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Ex9:
a) KMnO4 { potassium permanganate (K
+
+ MnO4
-
) C=0.2mol/L, added
(buret), violet}
FeSO4 { ferrous sulfate (Fe
2+
+ SO4
2-
) V=10mL, C=0.5mol/L, beaker, Light
green, acidified by H2SO4 }
The redox couples involved: MnO4
-
/ Mn
2+
and Fe
3+
/ Fe
2+
The spectator ions are potassium ion K
+
and sulfate ion SO4
2-
+VII II +II +II +III
The eq of the reaction: MnO4
-
+ Fe
2+
+ …. → Mn
2+
+ Fe
3+
+ …
X +4(-II)=-1 then x=-1+8=+7
Oxidation HR: ( Fe
2+
Fe
3+
+ 1e) x 5
Reduction HR: MnO4
-
Mn
2+
MnO4
-
Mn
2+
+ 4H2O
MnO4
-
+ 8H
+
Mn
2+
+ 4H2O
MnO4
-
+ 8H
+
+ 5e Mn
2+
+ 4H2O
Overall net ionic eq: ne lost=ne gained
1MnO4
-
+ 8H
+
+ 5Fe
2+
→ Mn
2+
+ 4H2O + 5Fe
3+
b) Deduce the chemical equation.
Remark: KMnO4 → K
+
+ MnO4
-
H2SO4 → 2H
+
+ SO4
2-
FeSO4 → Fe
2+
+ SO4
2-
Fe2(SO4)3 → 2Fe
3+
+ 3SO4
2-
What is the formula obtained by combining Fe
3+
and SO4
2-
?
It is Fe
2
(SO4)
3
Chemical equation:
KMnO4 + 4H2SO4 + 5FeSO4 → MnSO4 + 4H2O + 5/2 Fe
2
(SO4)
3
+1/2 K2SO4
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To avoid the presence of fractions as stoichiometric coefficient, we multiply by 2, we obtain:
2KMnO4 + 8H2SO4 + 10FeSO4 → 2MnSO4 + 8H2O + 5 Fe
2
(SO4)
3
+ K
2
SO4
Ionic equation
2(K
+
+ MnO4
-
)(aq) +8(2H
+
+SO4
2-
)(aq) + 10(Fe
2+
+SO4
2-
)(aq)→
2(Mn
2+
+ SO4
2-
)(aq) +8H2O(l)+ 5(2Fe
3+
+3SO4
2-
)(aq) +(2K
+
+SO4
2-
)(aq)
Net ionic equation:
2MnO4
-
(aq) +16H
+
(aq) + 10Fe
2+
(aq)→2Mn
2+
(aq) +8H2O(l)+ 10Fe
3+
(aq)
By dividing by 2 we obtain:
1MnO4
-
+ 8H
+
+ 5Fe
2+
→ Mn
2+
+ 4H2O + 5Fe
3+
c) Volume of the oxidizing solution (oxidizing agent or oxidant)?
The oxidizing solution is KMnO4 solution (since it undergoes reduction)
V of KMnO4 solution??
What are the different reactants in this reaction? KMnO4, FeSO4, H2SO4 (excess)
At equivalence, the reactants are in their SR: (based on the net ionic equation)
n(MnO4
-
)( added from the buret)/1= n(Fe
2+
)(present in the beaker)/5
n(KMnO4)/1=n(FeSO4)/5
C(KMnO4)xV(KMnO4)/1=C(FeSO4)xV(FeSO4)/5
V(KMnO4)=0.5mol/Lx0.01L/5x0.2mol/L=0.005L=5mL
KMnO4 { C=0.2mol/L, added (buret), violet , V =? }
FeSO4 { V=10mL, C=0.5mol/L, beaker, Light green to colorless, acidified by H2SO4 }
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d) How to detect equivalence?
It is detected by change of color from light green to violet( or when the violet color appears or
persists).
e) Determine the molar concentration of manganese ion Mn
2+
obtained in the
final solution.
