3)The following reaction, calculate H rxn(in kJ), S rxn(in J/K), and G rxn(in kJ
ID: 882669 • Letter: 3
Question
3)The following reaction, calculate Hrxn(in kJ), Srxn(in J/K), and Grxn(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 Ecell.
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.
Express your answers as chemical equations separated by a comma. IDENTIFY ALL OF THE PHASES in your answer.
Explanation / Answer
3) We need folowing data
HoNH4Cl= -314.4 KJ
SoNH4Cl= 94.6 J / mol K
HoHCl= -92.31 KJ
SoHCl= 186.8J / mol K
HoNH3= -45.94KJ
SoNH3= 192.3 J / mol K
Horxn = Hoproducts - Horeactants = -45.94 + (-92.13) - (-314.4) = 176.33 KJ / mole =
Sorxn = Soproducts - Soreactants = 186.8 + 192.3 - 94.6 = 284.5 J = 0.2845 KJ
Go = Ho - T So = 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 Ecell.
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)
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.