BALANCING REDOX doubt

balance the reactions:

(i) SO2 + Na2CrO4 + H2SO4 → Na2SO4 + Cr2(SO4)3 +H2O

(ii) C2H5OH + I2 + OH- → CHI3 + HCO2- + H2O + I-

I am facing the same problem in both the reactions

for eg taking (i) reaction

step I

S4+ → S6+ + 2e-

6e- + 2Cr6+ → (Cr3+)2

step II

combining the reactions in step I we get

3S4+ + 2Cr6+ → 3S6+ + (Cr3+)2

now when we write the S6+ on product side with its compounds in original reaction we have to distribute it in two compounds that are
Na2SO4 & Cr2(SO4)3 ........how can i know that how many moles of S6+ go to Na2SO4 and how many to other ??

Same is the problem with second reaction in which we have to distribute
C2+ in CHI3 & HCO2- and also distribute
I- in CHI3 & I-

plz explain how do we carry out such distributions and if any other way of balancing these reactions

5 Answers

1
pritishmasti ............... ·

somebody plz answer this

1357
Manish Shankar ·

1.
Reactant is 2Na2CrO4, So on product side it should be 2Na2SO4 (balancing of Na)

1
pritishmasti ............... ·

sir why are we looking to balance only Na
although your answer is right

if we use 2Na2SO4 then S will be unbalanced

23
qwerty ·

for 1st , after writing step 2 ,

u shud write the respective compounds of the species that are in step 2 only ,
for 3S6+ & 2Cr3+ , as they are in the same compound, u shud write that compound only ,bcz if u would hav written Na2SO4 , then while balancing Na atoms, u will be adding Na2CrO4 , but since Cr is already balanced , u shud not do that .
so
3SO2 + 2Na2CrO4 → Cr2(SO4)3

now balance Na atoms

3SO2 + 2Na2CrO4 → 2Na2SO4 + Cr2(SO4)3

now using H2SO4 balance SO4 grp of 2Na2SO4 only bcz u already balanced sulphur atoms of Cr2(SO4)3 while balancing with e-

so
3SO2 + 2Na2CrO4 + 2 H2SO4 → 2Na2SO4 + Cr2(SO4)3

balance H now and check if O atoms are also getting balanced simultaneously

3SO2 + 2Na2CrO4 + 2 H2SO4 → 2Na2SO4 + Cr2(SO4)3 + 2H2O

23
qwerty ·

2.
C2- → C2+ + 4e-

I2 + 2e- → 2 I-

so C2- + 2 I2 → C2+ + 4I-

multiply by 2

2C2- + 4 I2 → 2 C2+ + 8I-

i.e

2C2- + 4 I2 → (C2+ + 3I- ) + C2+ + 5 I-

write CHI3 for (C2+ + 3I- ) , HCO2- for C2+,

so

C2H5OH + 4 I2 → CHI3 + HCO2- + 5 I-

balance charge using OH-

C2H5OH + 4 I2 + 6OH- → CHI3 + HCO2- + 5 I-

now balance O atoms using H2O

C2H5OH + 4 I2 + 6OH- → CHI3 + HCO2- + 5 I- + 5 H2O

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