Flag Physical Chemistry> please please answer this question please...
question mark

please please answer this question please please sir
how many ml of 0.150 M Na2CrO4 will be required to oxidize 40 ml of 0.5 M Na2S2O3
Cr2O42- + S2O32- ------> Cr(OH)4- + So42-
a. 225 ml
b. 335 ml
c. 455 ml
d.555 ml

Mrunal Sonawane , 7 Years ago
Grade 12
anser 1 Answers
Sahani Kumar
 
 
 
Cr2O42-  +  S2O32-  ------> Cr(OH)4+ So42-
in the above equation sulphur (S) is being oxidised from S2+  to S6+ as follows
S2+ -------------->S6+ + 4e-         So n1=4
in the same way Cr is being reduced from Cr3+ to Cr2+ as follows  
Cr3+ + 1e- --------------->Cr2+            So n2= 1
Now we know that 
n1 * M1 * V1 = n2 * M2 * V2
4 * 0.5 * 40 = 1 * 0.150 * V2
V2 = (4 * 0.5 * 40)/(1 * 0.150) = 533 ml ~ 555 ml(which is almost nearer to the value 555 ml. )
Hence according to me the correct answer is 555 ml please let me know if this answer is correct. 
Regards 
Askiitian member
 
 
 
 
Last Activity: 7 Years ago
star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments