Question

Can someone explain step by step how to balance this redox reaction?

Answer 1

K2Cr2O7 + H2O + S = SO2 + KOH + Cr2O3

Answer 2

First find out oxidation numbers of elements.

K is +I and it stays same in reaction
O stays -II
H has +I

Pure substance S has oxidation number 0
In SO2 oxidation number of S is +IV , because O has -II

Cr has oxidation number +VI in K2CrO7

Sum of oxidation numbers in a compound is zero.
You can calculate oxidation number of Cr in this way.
MArk it is as X
2 K 2 Cr 7 O
2 x (+I) + 2 x X + 7 x (-II) = 0 -> X = 6 (+ VI)

MAybe you are familiar with this, but I explained
it anyway because that was a bit difficult compound.

Then in Cr2O3 odidation number of Cr is +III

Now you think:

S is oxidated from 0 to +IV
Cr is reduced from +VI to + III

Answer 3

Number of moving electrons in S is 4
Number of moving electrons of Cr is 3

You have to multiply s for 3
and Cr for 3, then there moves 12 electrons
from S to Cr.

Note there are Cr2 in compounds, so multiplier is 2
in front of substance.

Finally, you have to balance H2O and KOH

Answer 4

Sorry, you multiply Cr for 4, of course.

Answer 5

I got constants 2 ,2 , 3 , 3 , 4, 2 for reaction.

then check amount of every elements are same in both sides.