I need help on Chemistry question 3. In this experiment, the tests that separate

Question

I need help on Chemistry question

3. In this experiment, the tests that separate and identify the Group A metal ions produce AgCl(s) and HgNH2Cl(s), which are both white in color. How can we explain this based on the d electron count of the metal ions?

5. When we add 5% H2O2, an Oxidizing agent, to a mixture that may contain Mn(OH)2(s), Fe(OH)3(s), Ni(OH)2(s), Cr(OH)4^-(aq), and Zn(OH)4^2-(aq), only Mn^2+ and Cr^3+ will be oxidized (to Mn^4+ and Cr^6+, respectively).

a) From a table of standard recution potentials, we find: H2O2 + 2H+ 2e- --> 2H2O Ered = 1.763 V

CrO4^2- + 4H2O + 3e- --> Cr(OH)3 + 5OH- Ered = -0.13 V

Explain why the oxidation of Cr3+ to Cr6+O4^2- is spontaneous in the presence of hydrogen peroxide.

b) Given the above information, what can we deduce about the standard reduction potentials for the redox couples Fe^6+/Fe^3+ and Ni^4+/Ni^2+? Note: Zn^2+ is the highest oxidation state achieved by zinc.

8. When separating a precipitate from a supernatant, we are often instructed to "wash the precipitate to remove all traces of the supernatant liquid which can act as a contaminant." Why should we use only a minimal amount of deionized H2O for this step?

Thank you for your help! I really appreciate it! :)

Explanation / Answer

a) From a table of standard recution potentials, we find: H2O2 + 2H+ 2e- --> 2H2O Ered = 1.763 V

CrO4^2- + 4H2O + 3e- --> Cr(OH)3 + 5OH- Ered = -0.13 V

Explain why the oxidation of Cr3+ to Cr6+O4^2- is spontaneous in the presence of hydrogen peroxide.

Equation -1

H2O2 + 2H+ 2e- --> 2H2O Ered = 1.763 V

Equation-2

CrO4^2- + 4H2O + 3e- --> Cr(OH)3 + 5OH- Ered = -0.13 V

3*Equation_1 - 2*Equation_2 =

3H2O2 + 6H+ +2Cr(OH)3 + 10OH- = 2CrO4^2- + 14H2O

Ecell = 3*1.763 - 2*(-.13) = 5.549V

As Ecell is +ve above cell reaction is spontaneous.

b) Given the above information, what can we deduce about the standard reduction potentials for the redox couples Fe^6+/Fe^3+ and Ni^4+/Ni^2+? Note: Zn^2+ is the highest oxidation state achieved by zinc.

Fe+3 and Ni+2 does not oxidize in presence of H2O2.

Following are the reduction rxns

Equation -1

H2O2 + 2H+ 2e- --> 2H2O Ered = 1.763 V

Simplified Equation -1

Writing only reduced components

2O(-) + 2e- --> 2O(-2) Ered = 1.763 V

Equation-3a

Fe6+ + 3e- --> Fe3+ Ered = E3a

Equation-3b

Ni4+ + 2e- --> Ni2+ Ered = E3b

3*Equation_1 - 2*Equation_3a =

6O- + 2Fe(2+) --> 2Fe(3+) + 6O(-2)

Ecell = 3*1.763 - 2*(E3a) < 0

E3a > 2.6445

Similarly for Ni2+

Equation_1 - Equation_3b =

2O- + Ni(4+) --> Ni(2+) + 2O(-2)

Ecell = 1.763 - E3b < 0

E3b > 1.763

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