Question

4.In a electrochemical cell consisting of a silver electrode in a solution of AgNO3 and a gold electrode in a solution of Au(NO3)3 at standard conditions, the correct cell potential is

1. 0.70 V      b. 2.30 V

3. 0.90 V      d. -0.90 V

5. Predict which of the following reactions would exhibit an increase in entropy based on the nature of the reactants and products.

1. 2H2(g) + O2(g) → 2H2O(g)

2. 2NO2(g) → N2O4(g)

3. H+(aq) + F-(aq) → HF(aq)

4. BaF2(s) → Ba2+(aq) + 2F-(aq)

5. 2Hg(ℓ) + O2(g) → 2HgO(s)

6. The entropy change exhibited by the reaction below is -233 J/K∙mol at 25 °C.

C2H2(g) + 2H2(g) → C2H6(g)

Also available are the following thermodynamic data at 25 °C:

Substance

∆G°f (kJ/mol)

S° (J/K∙mol) C2H2(g) 209 201 C2H4(g) 68 219 C2H6(g) -33 230 H2O(g) -229 189 C2H5OH(ℓ) -175 161

Using this information, calculate the entropy of H2(g) at 25 °C.

1. 131 J/K∙mol     b. 204 J/K∙mol

3. 102 J/K∙mol     d. 262 J/K∙mol

5. 0 J/K∙mol

8.In an electrochemical cell consisting of a lead electrode in a solution of Pb(NO3)2 and a gold electrode in a solution of Au(NO3)3 at standard conditions

a. the gold electrode will be consumed

   b. the flow of electrons will be from the lead electrode to the gold electrode  

c. the lead electrode is the cathode

d. the cell potential is -1.63 V

9. For the following process with the associated thermodynamic data. At what temperature will it switch from being spontaneous to nonspontaneous or viceversa?

C(graphite) + H2O(g) → CO(g) + H2(g)

∆Hrxn = 131.3 kJ/mol; ∆Srxn = 133.6 J/K∙mol

1. 273 °C     b. 325 °C  

3. 710 °C     d. None, it is spontaneous at all T

Answer 1

4. The electrochemical cell reactions are:

Anode : Ag ---------> Ag^+ + e^- E°(Ag^+|Ag) = 0.80V

Cathode : Au^3+ + 3e^- -------> Au E°(Au^3+|Au) = 1.50V

Now, E°cell = E°cathode - E°anode

=> E°cell = 1.50V - 0.80V = 0.70 V

Hence, option-A that is 0.70V is the correct answer.

5. Entropy increases when something is dissolved, when more molecules are formed and when there is a increase in temperature.

solids have least entropy while gaseous molecules have highest entropy.

In the given options, the reaction BaF2 (s) ------> Ba^2+ (aq) + 2F^- (aq) will increase the entropy because 2 molecules in the are formed in the aqueous state from 1 molecule in the solid state.

Hence, option-d i.e, BaF2(s) ------> Ba^2+ (aq) + 2F^- (aq) is the correct answer.

6. The S° of the Reaction is given by:

S° reaction = S°(C2H6, g) - [ S°(C2H2, g) - 2* S° (H2, g) ]

=> -233 J/K.mol = -230J/K.mol -[ 201 J/K.mol - 2 S°(H2, g) ]

=> 2* S° (H2,g) = -452J/K.mol + 201 J/K.mol

=> S° (H2,g) = -131 J/K.mol

Hence, option-a ie, 131J/K.mol is the correct answer.

8. The standard electrochemical is:

Pb(s)|Pb^2+(aq) || Au^3+(aq)|Au(s)

The cell reactions are:

Anode: Pb (s) -----> Pb^2+ (aq) + 2e^- E° (Pb^2+|Pb) = -0.13V

Cathode : Au^3+ (aq) + 3e^- -----> Au(s) E°(Au^3+|Au) = 1.50V

Thus, E°cell = E°cathode - E°anode

=> E°cell = 1.50V - (-0.13V) = +1.63V

Electron will flow from lead to gold and gold will be deposited at the cathode.

Hence, the correct statement is: electron will from flow from lead electrode to the gold electrode. Option-B is the correct answer.

9. We know that ∆Grxn = ∆Hrxn - T∆Srxn

The reaction is spontaneous when the value of ∆G rxn becomes negative.

Consider ∆Grxn =0 then,

0 = ∆H - T∆S

=> 0 = 131.3 kJ/mol - T* 133.6 J/K.mol

=> 0 = 131300 J/mol - T* 133.6 J/K.mol

Solving, we get, T= 982.78 K

or, T= (982.78 - 273 )°C

=> T = 709 °C (approx)

Now, above this reaction, the reaction will be spontaneous as ∆G becomes negative.

Thus, out of the given options, 710 °C is higher in value than the 709°C.

Hence, option-c i.e, 710°C is the correct answer.