RelativeResourceManager(14) - by 1 and the reducing agent...

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14 of 17 A. Balancing Redox Equations To balance a redox equation follow these steps: Example: Take the reaction of K 2 Cr 2 O7 with FeSO 4 in the presence of H 2 SO 4 : K 2 Cr 2 O 7 + H 2 SO 4 + FeSO 4  Cr 2 (SO 4 ) 3 + Fe 2 (SO 4 ) 3 + K 2 SO 4 + H 2 O 1. First assign oxidation numbers to each atom: +1 +6 -2 +1 +6 -2 +2 +6 -2 K 2 Cr 2 O 7 + H 2 S O 4 + Fe S O 4 +3 +6 -2 +3 +6 -2 +1 +6 -2 +1 -2 Cr 2 (S O 4 ) 3 + Fe 2 (S O 4 ) 3 + K 2 S O 4 + H 2 O 2. Identify the oxidizing and reducing agents by the change in their oxidation number and set up a partial equation for each. a. The oxidizing agent in this reaction is K 2 Cr 2 O 7 because the Cr 6 + ion gains electrons and is reduced to Cr 3 +. +6 +3 K 2 Cr 2 O 7 + 6e -  Cr 2 (SO 4 ) 3 b. The reducing agent is FeSO 4 because the Fe 2 + ion loses an electron and is oxidized to Fe 3 +. +2 +3 FeSO 4 - 1e -  1/2Fe 2 (SO 4 ) 3 3. Balance the number of electrons exchanged in the reaction by multiplying the oxidizing agent
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Unformatted text preview: by 1 and the reducing agent by 6. The equation then becomes: K 2 Cr 2 O 7 + H 2 SO 4 + 6FeSO 4 Cr 2 (SO 4 ) 3 + 3Fe 2 (SO 4 ) 3 + K 2 SO 4 + H 2 O 4. The coefficients of H 2 SO 4 and H 2 O are still unknown. If we add the number of sulfates (SO 4 ) on each side of the equation, we see that the coefficient for H 2 SO 4 must be 7 and therefore 7 for H 2 O. The balanced equation is: K 2 Cr 2 O 7 + 7H 2 SO 4 + 6FeSO 4 Cr 2 (SO 4 ) 3 + 3Fe 2 (SO 4 ) 3 + K 2 SO 4 + 7H 2 O B. Determination of Equivalent Weights of Oxidizing and Reducing Agents. The equivalent weight of an oxidizing or reducing agent is equal to the molecular weight of that substance divided by the total electron change per formula unit....
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This note was uploaded on 06/21/2011 for the course NRES 201 taught by Professor Olson,k during the Spring '08 term at University of Illinois, Urbana Champaign.

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