Can someone help me to balance this redox equation?
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cloudian
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Can someone help me to balance this redox equation?


Cr2S3 + K2CrO4 + H2SO4 = K2Cr2O7 + Cr2(SO4)3 + K2SO4 + H2O
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ChenBeier
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What did you try so far?
What are the redoxpairs
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cloudian
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https://prnt.sc/m5bkdv
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ChenBeier
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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 H2O

Balance Charge gives

CrO4 2- + 8 H+ +3 e- => Cr 3+ + 4 H2O

Oxidation S2- + 4 H2O => SO4 2- + 8 H+


Balance Charge gives

S2- + 4 H2O => SO4 2- + 8 H+ + 8 e-

KGV of elektrones is 24

You get

8 CrO4 2- + 64 H+ + 24 e- => 8 Cr 3+ + 32 H2O

3 S2- + 12 H2O =>3 SO4 2- + 24 H+ + 24 e-


Addition

8 CrO4 2- + 40 H+ + 3 S 2- => 8 Cr 3+ + 3 SO4 2- +20 H2O

That is the basic reaction.

Now add of 2 Cr3+


8 CrO4 2- + 40 H+ + Cr2S3 => 10 Cr 3+ + 3 SO4 2- +20 H2O

add K+ and SO4 2-


8 K2CrO4 + 20 H2SO4 + Cr2S3 => 5 Cr2 (SO4)3 + 8 K2SO4 + 20 H2O

Add 2 more K2CrO4 + one H2SO4 to get dichromate.


10 K2CrO4 + 21 H2SO4 + Cr2S3 => K2Cr2O7 + 5 Cr2 (SO4)3 + 9 K2SO4 + 21 H2O

finished
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