Chapter 21 Lecture 4

Chapter 21 Lecture 4 - Writing Spontaneous Redox Reactions...

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Writing Spontaneous Redox Reactions and Ranking Oxidizing and Reducing Agents by Strength SAMPLE (a) Combine the following three half-reactions into three balanced PROBLEM: equations (A, B, and C) for spontaneous reactions, and calculate E o cell for each. (b) Rank the relative strengths of the oxidizing and reducing agents: E o = 0.96 V (1) NO 3 - ( aq ) + 4H + ( aq ) + 3e - NO( g ) + 2H 2 O( l ) E o = -0.23 V (2) N 2 ( g ) + 5H + ( aq ) + 4e - N 2 H 5 + ( aq ) LAN: E o = 1.23 V (3) MnO 2 ( s ) +4H + ( aq ) + 2e - Mn 2+ ( aq ) + 2H 2 O( l ) ut the equations together in varying combinations so as to produce PLAN: Put the equations together in varying combinations so as to produce (+) E o cell for the combination. Since the reactions are written as reductions, remember that as you reverse one reaction for an xidation alance the number of electrons gained and lost without oxidation. Balance the number of electrons gained and lost without changing the E o . In ranking the strengths, compare the combinations in terms of E o cell . 21-42 continue on next slide…
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E o = 0.96 V (1) NO 3 - ( aq ) + 4H + ( aq ) + 3e - NO( g ) + 2H 2 O( l ) E o = -0.23 V (2) N 2 ( g ) + 5H + ( aq ) + 4e - N 2 H + ( aq ) SOLUTION: ) NO - q + 4H + q + 3e - O( + 2H ( o = 0.96 V E o = 1.23 V (3) MnO 2 ( s ) +4H + ( aq ) + 2e - Mn 2+ ( aq ) + 2H 2 O( l ) (a) (1) NO 3 ( aq ) 4H ( aq ) 3e NO( g ) 2H 2 O( l ) E 0.96 V E o ell = 1.19 V (2) N 2 H 5 + ( aq ) N 2 ( g ) + 5H + ( aq ) + 4e - E o = - 0.23 V Rev (1) NO 3 - ( aq ) + 4H + ( aq ) + 3e - NO( g ) + 2H 2 O( l ) (2) N 2 H 5 + ( aq ) N 2 ( g ) + 5H + ( aq ) + 4e - x4 x3 cell o = 96V ) NO( +2H ( N O - q +4H + q +3e - ev 4NO 3 - ( aq ) + 3N 2 H 5 + ( aq ) + H + ( aq ) 4NO( g ) + 3N 2 ( g ) + 8H 2 O( l ) (A) E o = 1.23 V (3) MnO 2 ( s ) +4H + ( aq ) + 2e - Mn 2+ ( aq ) + 2H 2
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Chapter 21 Lecture 4 - Writing Spontaneous Redox Reactions...

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