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cloudian
Newbie
Posts: 2 Joined: Thu Jan 10, 2019 2:22 am
Location: Turkey
Post
by cloudian » Thu Jan 10, 2019 2:27 am
Code: Select all
Cr2S3 + K2CrO4 + H2SO4 = K2Cr2O7 + Cr2(SO4)3 + K2SO4 + H2O
ChenBeier
Distinguished Member
Posts: 1569 Joined: Wed Sep 27, 2017 7:25 am
Location: Berlin, Germany
Post
by ChenBeier » Thu Jan 10, 2019 3:18 am
What did you try so far?
What are the redoxpairs
cloudian
Newbie
Posts: 2 Joined: Thu Jan 10, 2019 2:22 am
Location: Turkey
Post
by cloudian » Thu Jan 10, 2019 4:16 am
ChenBeier
Distinguished Member
Posts: 1569 Joined: Wed Sep 27, 2017 7:25 am
Location: Berlin, Germany
Post
by ChenBeier » Thu Jan 10, 2019 4:44 am
That is not the way how to do
One redoxpair is CrO4 2- and Cr3 +
The second one is S2 - and SO4 2-.
In acidic condition the oxidation needs water to produce H+
The reduction consumes H+ and produces water.
In this case for Reduction
CrO4 2- + 8 H+ => Cr 3+ + 4 H2 O
Balance Charge gives
CrO4 2- + 8 H+ +3 e- => Cr 3+ + 4 H2 O
Oxidation S2 - + 4 H2 O => SO4 2- + 8 H+
Balance Charge gives
S2 - + 4 H2 O => SO4 2- + 8 H+ + 8 e-
KGV of elektrones is 24
You get
8 CrO4 2- + 64 H+ + 24 e- => 8 Cr 3+ + 32 H2 O
3 S2 - + 12 H2 O =>3 SO4 2- + 24 H+ + 24 e-
Addition
8 CrO4 2- + 40 H+ + 3 S 2- => 8 Cr 3+ + 3 SO4 2- +20 H2 O
That is the basic reaction.
Now add of 2 Cr3 +
8 CrO4 2- + 40 H+ + Cr2 S3 => 10 Cr 3+ + 3 SO4 2- +20 H2 O
add K+ and SO4 2-
8 K2 CrO4 + 20 H2 SO4 + Cr2 S3 => 5 Cr2 (SO4 )3 + 8 K2 SO4 + 20 H2 O
Add 2 more K2 CrO4 + one H2 SO4 to get dichromate.
10 K2 CrO4 + 21 H2 SO4 + Cr2 S3 => K2 Cr2 O7 + 5 Cr2 (SO4 )3 + 9 K2 SO4 + 21 H2 O
finished