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Chapt 20_10_ed_stu_07

# Chapt 20_10_ed_stu_07 - Redox Equations involve the...

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Redox Equations involve the transfer of one or more electrons OXIDATION - loss of electrons - increase in oxidation number REDUCTION - gain of electrons - decrease in oxidation number

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OXIDIZING AGENT - causes something to be oxidized - gets reduced in the process. - is an electron acceptor. REDUCING AGENT - causes something to be reduce - gets oxidized in the process. - is an electron donor.
Rules For Balancing Redox Equations by the Half-Reaction Method 1) Break the reaction down into two half-reactions. 2) Balance all atoms except for H’s and O’s. (This may involve adding the expected redox product in some cases.) 3) Balance each half reaction by adding H 1+ 's and H 2 O‘s as needed to balance the atoms in the half reaction. If the reaction is in ACID skip to step 5. 4) For reactions in BASE, convert all H 1+ ’s ions to H 2 O by adding an equal number of OH 1- ions to that of the H 1+ ions to both side of the equation. The OH 1- ’s ions will react with the H 1+ ’s ions converting them H 2 O.

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5) Balance the charges in the half-reactions by adding electrons. 6) Multiply the half-reaction by the least common multiple such that THE NUMBER OF ELECTRONS LOST IS EQUAL TO THE NUMBER OF ELECTRONS GAINED. 7) Add the two half-reactions together, canceling out any H 1+ 's, OH 1- 's and H 2 O's that appear on both sides of the equation.
Fe 2+ + Cr 2 O 7 2- R Fe 3+ + Cr 3+ (acid) Fe 2+ + Cr 2 O 7 2- R Fe 3+ + Cr 3+ (acid) Fe 2+ R Fe 3+ + e 1- Cr 2 O 7 2- R Cr 3+ Cr 2 O 7 2- R 2 Cr 3+ Cr 2 O 7 2- R 2 Cr 3+ + 7 H 2 O 14 H 1+ + Cr 2 O 7 2- R 2 Cr 3+ + 7 H 2 O 14 H 1+ + Cr 2 O 7 2- + 6 e 1- R 2 Cr 3+ + 7 H 2 O (+12) (+6) 6 (Fe 2+ R Fe 3+ + e 1- ) 14H 1+ + 6 Fe 2+ + Cr 2 O 7 2- R 6 Fe 3+ + 2 Cr 3+ + 7 H 2 O Check: both the number of each atom on each side must be balanced and the total charges on each side must be equal.

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Mn 2+ + BiO 3 1- R MnO 4 1- + Bi 3+ (acid) Mn 2+ + BiO 3 1- R MnO 4 1 - + Bi 3+ (acid) Mn 2+ R MnO 4 1- Mn 2+ + 4 H 2 O R MnO 4 1- Mn 2+ + 4 H 2 O R MnO 4 1- + 8 H 1+ Mn 2+ + 4 H 2 O R MnO 4 1- + 8 H 1+ + 5 e 1- BiO 3 1- R Bi 3+ BiO 3 1- R Bi 3+ + 3 H 2 O BiO 3 1- + 6 H 1+ R Bi 3+ + 3 H 2 O BiO 3 1- + 6 H 1+ + 2 e 1- R Bi 3+ + 3 H 2 O
Mn 2+ + 4 H 2 O R MnO 4 1- + 8 H 1+ + 5 e 1 - BiO 3 1- + 6 H 1+ + 2 e 1- R Bi 3+ + 3 H 2 O 2( Mn 2+ + 4 H 2 O R MnO 4 1- + 8 H 1+ + 5 e 1- ) 5 ( BiO 3 1- + 6 H 1+ + 2 e 1- R Bi 3+ + 3 H 2 O ) Now have 10 e 1- on each side 8 H 2 O + 2 Mn 2+ + 30 H 1+ + 5 BiO 3 1- R 5 Bi 3+ + 2 MnO 4 1- + 16 H 1+ + 15 H 2 O 8 H 2 O + 2 Mn 2+ + 30 H 1+ 5 BiO 3 1- R 5 Bi 3+ +2 MnO 4 1- + 16 H 1+ + 15 H 2 O 2 Mn 2+ 14 H 1+ + 5 BiO 3 1- R 5 Bi 3+ + 2 MnO 4 1- + 7 H 2 O

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Reactions in base AsO 3 3- + I 2 R AsO 4 3- + I 1- (base) AsO 3 3- + I 2 R AsO 4 3- + I 1- (base) I 2 R I 1- I 2 R 2 I 1- I 2 + 2 e 1- R 2 I 1- AsO 3 3- R AsO 4 3- H 2 O + AsO 3 3- R AsO 4 3- H 2 O + AsO 3 3- R AsO 4 3- + 2 H 1+ add 2 OH 1- ’s to each side of the equation 2 OH 1- + H 2 O + AsO 3 3- R AsO 4 3- + 2 H 1+ + 2 OH 1- 2 H 2 O 2 OH 1- + AsO 3 3- R AsO 4 3- + H 2 O
2 OH 1- + AsO 3 3- R AsO 4 3- + H 2 O 2 OH 1- + AsO 3 3- R AsO 4 3- + H 2 O + 2e + 1- I 2 + 2 e 1- R 2 I 1- 2 OH 1- + AsO 3 3- + I 2 R 2 I 1- + AsO 4 3- + H 2 O

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CrO 4 2- + HSnO 2 1- R HSnO 3 1- + CrO 2 1- (base) CrO 4 2- + HSnO 2 1- R HSnO 3 1- + CrO 2 1- (base) HSnO 2 1- R HSnO 3 1- HSnO 2 1- + H 2 O R HSnO 3 1- HSnO 2 1- + H 2 O R HSnO 3 1- + 2 H 1+ 2 OH 1- + HSnO 2 1- + H 2 O R HSnO 3 1- + 2 H 1+ + 2 OH 1- 2 OH 1- + HSnO 2 1- + H 2
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