

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

Answer 25

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

Answer 26

first u say den i ll solve

Answer 27

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

Answer 28

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

Answer 29

but charges are not balanced so what to do?

Answer 30

@heena u der?

Answer 31

give me a sec

Answer 32

kk

Answer 33

srry wait

Answer 34

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

Answer 35

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

Answer 36

and in ure equation charges are not balanced

Answer 37

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

Answer 38

no but my sir wants masses and chrges both balanced

Answer 39

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

Answer 40

Hello :)

Answer 41

ok u done this with ion electron method

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

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

Answer 46

gimme some time to write , please?

Answer 47

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

Answer 48

@heena bye@callisto thnx a lot and bye!

Answer 49

People are leaving.. :(

Answer 50

Do you understand...?

Answer 51

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

Answer 52

Good :) Do you understand?

Answer 53

yes mam :)

Answer 54

Great! and my mission's completed?

Answer 55

:) are u good in bio i need help?

Answer 56

i don't study Biology, sorry :(

Answer 57

no prob :) u knw hydrocarbon?