Balancing Redox Equations in Acidic Solutions. 6 2Cr H 2 O. 6 2Cr 3+ Example 1: Balance the following equation for the reaction of Cr 2 O 7
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1 Balancing Redox Equations in Acidic Solutions Example 1: Balance the following equation for the reaction of Cr 2 O 7 2 with Fe 2+ in an acidic solution. Cr 2 O Fe 2+ 6 Cr 3+ + Fe 3+ Step 1: Divide into halfreactions 2 Cr 2 O 6 7 Cr3+ Fe 2+ 6 Fe 3+ Step 2: Balance atoms other than H & O Cr 2 O Cr 3+ Fe 2+ 6 Fe 3+ Step 3: Balance oxygen atoms by adding H 2 O to the side that needs O. 2 Cr 2 O 7 6 2Cr 3+ Fe 2+ 6 Fe H 2 O Step 4: Balance hydrogen by adding by adding H + to the side that needs H 14H Cr 2 O 7 6 2Cr 3+ Fe 2+ 6 Fe H 2 O Step 5: Balance the charge by adding electrons 14H Cr 2 O 7 6 2Cr H 2 O = +12 = Add 6 e to the left to make both sides equal. 2. Remember electrons are negative therefore add 6 6e + 14H + + Cr 2 O Cr H 2 O (Fe 2+ 6 Fe e) x 6 Step 6: Make electrons gained equal to the electrons lost and then add the two halfreactions. 6e + 14H + + Cr 2 O Cr H 2 O 6Fe 2+ à 6Fe e electrons cancel! 6Fe 2+ (aq) + 14H + 2 (aq) + Cr 2 O 7 (aq) 6 6Fe 3+ (aq)+ 2Cr 3+ (aq) + 7H 2 O (l)
2 Example 2: Balance the following equation in acidic solution. (aq) + H 2 C 2 O 4 (aq) à Mn 2+ (aq) + CO 2 (g) 1. Divide into halfreactions 2. Balance 6 Mn 2+ H 2 C 2 O 4 6 2CO 2 ( 6 Mn 2+ H 2 C 2 O 4 6 2CO 2 3. Balance Oxygen 6 Mn H 2 O H 2 C 2 O 4 6 2CO 2 (already balanced) 4. Balance hydrogen by adding by adding H + to the side that needs H 8H + + MnO4 (aq) 6 Mn 2+ (aq) + 4H 2 O (+8) (1) From +7 to +2 = 5 electrons gained H 2 C 2 O 4 6 2CO 2 + 2H From 0 to +2 = electrons lost 5. a& 6. Balance the charge by adding electrons and make electrons gained equal to the electrons lost and then add the two halfreactions. (5 e + 8H + + MnO4 (aq) 6 Mn 2+ (aq) + 4H 2 O ) x 2 (H 2 C 2 O 4 (aq) 6 2CO 2 (g) + 2H + (aq) + 2e ) x e + 16H + (aq) + 2MnO4 (aq) 6 2Mn 2+ (aq) + 8H 2 O 5H 2 C 2 O CO H e Overall : 6H + (aq) + 2MnO4 (aq) + 5H 2 C 2 O 4 (aq) 6 2Mn 2+ (aq) + 10CO 2(g) + 8H 2 O (l)
3 Example 3: Balance the following equation in acidic solution + H 2 SO 3 6 SO Mn 2+ 6 Mn 2+ H 2 SO 3 6 SO Mn H 2 O H 2 SO 3 + H 2 O 6 SO 4 2 8H Mn H 2 O H 2 SO 3 + H 2 O 6 SO H + 8H Mn H 2 O So we need to add 5 e! H 2 SO 3 + H 2 O 6 SO H = 2+ So we need to add 2 e! (5 e + 8H Mn H 2 O) x 2 (H 2 SO 3 + H 2 O 6 SO H + + 2e) x 5 10e + 16H Mn (3)H 2 O 5H 2 SO 3 + 5H 2 O 6 5SO (4)H e Overall: 5H 2 SO 3 (aq) + 2 (aq) 6 2Mn 2+ (aq) + 5SO 4 2 (aq) + 3H 2 O (l) + 4H + (aq)
4 Example 4: Balance the following equation in acidic solution Zn (s) + NO 3 (aq) à Zn 2+ (aq)+ NO(g) Zn 6 Zn 2+ NO 3 6 NO Zn 6 Zn 2+ nothing to balance NO 3 6 NO + 2H 2 O Zn (s) 6 Zn 2+ (aq) nothing to balance 4H + + NO 3 6 NO + 2H 2 O Zn (s) 6 Zn 2+ +2e 4H + + NO 3 6 NO + 2H 2 O = e + 4H + + NO 3 6 NO + 2H 2 O (Zn (s) à Zn 2+ (aq) +2e) x 3 (3e + 4H + + NO 3 6 NO + 2H 2 O) x 2 3Zn 6 3Zn e 6e + 8H + + 2NO 3 6 2NO + 4H 2 O Overall: 3Zn (s) + 8H + (aq) + 2NO 3 (aq) 6 3Zn 2+ (aq) + 2NO (g) + 4H 2 O (l)
