Question

# 3)The following reaction, calculate ΔH∘rxn(in kJ), ΔS∘rxn(in J/K), and ΔG∘rxn(in kJ) at 25 ∘C. NH4Cl(s)→HCl(g)+NH3(g) 4)...

3)The following reaction, calculate ΔH∘rxn(in kJ), ΔS∘rxn(in J/K), and ΔG∘rxn(in kJ) at 25 ∘C.

NH4Cl(s)→HCl(g)+NH3(g)

4) Electrons flow from right to left (anode to cathode). Anode=Cr(s) in 1M Cr3+
cathode= Fe(s) in 1M Fe3+. A salt bridge containing KNO3(aq) between the beakers.

a) Write a balanced equation for the overall reaction.

b)Calculate E∘cell.

5)Electrons flow from Right to left (anode to cathode). Anode=Ni(s) in Ni2+
cathode= Cd(s) in Cd2+

a)Indicate the half-reaction occurring at Anode and cathode

b)Calculate the minimum voltage necessary to drive the reaction.

6)Consider an electrolysis cell that might be used to electroplate copper onto other metal surfaces. Anode=Cu, Cathode=M

Show the reactions that occur at each.

3) We need folowing data

ΔHoNH4Cl= -314.4 KJ

ΔSoNH4Cl= 94.6 J / mol K

ΔHoHCl= -92.31 KJ

ΔSoHCl= 186.8J / mol K

ΔHoNH3= -45.94KJ

ΔSoNH3= 192.3 J / mol K

ΔHorxn = ΔHoproducts - ΔHoreactants = -45.94 + (-92.13) - (-314.4) = 176.33 KJ / mole =

ΔSorxn = ΔSoproducts - ΔSoreactants = 186.8 + 192.3 - 94.6 = 284.5 J = 0.2845 KJ

ΔGo = ΔHo - T ΔSo = 176.33 - 298 X 0.2845 = 91.546 KJ

4) Anode=Cr(s) in 1M Cr3+
cathode= Fe(s) in 1M Fe3+. A salt bridge containing KNO3(aq) between the beakers.

a) Write a balanced equation for the overall reaction.

Answer : Fe+3(aq) + Cr(s) --> Cr+3(aq) + Fe(s)

b)Calculate E∘cell.

E0cell = E0cathode - E0anode = -0.04 -(-0.74) = 0.78V

5) Half anodic reaction : Ni(s) --> Ni+2 (aq) + 2e

Half cathodic reaction : Cd+2 + 2e --> Cd(s)

The minimum voltage should be greater than zero.

6) anode

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

Cathode will be one with high electrode potential

like : silver

reaction will be

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

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