Chapter-18-Tro

Chapter-18-Tro - Chapter 18: Electrochemistry Review of...

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Unformatted text preview: Chapter 18: Electrochemistry Review of Oxidation-reduction concepts : Oxidation is the loss of electrons and reduction is the gain of electrons . An oxidizing agent is reduced and reducing agent is oxidized . Example: Mn 7+ ( aq ) + 5 Fe 2+ ( aq ) =Mn 2+ ( aq ) + 5 Fe 3+ ( aq ) Mn 7+ with oxidation number of +7 becomes Mn 2+ gaining 5 electrons. So it is reduced and is the oxidizing agent. Five Fe 2+ ions lose one electron each (to become Fe 3+ ) so Fe 2+ is oxidized and is the reducing agent. Review of balancing chemical equations : In balancing the chemical equations, we adjusted the stoichiometric coefficients to ensure that material is balanced. That is, the number of atoms of a given kind on the left hand side of the reaction equals that number on the right hand side of the chemical equation. Now we will discuss a different method for balancing the redox equations using half-reaction ( or half-equation ) method. Balancing Redox equations in acidic solution : (1). Divide the unbalanced Redox equation into two parts: (a). Oxidation half-reaction; (b). Reduction half-reaction Each half-reaction contains the oxidized and reduced forms of one of the species. In identifying these species we look at the species that contain atoms other than O and H. Example: Cr 2 O 7 2- (aq) + I- (aq) = Cr 3+ (aq) + I 2 (s) For this reaction two half-reactions are : Cr 2 O 7 2- (aq) = Cr 3+ (aq); I- (aq) = I 2 (s) (2).Balance these two half equations for atoms other than H and O. Cr 2 O 7 2- (aq) = 2 Cr 3+ (aq); 2 I- (aq) = I 2 (s) (3). Balance O atoms by adding enough H 2 O molecules Cr 2 O 7 2- (aq) = 2Cr 3+ (aq)+ 7 H 2 O(l); 2I- (aq) = I 2 (s) (4). For acidic solution , balance H atoms by adding enough H + (5). Balance the charges by adding enough electrons 6e- + 14H + (aq) + Cr 2 O 7 2- (aq) = 2Cr 3+ (aq)+ 7 H 2 O(l) 2I- (aq) = I 2 (s) + 2e- (6). Multiply both half equations by numbers appropriate to make the number of electrons in these equations equal. 6 I- (aq) = 3 I 2 (s) + 6 e- (7). Combine the half-equations to get the total equation. 6e- + 14H + (aq) + Cr 2 O 7 2- (aq) = 2Cr 3+ (aq)+ 7 H 2 O(l) 6I- (aq) =3I 2 (s) + 6e- 14H + (aq) + Cr 2 O 7 2- (aq)+ 6I- (aq) = 2Cr 3+ (aq)+ 7 H 2 O(l)+ 3I 2 (s) .. (8). Check that atoms and charges balance. Balancing redox reactions in basic solution : balance the equation, MnO 4- (aq)+C 2 O 4 2- (aq) =MnO 2 (s) + CO 3 2- (aq) (a). Divide the reaction into two halves Half-reaction: 1 Half-reaction: 2 MnO 4- (aq)= MnO 2 (s) C 2 O 4 2- (aq) = CO 3 2- (aq) (b). Balance the atoms other than H and O MnO 4- (aq)= MnO 2 (s) C 2 O 4 2- (aq) = 2 CO 3 2- (aq) (c). Balance O atoms by adding enough H 2 O molecules MnO 4- (aq)= MnO 2 (s) + 2 H 2 O(l); 2 H 2 O(l) +C 2 O 4 2- (aq) = 2CO 3 2- (aq) (d). For basic solutions , balance H atoms by adding one H...
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This note was uploaded on 03/06/2010 for the course CHEM 102B taught by Professor A during the Spring '09 term at Vanderbilt.

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Chapter-18-Tro - Chapter 18: Electrochemistry Review of...

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