5 Balancing Redox Equations in Basic Solutions The easiest way to balance these equations is to pretend the solution is acidic and add 3 more steps Example 1: Balance the following equation SO 3 2 (aq) + (aq) 6 SO 3 4 (aq) + MnO 2 (aq) (SO SO H + + 2e ) x 3 (3 e + 4H MnO 2 + 2H 2 O) x 2 3H 2 O + 3SO SO H + + 6e 6e (2)H + 6 2MnO 2 + 4(1)H 2 O 2 Overall (in acid): 3SO 3 (aq) + 2H + 2 (aq) + 2 (aq) 6 3SO 4 (aq) + MnO 2 (aq) + H 2 O (l) Step 7: Add the same # of OH as there are H+ to both sides of the equation 2 OH + 3SO H SO MnO 2 + H 2 O + 2 OH Step 8: Combine OH and H + to form H 2 O 2H 2 O + 3SO SO MnO 2 + H 2 O + 2 OH Step 9: Cancel any H 2 O that you can H 2 O + 3SO SO MnO OH Example 2: Balance the following redox equations in basic solution ClO (aq) + CrO 2 (aq) 6 Cl (aq) + CrO 4 2 (aq) (2e + 2H + + ClO 6 Cl + H 2 O) x 3 (2H 2 O + CrO 2 6 CrO H + + 3e) x 2 6e + 6H + + 3ClO 6 3Cl + 3H 2 O 2CrO (1)H 2 O 6 2CrO 4 + 8(2)H + + 6e Overall: (in acid) 2CrO 2 (aq) + 3ClO (aq) + H 2 O (l) 6 3Cl 2 (aq) + 2CrO 4 (aq) + 2H + (aq)
6 Step 7: Add the same # of OH as there are H+ to both sides of the equation 2 OH + 2CrO 2 + 3ClO + H 2 O 6 3Cl + 2CrO H OH Step 8: Combine OH and H + to form H 2 O 2 OH + 2CrO 2 + 3ClO + H 2 O 6 3Cl + 2CrO H 2 O Step 9: Cancel any H 2 O that you can 2 OH (aq) + 2CrO 2 (aq) + 3ClO (aq) 6 3Cl (aq) + 2CrO 4 2 (aq) + H 2 O(l) Example 3: Balance the following redox equation in basic solution Bi(OH) 3 (s) + SnO 2 2 (aq) 6 Bi (s) + SnO 3 2(aq) (3e + 3H + + Bi(OH) 3 6 Bi + 3H 2 O) x 2 (H 2 O + SnO SnO H + + 2e ) x 3 6e + 6H + + 2Bi(OH) 3 6 2Bi + 6H 2 O 3H 2 O + 3SnO SnO H + + 6e Overall 2Bi(OH) 3 (s) + 3SnO 2 2 (aq) 6 2Bi (s) + 3SnO 3 2(aq) + H 2 O (l) Example 4: Since there is no H + in the equation it is already balanced for a basic solution. Balance the following equation in basic solution MnO 2 (aq) + ClO 3 (aq) 6 (aq) + Cl (aq) (2H 2 O + MnO H + + 3e) x 2 6e + 6H + + ClO 3 6 Cl + 3H 2 O 2MnO 2 + 4H 2 O H + + 6e 6e + 6H + + ClO 3 6 Cl + 3H 2 O Overall: (in acid) 2MnO 2 (aq) + ClO 3 (aq) + H 2 O (l) 6 2 (aq) + Cl (aq) + 2H + (aq) Step 7: Add the same # of OH as there are H+ to both sides of the equation 2OH (aq) + 2MnO 2 (aq) + ClO 3 (aq) + H 2 O (l) 6 2 (aq) + Cl (aq) + 2H + (aq) + 2OH (aq)
7 Step 8: Combine OH and H + to form H 2 O 2OH (aq) + 2MnO 2 (aq) + ClO 3 (aq) + H 2 O (l) 6 2 (aq) + Cl (aq) + 2H 2 O (l) Step 9: Cancel any H 2 O that you can 2OH (aq) + 2MnO 2 (aq) + ClO 3 (aq) 6 2 (aq) + Cl (aq) + H 2 O (l)
8 Chemistry 3202 Balancing Equations in Acidic or Basic Solutions 1. Balance the following reactions in acid solution. a. BrO 3 (aq) + SO 2 (g) 6 SO 4 2(aq) + Br 2 (l) b. Cr 2 O 7 2(aq) + S (s) 6 SO 2 (g) + Cr 2 O 3 (s) c. I (s) + NO 3 (aq) 6 IO 3 (aq) + NO (g) d. C 3 H 7 OH (aq) + Cr 2 O 7 2 (aq) 6C 2 H 5 COOH (aq) + Cr 3+ (aq) e. H 2 C 2 O 4 (aq) + (aq) 6 Mn 2+ (aq) + CO 2 (g) f. Fe 2+ (aq0 + Cr 2 O 7 2(aq) 6 Fe 3+ (aq) +Cr 3+ (aq) g. (aq) + SCN (aq) 6 Mn 2+ (aq) + SO 4 2(aq) + CO 2 (g) + N 2 (g) 2. Balance the following reactions in basic solution. a. Ni (s) + NO 3 (aq) 6 NH 4 +(aq) + Ni 2+ (aq) b. Cr 2 O 7 2 (aq) + AsO 2 (aq) 6 Cr 3+ (aq) + AsO 4 3(aq) c. ClO 3 (aq) + C 2 H 4 O (aq) 6 CO 2 (g) + Cl (aq) d. P 4 (s) + IO 3 (aq) 6 H 2 PO 4 2(aq) + I (aq) e. CrO 4 2 (aq) + Sn 2+ (aq) 6 Sn 4+ (aq) + CrO 2 (aq) f. C 3 H 8 O 3 (aq) + (aq) 6 CO 3 2 (aq) + 2(aq) g. Cr(OH) 6 3 (aq) + BrO (aq) 6 CrO 4 2(aq) + Br (aq) 3. Page : # s (new) 4. Page 496: # 7 (old)
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