Gibb's Energy and Nernst Equation
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On the basis of information available from the reaction:
43Al+O2→23Al2O3ΔG=−827 KJ/mol of O2. The minimum e.m.f. required to carry out
electrolysis of Al2O3 is
4.28V
6.42V
8.56V
2.14V
How does concentration affect EMF?
Cu2+(aq)+2Cl−(aq)→Cu(s)+Cl2(g)
E∘cell is given as −1.02 V.
Which of the following is correct?
- The given reaction is spontaneous
- Reaction will occur whenever Cu2+and Cl− are brought together in an aqueous solution.
- The given reaction is non-spontaneous
- The reaction is in equilibrium
- 0.44V
- -0.34V
- -0.44V
- 0.34V
For an electrochemical cell, . The ratio when this cell attains equilibrium is ______.
Ni2+(aq)+2e−→Ni(s);Eo=−0.28 V
Mg2+(aq)+2e−→Mg(s);Eo=−2.37 V
The standard cell potential for a voltaic cell constructed using the two half reaction is
- - 98430 J
- 98430 J
- - 49215 J
- 96500 J
2Fe3+(aq) + 2I−(aq) → 2Fe2+(aq) + I2(aq)E°cell=0.24 V at 298 K The standard Gibbs energy △Go of the cell reaction is:
[Given that Faraday constant F = 96500 C mol–1]
- – 46.32 kJ mol−1
- – 23.16 kJ mol−1
- 46.32 kJ mol−1
- 23.16 kJ mol−1
Given standard electrode potentials
Fe+++2e−→Fe; E∘=−0.440 V
Fe++++3e−→Fe; E∘=−0.036 V
The standard electrode potential (E∘) for Fe++++e−→Fe++ is
- 0.476 V
- 0.404 V
+ 0.404 V
+ 0.772 V
2Fe3+(aq) + 2I−(aq) → 2Fe2+(aq) + I2(aq)E°cell=0.24 V at 298 K The standard Gibbs energy △Go of the cell reaction is:
[Given that Faraday constant F = 96500 C mol–1]
- – 46.32 kJ mol−1
- – 23.16 kJ mol−1
- 46.32 kJ mol−1
- 23.16 kJ mol−1
Zn | Zn2+ (C1) || Zn2+ (C2) | Zn. For this cell G is negative if
C1 = C2
C1> C2
C2> C1
None
The electrical work done during the reaction at 298 K:
2Hg(l)+Cl2(g)→Hg2Cl2(s), isGiven that E∘Cl2Cl−=1.36V;E∘Hg2Cl2Hg, Cl−=0.27V;PCl2=1 atm,
420.74 kJ mol−1
110.37 kJ mol−1
210.37 kJ mol−1
105.185 kJ mol−1
- True
- False
- 0.34V
- 0.44V
- -0.34V
- -0.44V
- −322 kJ mol−1
- −161 kJ mol−1
- −152 kJ mol−1
- 76 kJ mol−1
Zn(s)|ZnSO4(aq)||CuSO4(aq))|Cu(s)
When the concentration of Zn2+ is 10 times the concentration of Cu2+, the expression for ΔG (inJmol−1) is
[F is Faraday constant; R Is gas constant; T is temperature ; E∘(cell)=1.1v]
- 2.303 RT + 1.1 F
- 1.1 F
- 2.303 RT - 2.2 F
- -2.2 F
E∘Fe2+/Fe=−0.44 V and E∘Fe3+/Fe2+=0.77 V
E∘Fe3+/Fe is:
- 1.21 V
- 0.33 V
- −0.036 V
- 0.036 V
- E∘cell>0, ΔG∘<0
- E∘cell<0, ΔG∘>0
- E∘cell>0, ΔG∘>0
- E∘cell=0, ΔG∘=0
Electra is almost bored now. Stupid Nernst making her solve so many problems.
But wait, this is her last one. If she completes this, she is through as an apprentice!Quick. Tell her what is the Gibb's energy of a Daniel Cell.
Use F=96487 C mol−1
- 21.227 J mol−1
- 21.227 mJ mol−1
- 21.227 kJ mol−1
- 21.227 mol−1
Given standard electrode potentials
Fe+++2e−→Fe; E∘=−0.440 V
Fe++++3e−→Fe; E∘=−0.036 V
The standard electrode potential (E∘) for Fe++++e−→Fe++ is
- 0.476 V
- 0.404 V
+ 0.404 V
+ 0.772 V