Question
Please explain the answers thoroughly as well as allthe necessary steps arriving at the answers. Thanks.
Balance the following equation for oxidation reductionreaction: Use the method for halfreactions; the reaction in acidic medium: CuS + NO3- ----->Cu+2 + NO +S8 Please explain the answers thoroughly as well as allthe necessary steps arriving at the answers. Thanks. Balance the following equation for oxidation reductionreaction: Use the method for halfreactions; the reaction in acidic medium: CuS + NO3- ----->Cu+2 + NO +S8
Explanation / Answer
The net ionc equation for the given reaction is - S-2 +NO3- --------> NO + S8 Thus the balanced half reactions are - Oxidation: S-2 ---------> S8 Reduction: NO3- --------> NO Balance the O, H and main elements then we get - Oxidation: 8 S-2 ---------> S8 Reduction: NO3- +4 H+ --------> NO + 2 H2O Balance the charge - Oxidation: 8 S-2 ---------> S8 + 16 e Reduction: NO3- +4 H+ + 3 e --------> NO + 2 H2O multiply the reduction with 16 and oxidation with 3,then add both reactions, then we get the overall balanced equationas - 24S-2 ---------> 3 S8+ 48 e 16 NO3- + 64H+ + 48 e --------> 16 NO +32H2O ---------------------------------------------------------- 24 S-2 + 16NO3- + 64 H+----------> 3 S8 + 16 NO +32H2O --------------------------------------------------------------------- But we left the Cu+2 as a spectator ion,hence - The overall balanced equation is - 24 CuS + 16 NO3- + 64 H+ ----------> 3 S8 + 16 NO+32 H2O + 24 Cu+2 Reduction: NO3- --------> NO Balance the O, H and main elements then we get - Oxidation: 8 S-2 ---------> S8 Reduction: NO3- +4 H+ --------> NO + 2 H2O Balance the charge - Oxidation: 8 S-2 ---------> S8 + 16 e Reduction: NO3- +4 H+ + 3 e --------> NO + 2 H2O multiply the reduction with 16 and oxidation with 3,then add both reactions, then we get the overall balanced equationas - 24S-2 ---------> 3 S8+ 48 e 16 NO3- + 64H+ + 48 e --------> 16 NO +32H2O ---------------------------------------------------------- 24 S-2 + 16NO3- + 64 H+----------> 3 S8 + 16 NO +32H2O --------------------------------------------------------------------- But we left the Cu+2 as a spectator ion,hence - The overall balanced equation is - 24 CuS + 16 NO3- + 64 H+ ----------> 3 S8 + 16 NO+32 H2O + 24 Cu+2 Reduction: NO3- +4 H+ --------> NO + 2 H2O Balance the charge - Oxidation: 8 S-2 ---------> S8 + 16 e Reduction: NO3- +4 H+ + 3 e --------> NO + 2 H2O multiply the reduction with 16 and oxidation with 3,then add both reactions, then we get the overall balanced equationas - 24S-2 ---------> 3 S8+ 48 e 16 NO3- + 64H+ + 48 e --------> 16 NO +32H2O ---------------------------------------------------------- 24 S-2 + 16NO3- + 64 H+----------> 3 S8 + 16 NO +32H2O --------------------------------------------------------------------- But we left the Cu+2 as a spectator ion,hence - The overall balanced equation is - 24 CuS + 16 NO3- + 64 H+ ----------> 3 S8 + 16 NO+32 H2O + 24 Cu+2 Reduction: NO3- +4 H+ + 3 e --------> NO + 2 H2O multiply the reduction with 16 and oxidation with 3,then add both reactions, then we get the overall balanced equationas - 24S-2 ---------> 3 S8+ 48 e 16 NO3- + 64H+ + 48 e --------> 16 NO +32H2O ---------------------------------------------------------- 24 S-2 + 16NO3- + 64 H+----------> 3 S8 + 16 NO +32H2O --------------------------------------------------------------------- But we left the Cu+2 as a spectator ion,hence - The overall balanced equation is - 24 CuS + 16 NO3- + 64 H+ ----------> 3 S8 + 16 NO+32 H2O + 24 Cu+2