Gibb's Free Energy
Trending Questions
(JEE MAIN 2020)
- −0.182 V
- +0.158 V
- 0.182 V
- −0.158 V
Cu2+(aq)+2Ag(s)→Cu(s)+2Ag+(aq)
Given:E0Cu2+/Cu=0.34V, E0Ag/Ag+=−0.8V
- -89 kJ
- 89 kJ
- -44.5 kJ
- 44.5 kJ
(1F=96500C mol−1).
Identify the following equation
Give examples of slow reaction
Cu(s)+2Ag+(1×10−3M)→Cu2+(0.250 M)+2 Ag(s)
E0cell=2.97 V
Ecell for the above reaction is ____V. (Nearest integer)
[Given : log 2.=0.3979, T=298 K]
[ P is pairing energy ]
- 0.6+P
- −0.4+P
- 1.2+2P
- 2.4+4P
Sn, Al, Cr, Mn, Pb, Ca, Na, Zn
Pt(s)|Fe2+(aq, 0.1 M), Fe3+(aq, 0.01 M);E0Fe3+/Fe2+=+0.77 V
- EFe3+/Fe2+=0.77 V
- EFe3+/Fe2+=−0.83 V
- EFe3+/Fe2+=0.711 V
- EFe3+/Fe2+=0.83 V
- 14 J
- 27 J
- -14 J
- -27 J
[X] +Zn→[Y]+Au
Identify the compounds [X] and [Y].
- [X]=[Au(CN)2]−, [Y]=[Zn(CN)4]2−
- [X]=[Au(CN)4]3−, [Y]=[Zn(CN)4]2−
- [X]=[Au(CN)2]−, [Y]=[Zn(CN)6]4−
- [X]=[Au(CN)4], [Y]=[Zn(CN)6]2−
H2(g)+2AgCl(s)→2Ag(s)+2HCl(aq)
- 5.2×108
- 5.2×106
- 5.2×103
- 2.8×107
(Given, log4=0.6)
- 4
- 2
- 3
- 6
(I) Leaching
(II) Reduction
(III) Complexation
Out of the given processes the one(s) which are used in the cyanide process is/are
- Only (I) and (III)
- (I), (II) and (III)
- Only (I) and (III)
- Only (II) and (III)
Hydrogen peroxide decomposes into water and oxygen. The uncatalyzed reaction has activation energy of . The value in the presence of acetanilide is and in the presence of it is . What conclusion can you draw from the above observations?
The metal extracted by leaching with cyanide is:
The emf of the given cell is found to be 0.42 V at 25∘C. Calculate the standard potential of the half reaction.
M3+(aq) + 3e−→M(s)
Given:
E0Ag+/Ag = 0.80 V
- −0.041 V
- −0.723 V
- 0.315 V
- 0.124 V
For a thermodynamically spontaneous process the change in free energy is
- Cr2+ ion is more stable than Fe2+.
- Cr3+ ion with d3 configuration has favourable crystal field stabilisation energy.
- Cr3+ has half-filled configuration and hence more stable
- Fe3+ in aqueous solution is more stable than Cr3+.
- Fe2+ ion with d6 configuration has favourable crystal field stabilisation energy.
Find the value of the E0Ag+/Ag
- 0.8 V
- 1.2 V
- 0.34 V
- 1.4 V
MnO2→Mn3+ E∘=0.95V
Mn+3→Mn2+ E∘=1.51V
The standard potential of MnO2→Mn2+ is :
- −0.56V
- −2.46V
- −1.23V
- 1.23V
Calculate the EMF of the cell.
Given:
E0Ag+/Ag=0.80 V
E0Cu2+/Cu=0.34 V
- Ecell= 0.301 V
- Ecell= 0.383 V
- Ecell= 0.45 V
- Ecell= 1.04 V
The E∘ in the given diagram is
0.52 V
0.61 V
0.74 V
0.83 V
Write the chemical reactions involved in the extraction of gold by cyanide process. Also give the role of zinc in the extraction.
are extracted respectively from their chief ore by
(a) Carbon reduction and self reduction
(b) Self reduction and carbon reduction
(c) Electrolysis and self reduction
(d) Self reduction and electrolysis
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
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
(i) SO2 C12(g)Δ−→SO2(g)+C12(g)
(ii) nCH2=CH2Catalyst−−−−−→[−CH2−CH2−]n(s)
(iii) I2(s)Δ−−−→1atmI2(v)
(iv) Adiabatic reversible expansion of an ideal gas
- +, -, 0, 0
- +, -, 0, +
- -, +, +, 0
- +, -, +, 0
- −0.6△o and −0.8△t
- −0.4△o and −0.8△t
- −2.4△o and −1.2△t
- −0.4△o and −1.2△t
Ag(s)|AgNO3(aq, C1 M)||AgNO3(aq, C2=0.2 M)|Ag(s)
The emf of the cell is 0.8 V
Take
Antilog (13.5)=3.16×1013
- C1=3.16×10−13
- C1=3.16×1013
- C1=6.32×1013
- C1=0.63×10−14