[Mn
2+
]=?= n/V
Acc to SR: n(MnO4
-
)/1=n(Mn
2+
)/1
then n(Mn
2+
)=n(MnO4
-
)
n(Mn
2+
)=n(KMnO4)= C xV = 0.2mol/L x 0.005L=0.001mol
the volume of the final solution is:
V= V(KMnO4) + V(FeSO4)= 5mL+10mL=15mL (we neglect the volume of
H2SO4)
[Mn
2+
]= n/V= 0.001/15x10
-3
= 0.06 mol/L

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Ex9: a) KMnO4 { potassium permanganate (K+ + MnO4-) C=0.2mol/L, added (buret), violet} FeSO4 { ferrous sulfate (Fe2+ + SO42-) V=10mL, C=0.5mol/L, beaker, Light green, acidified by H2SO4 } The redox couples involved: MnO4-/ Mn2+ and Fe3+/ Fe2+ The spectator ions are potassium ion K+ and sulfate ion SO42+VII –II +II +II +III The eq of the reaction: MnO4- + Fe2+ + …. → Mn2+ + Fe3+ + … X +4(-II)=-1 then x=-1+8=+7 Oxidation HR: ( Fe2+ Reduction HR: MnO4MnO4MnO4- + 8H+ MnO4- + 8H+ + 5e → → → → → Fe3+ + 1e) x 5 Mn2+ Mn2+ + 4H2O Mn2+ + 4H2O Mn2+ + 4H2O Overall net ionic eq: ne lost=ne gained 1MnO4- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+ b) Deduce the chemical equation. Remark: KMnO4 → K+ + MnO4H2SO4 → 2H+ + SO42FeSO4 → Fe2+ + SO42Fe2(SO4)3 → 2Fe3+ + 3SO42• What is the formula obtained by combining Fe3+ and SO42- ? It is Fe2(SO4)3 • Chemical equation: KMnO4 + 4H2SO4 + 5FeSO4 → MnSO4 + 4H2O + 5/2 Fe2(SO4)3 +1/2 K2SO4 To avoid the presence of fractions as stoichiometric coefficient, we multiply by 2, we obtain: 2KMnO4 + 8H2SO4 + 10FeSO4 → 2MnSO4 + 8H2O + 5 Fe2(SO4)3 + K2SO4 • Ionic equation 2(K+ + MnO4-)(aq) +8(2H+ +SO42-)(aq) + 10(Fe2++SO42-)(aq)→ 2(Mn2+ + SO42-)(aq) +8H2O(l)+ 5(2Fe3+ +3SO42-)(aq) +(2K+ +SO42-)(aq) • Net ionic equation: 2MnO4- (aq) +16H+(aq) + 10Fe2+(aq)→2Mn2+(aq) +8H2O(l)+ 10Fe3+ (aq) By dividing by 2 we obtain: 1MnO4- + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+ c) Volume of the oxidizing solution (oxidizing agent or oxidant)? The oxidizing solution is KMnO4 solution (since it undergoes reduction) V of KMnO4 solution?? KMnO4 { C=0.2mol/L, added (buret), violet , V =? } FeSO4 { V=10mL, C=0.5mol/L, beaker, Light green to colorless, acidified by H2SO4 } What are the different reactants in this reaction? KMnO4, FeSO4, H2SO4 (excess) At equivalence, the reactants are in their SR: (based on the net ionic equation) n(MnO4-)( added from the buret)/1= n(Fe2+)(present in the beaker)/5 n(KMnO4)/1=n(FeSO4)/5 C(KMnO4)xV(KMnO4)/1=C(FeSO4)xV(FeSO4)/5 V(KMnO4)=0.5mol/Lx0.01L/5x0.2mol/L=0.005L=5mL d) How to detect equivalence? It is detected by change of color from light green to violet( or when the violet color appears or persists). e) Determine the molar concentration of manganese ion Mn2+ obtained in the final solution. [Mn2+]=?= n/V Acc to SR: n(MnO4-)/1=n(Mn2+)/1 then n(Mn2+)=n(MnO4-) n(Mn2+)=n(KMnO4)= C xV = 0.2mol/L x 0.005L=0.001mol the volume of the final solution is: V= V(KMnO4) + V(FeSO4)= 5mL+10mL=15mL (we neglect the volume of H2SO4) [Mn2+]= n/V= 0.001/15x10-3= 0.06 mol/L Name: Description: ...
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