1. Which of the reactions below represents oxidation-reduction? I: HBr (g) + Al

Question

1. Which of the reactions below represents oxidation-reduction?

I: HBr (g) + Al (s) —> H2 (g) + AlBr3 (s)

II: HClO4 (aq) + Ti (OH)2 (aq) —> HClO (aq) + TiO2 (s) + 2H2O (l)

III: Na2CO3 (aq) + HCl (aq) —> H2O (l) + CO2 (g) + NaCl (aq)

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2. Which compound is the oxidant (oxidizing agent) in the reaction below?

HCl (aq) + KMnO4 (aq) + FeCl2 (aq) —> MnCl2 (aq) + FeCl3 (aq) + H2O (l) + KCl (aq)

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3. A cell is composed of Zn metal in 0.10 M Zn2+(aq) solution in one half-cell and Cu metal in 0.10 M Cu2+(aq) solution in the other half-cell. The metal plates are connected by a wire, and the solutions are connected by a salt bridge in the standard cell manner. After a given amount of time, it is expected that:

a) Zinc metal dissolves away

b) Copper metal dissolves away

c) Zinc cation precipitates out as zinc metal

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4. What is the voltage an electrochemical cell with an anode of:

zinc metal in 0.01 M Zn(NO3)2(aq) and a cathode of zinc metal in 1.00 M AgNO3 (aq)?

Explanation / Answer

1)

a) HBr + Al ---> H2 + AlBr3

in this reaction

H+ is reduced to H2

and

Al is oxidized to Al+3

so

this is a oxidation-reduction reaction

b)

HClO4 (aq) + Ti (OH)2 (aq) —> HClO (aq) + TiO2 (s) + 2H2O

in this reaction

Ti is oxidized to Ti+4

and

Cl+7 is reduced to Cl+1

so

this is a oxidation -reduction reaction

c)

Na2CO3 (aq) + HCl (aq) —> H2O (l) + CO2 (g) + NaCl (aq)

in this reaction

reactants ---> Na+ , C+4 , 0-2 , H+ , Cl-

products --> H+ , 0-2 , C+4 , Na+ , Cl-

no change in oxidation numbers

so

no oxidation and reduction

2)

we know that

oxidant is the one which undergoes reduction

now

consider the given reaction

HCl (aq) + KMnO4 (aq) + FeCl2 (aq) —> MnCl2 (aq) + FeCl3 (aq) + H2O (l) + KCl (aq)

in this reaction

Mn+7 is reduced to Mn+2

and

Fe+2 is oxidized to Fe+3

so

KMnO4 is the one which undergoes reduction

so

KMnO4 is the oxidant

3)

consider the reduction potentials

Eo Cu+2/Cu = 0.34 V

Eo Zn+2/Zn = -0.76

we

higher reduction potential undergoes reduction

and

the one with lower reduction potential undergoes oxidation

so

Cu+2 + 2e- ---> Cu (s)

Zn (s) ---> Zn+2 + 2e-

so

zinc metal dissolves away

4)

we know that

Eo cell = Eo cathode - Eo anode

Eo cell = Eo Ag+/Ag - Eo Zn+2/Zn

Eo cell = 0.8 - (-0.76)

Eo cell = 1.56 V

so

voltage is 1.56 V

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