A certain electrochemical cell has for its cell reaction: Zn + HgO rightarrow Zn

Question

A certain electrochemical cell has for its cell reaction: Zn + HgO rightarrow ZnO + Hg Which is the half-reaction occuring at the anode HgO + 2e^- rightarrow Hg + O^2- Zn^2+ + 2e^- rightarrow Zn Zn rightarrow Zn^2+ + 2e^- ZnO + 2e^- rightarrow Zn In the following half equation, which is the oxidizing agent NO_3^-(aq) + 4H^+(aq) + 3e^- rightarrow NO(g) + 2H_2O NO_3^- H^+ e^- NO H_2O A metal object is to be gold-plated by an electrolytic procedure using aqueous AuCl_3 electrolyte. Calculate the number of moles of gold deposited in 3.0 min by a constant current of 10.A. 6.2 * 10^-3 mol 9.3 * 10^-3 mol 1.8 * 10^-2 mol 3.5 * 10^-5 mol 160 mol What is the free energy change for the reaction SiO_2(s) + Pb(s) rightarrow PbO_2(s) + Si(s) delta G degree_f(PbO_2) = -217 kJ/mol Delta G degree_f(SiO_2) = -856 kJ/mol The molar heats of fusion and vaporization of benzene are 10.9 kJ/mol and 31.0 kJ/mol respectively. Calculate the entropy changes for the solid rightarrow liquid and liquid rightarrow vapor transitions for benzene. At 1 atm pressure, benzene melts at 5.5 degree C and boils at 80.1 degree C

Explanation / Answer

1). Given reaction

SiO2 (s) + Pb (s)--- > PbO2(s) + Si(s)

Solution: We have to find free energy of above reaction.

Free energy change = Delta G product – delta G reaction

Delta G = Delta G (PbO2(s)) + Delta G Si(s) - [( Delta G SiO2(s) + delta G (Pb(s) ) ) ]

Delta G of elements in their pure state have zero delta G.

Delta Grxn = ( -217 kJ/mol) – ( - 856 kJ/mol )

= [-217 + 856 ] kJ

= 639 kJ

2).

Given:

Molar heat of fusion of benzene = 10.9 kJ/mol

And molar heat of vaporization = 31.0 kJ /mol

We have to calculate change in entropy for solid to liquid and liquid to vapor.

Melting point of benzene =5.5 0C , and boiling point at 80.1 0C

We use following formula to find out delta S of both process.

Delta S = Delta H/ T

Here Delta H is heat of reaction. And T is in K .

Lets convert T in K

Melting T= 5.5 0C + 273.15 = 278.65 K

Boiling T = 80.10C + 273.15 = 353.25 K

Lets use these T and given values of heat for two transition, we find Delta S for both process.

Delta S for transition of solid benzene to liquid:

Delta S = 10.9 E3 J per mol / 278.65 K = 39.12 J / K mol

Delta S for liquid to vapor

Delta S = 31.0 E 3 J per mol / 353.25 K = 87.76 J / K mol

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