RedOx Lab
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Need 6 groups to prepare solutions
Each group will prepare 50mL of a 0.1M solution
Solid sample or Stock solution ?????
Make your calculations and check them with Mr.
Bookwalter
• Clean and fill reagent bottles with your solution
MgSO4
CuSO4
AgNO3
ZnSO4
FeSO4
H2SO4
Results – Bookwalter
NR
Gas
NR
NR
NR
NR
Gas
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Gas
NR
NR
NR
NR
NR
RedOx reactions
(Single Replacement)
0
0
+2 -2
2 Mg (s) + O2 (g)  2 MgO (s)
Mg = oxidized  more positive
(reducing agent)
O2 = reduced  more negative
(oxidizing agent)
Oxidation of Metals by Acids and Salts
0
+1
+2
0
Zn (s) + 2 HCl (aq)  ZnCl2 (aq) + H2 (g)
Zn = oxidized  more positive
(reducing agent)
H = reduced  more negative
(oxidizing agent)
Oxidation of Metals by Acids and Salts
+1
0
+2
0
H2SO4 (aq) + Fe (s)  FeSO4 (aq) + H2 (g)
Fe = oxidized  more positive
(reducing agent)
H = reduced  more negative
(oxidizing agent)
Oxidation of Metals by Acids and Salts
Ni(NO3)2 (aq) + Fe (s)  Ni (s) + Fe(NO3)2 (aq)
Net Ionic:
+2
0
0
+2
Ni+2 (aq) + Fe (s)  Ni (s) + Fe+2 (aq)
Fe = oxidized  more positive
(reducing agent)
Ni = reduced  more negative
(oxidizing agent)
Activity Series
• Use to predict if a reaction will occur
• Single Replacement
• Any metal can be OXIDIZED by any metal
below it on the series
Examples
Cu (s) + 2AgNO3 (aq)  Cu(NO3)2
Net Ionic:
0
+1
+2
(aq) +
2 Ag (s)
0
Cu (s) + 2Ag+1(aq)  Cu+2 (aq) + 2Ag(s)
Cu = oxidized  more positive
(reducing agent)
Ag = reduced  more negative
(oxidizing agent)
Examples
Cu (s) + 2FeNO3 (aq)  No Reaction
Based on the Activity Series Cu can’t be oxidized by Fe
Meaning that Cu is below Fe on the activity series
Examples
Ni (s) + 2 HCl (aq)  NiCl2
Net Ionic:
0
+1
+2
(aq) +
H2 (g)
0
Ni (s) + 2H+1(aq)  Ni+2 (aq) + H2 (g)
Ni = oxidized  more positive
(reducing agent)
H = reduced  more negative
(oxidizing agent)
Types of RedOx reactions
• Single replacement in acid
– Mg(s) + HCl(aq) 
• Single replacement in H2O
– Ca(s) + H2O(l) 
• Single replacement in aqueous salt
– Cu(s) + AgNO3(aq) 
• Single replacement of nonmetals
– F2(g) + LiBr(aq) 
E. Solution Stoichiometry and
Titrations
• 1. Stoichiometry
Ex. How many grams of H2O form when 25.0 mL of
0.100M HNO3 is neutralized by NaOH?
0.100 mole HNO3
0.025 L
= 2.5 x 10-3 mole HNO3
L
2.5 x 10-3 mole
X grams = 0.045 grams H2O
HNO3(aq) + NaOH (aq)  H2O(l) + NaNO3 (aq)
1 mole
1 mole
18 grams
2. Titrations
• Use a known standard solution to react with an
unknown concentration of solution.
• Equivalence point = same # of moles of
each reactant
• Indicator = a chemical that determines the end
point of reaction (Ie. Equivalence point)
– The indicator will change color
– Phenolphtalein:
CLEAR = ACID
RED = BASE
The end point is neutralized!!!