Gibb's Free Energy
Trending Questions
Q. If E∘Au+/Au is 1.69 V & E∘Au3+/Au is 1.40 V, then E∘Au3+/Au+ will be:
- 0.19 V
- 2.945 V
- 1.255 V
- None
Q. The free energy change when 1 mole of NaCl is dissolved in water at 25oC is −x kJ mol−1 Lattice energy of NaCl = 777.8 kJ mol−1; △S for dissolution =0.043 kJ mol−1 and hydration energy of NaCl=−774.1 kJ mol−1. Value of x (rounded off to nearest integer) is
Q. If E∘Au+/Au is 1.69 V & E∘Au3+/Au is 1.40 V, then E∘Au3+/Au+ will be:
- 0.19 V
- 2.945 V
- 1.255 V
- None
Q. E∘(SRP) of different half cells are given:-
E∘Cu2+/Cu=0.34 volt;E∘Zn2+/Zn=−0.76 volt
E∘Ag+/Ag=0.8 volt;E∘Mg2+/Mg=−2.37 volt
Which of the following has the most negative ΔG∘ value?
E∘Cu2+/Cu=0.34 volt;E∘Zn2+/Zn=−0.76 volt
E∘Ag+/Ag=0.8 volt;E∘Mg2+/Mg=−2.37 volt
Which of the following has the most negative ΔG∘ value?
- Zn(s)|Zn2+(1M)||Mg2+(1M)|Mg(s)
- Zn(s)|Zn2+(1M)||Ag+(1M))|Ag(s)
- Cu(s)|Cu2+(1M)||Ag+(1M)|Ag(s)
- Ag(s)|Ag+(1M)||Mg2+(1M)|Mg(s)
Q. The Gibbs energy for the decomposition of Al2O3 at 5000C is as follows:
23Al2O3→43Al+O2
ΔrxnG=+960 kJmol−1.The potential difference needed for the electrolytic reduction of aluminum oxide(Al2O3) at 500oC is:
23Al2O3→43Al+O2
ΔrxnG=+960 kJmol−1.The potential difference needed for the electrolytic reduction of aluminum oxide(Al2O3) at 500oC is:
- 4.5 V
- 3.0 V
- 2.5 V
- 5.0 V
Q. d(ΔG)=−(ΔSreaction).dT ΔG=ΔH−TΔS ΔG=ΔH+T(d(ΔG)dT)P
At 300 K, ΔH for the reaction,
Zn(S)+2AgCl(S)→ZnCl2(aq)+2Ag(S)
is −218 kJ/mol while the e.m.f of the cell is 1.015 V. (dEdT)P of the cell is:
At 300 K, ΔH for the reaction,
Zn(S)+2AgCl(S)→ZnCl2(aq)+2Ag(S)
is −218 kJ/mol while the e.m.f of the cell is 1.015 V. (dEdT)P of the cell is:
- −4.2×10−4 VK−1
- −3.81×10−4 VK−1
- 7.62×10−4 VK−1
- 0.11 VK−1
Q. If the E∘cell for a given reaction has a negative value, which of the following gives the correct relationships for the values of ΔG∘ and Keq ?
- ΔG∘<0;Keq>1
- ΔG∘<0;Keq<1
- ΔG∘>0;Keq<1
- ΔG∘>0;Keq>1
Q. d(ΔG)=−(ΔSreaction).dT ΔG=ΔH−TΔS ΔG=ΔH+T(d(ΔG)dT)P
Calculate ΔS for the given cell reaction in the previous question.
Calculate ΔS for the given cell reaction in the previous question.
- −73.53 JK−1mol−1
- 83.53 JK−1mol−1
- 100 JK−1mol−1
- None of these
Q. d(ΔG)=−(ΔSreaction).dT ΔG=ΔH−TΔS ΔG=ΔH+T(d(ΔG)dT)P
Calculate ΔS for the given cell reaction in the previous question.
Calculate ΔS for the given cell reaction in the previous question.
- −73.53 JK−1mol−1
- 83.53 JK−1mol−1
- 100 JK−1mol−1
- None of these
Q.
For a thermodynamically spontaneous process the change in free energy is
Q.
Calculate the free energy change when 1 mol of NaCl is dissolved in water at 298 K. given:
(i) Lattice energy of NaCl = -778 kJ mol−1
(ii) Hydration energy of NaCl = -778 kJ mol−1
(iii) Entropy change at 298 K = 43 J mol−1
-9.11 kJ
-11.9 kJ
-10.9 kJ
None of these
Q. d(ΔG)=−(ΔSreaction).dT
ΔG=ΔH−TΔS
ΔG=ΔH+T(d(ΔG)dT)P
The temperature coefficient of the e.m.f of a cell, (dEdT)P is given by:
ΔG=ΔH−TΔS
ΔG=ΔH+T(d(ΔG)dT)P
The temperature coefficient of the e.m.f of a cell, (dEdT)P is given by:
- nFΔS
- ΔSnF
- ΔSnFT
- −nFE
Q. What is the ΔG∘ for the following reaction?
Cu2+(aq)+2Ag(s)→Cu(s)+2Ag+(aq)
Given:E0Cu2+/Cu=0.34V, E0Ag/Ag+=−0.8V
Cu2+(aq)+2Ag(s)→Cu(s)+2Ag+(aq)
Given:E0Cu2+/Cu=0.34V, E0Ag/Ag+=−0.8V
- -44.5 kJ
- 44.5 kJ
- -89 kJ
- 89 kJ
Q. The temperature coefficient of emf of a cell designed as Zn|Zn2+(1M)||Cu2+(0.1M)|Cu is −1.4×10−4V per degree. Change in entropy for the reaction Zn+Cu2+0.1M⇌Zn2+1M+Cu at 27∘C is:(E∘cell=1.1V)
- 14 J
- 27 J
- -14 J
- -27 J
Q. The carbon reduction method is mainly applicable for the extraction of how many of the following metals?
Sn, Al, Cr, Mn, Pb, Ca, Na, Zn
Sn, Al, Cr, Mn, Pb, Ca, Na, Zn
Q. For the reaction, Ag+(aq)+Cl−(aq)⇌AgCl(s)
Given:
SpeciesΔG∘f(kJ/mol)Ag+(aq)+77Cl−(aq)−129AgCl(s)−109
Calculate E∘cell at 298 K.
Also find the solubility product of AgCl.
Given:
SpeciesΔG∘f(kJ/mol)Ag+(aq)+77Cl−(aq)−129AgCl(s)−109
Calculate E∘cell at 298 K.
Also find the solubility product of AgCl.
- 0.59 V and Ksp=10−10
- 0.79 V and Ksp=3×10−5
- 0.36 V and Ksp=10−14
- −0.59 V and Ksp=10−10