Question

Q.17Balance the following redox reaction by any method showing all the steps:
Cr₂O7² m + Fe (aut
→Cri
(aq)
+ Pen
cap (Acidic medium)
MnO, +
and
MnO2 + 1; (Basic medium)​

Answer 1

Explanation:

Answer

(i) K2Cr2O7+KI+H2SO4→K2SO4+Cr2(SO4)3+I2+H2O

Ref image

Equate increase and decrease of oxidation number

K2Cr2O7+6KI+H2SO4→K2SO4+Cr2(SO4)3+3I2+H2O

Equate number of potassium and then SO42− and then hydropen 

K2Cr2O7+6KI+7H2SO4→4K2SO4+Cr2

THANK YOU

Answer 2

Explanation:

The oxidation half reaction is SO2(g)+2H2O(l)→SO42−+4H+(aq)+2e−.

The reduction half reaction is Cr2O72−(aq)+14H+(aq)+6e−→2Cr3+(aq)+7H2O(l).

The oxidation half reaction is multiplied by 3 and added to the reduction half reaction to obtain the balanced redox reaction.

Cr2O72−+3SO2(g)+2H+(aq)→2Cr3+(aq)+3SO42−(aq)+H2O(l)