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Newbie

Joined: 10 Jan 2019
Posts: 2
Location: Turkey

Can someone help me to balance this redox equation?

Cr2S3 + K2CrO4 + H2SO4 = K2Cr2O7 + Cr2(SO4)3 + K2SO4 + H2O

Sr. Staff Member

Joined: 27 Sep 2017
Posts: 244
Location: Berlin, Germany
 What did you try so far? What are the redoxpairs

Newbie

Joined: 10 Jan 2019
Posts: 2
Location: Turkey

Sr. Staff Member

Joined: 27 Sep 2017
Posts: 244
Location: Berlin, Germany
 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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