Gibb's Energy and Nernst Equation
Trending Questions
Q. The electrochemical cell shown below is a concentration cell.M|M2+ (Saturated solution of a sparingly soluble salt, MX2)||M2+(0.001 mol dm−3)M. The emf of the cell depends on the difference in concentration of M2+ ions at the two electrodes. The emf of the cell at 298K is 0.059
The value of ΔG(kJmol−1) for the given cell is: (take1F=96500 C mol−1)
The value of ΔG(kJmol−1) for the given cell is: (take1F=96500 C mol−1)
- 5.7
- 11.4
- −5.7
- −11.4
Q. The half-cell reactions for rusting of iron are,
2H++12O2+2e−→2H2O, E∘=+1.23V,
Fe2++2e−→Fe(s); E∘=−0.44V.
ΔG∘ (in kJ) for the reaction is :
2H++12O2+2e−→2H2O, E∘=+1.23V,
Fe2++2e−→Fe(s); E∘=−0.44V.
ΔG∘ (in kJ) for the reaction is :
- – 322
- – 122
- – 176
- − 76
Q. Calculate standard free energy change for the reaction 12Cu(s)+12Cl2(g)⇌12Cu2++Cl−taking place at in a cell whose standard e.m.f. is 1.02 volts
- – 98430 J
- 98430 J
- 96500 J
- – 49215 J
Q.
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
Q. The rusting of iron takes place as follows 2H++2e−+12O2→H2O(l); E∘=+1.23 V Fe2++2e−→Fe(s); E∘=−0.44 V Calculate ΔG∘ for the net process
- -322 kJ mol-1
- -161 kJ mol-1
- -152 kJ mol-1
- -76 kJ mol-1
Q.
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
2.14V
6.42V
8.56V
Q. For the following cell:
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]
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 - 2.2 F
- 1.1 F
- -2.2 F
- 2.303 RT + 1.1 F
Q. Cool! Electra has mastered Equilibrium constants and has moved on to Gibb's energy.
Electra was thinking, "When we calculate the Gibb's energy for a SHE using electrode potential, we get 0.
So is it right to say that this half-cell ‘has no change in Gibb's energy’ ?"
What will you tell her?
Electra was thinking, "When we calculate the Gibb's energy for a SHE using electrode potential, we get 0.
So is it right to say that this half-cell ‘has no change in Gibb's energy’ ?"
What will you tell her?
- It's true Electra
- Gibb’s energy cannot be applied here. Foolish Electra.
- Half-cells don’t have electrode potential.
- No way, that's wrong
Q.
Quick. Tell her what is the Gibb's energy of a Daniel Cell.
Use F=96487 C mol−1
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 kJ mol−1
- 21.227 mJ mol−1
- 21.227 mol−1
Q. Electra - "Hmm, this looks tricky"
At 25∘C,
Cr3++e−→Cr2+, given E0=−0.424 VCr2++2e−→Cr(s), given E0=−0.9 VFind E0 at 25∘C for Cr3+(aq)+3e−→Cr(s)
Electra was completely stumped. Seems like you need a way to add electrode potentials. Please tell her the answer.
At 25∘C,
Cr3++e−→Cr2+, given E0=−0.424 VCr2++2e−→Cr(s), given E0=−0.9 VFind E0 at 25∘C for Cr3+(aq)+3e−→Cr(s)
Electra was completely stumped. Seems like you need a way to add electrode potentials. Please tell her the answer.
- 0.741 V
- -1.324 V
- - 0.741 V
-0.476 V