Equilibrium Constant from Nernst Equation
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Q. The emf of the cell Zn|Zn2+(0.1M) || Fe2+|Fe(0.01M) is 0.2905 V. Equilibrium constant for the cell reaction is
Q.
Electra needs to answer this question to master this topic.
Help her calculate the equilibrium constant of the following reaction:
Cu(s)+2Ag+(aq)→Cu2+(aq)+2Ag(s)
E0cell=0.46V
The options below are logK values where K is the equilibrium constant.
- 15.93
- 19.5
- 15.59
15.45
Q. Electra has mastered the basics of Nernst equation. She is now studying equilibrium constants relation with Nernst equation.
Electra was performing experiments with Nernst with a Daniel Cell. She noticed that when the reaction reaches equilibrium, the cell potential goes to 0.
She asked Nernst why this happened. His reply was
Electra was performing experiments with Nernst with a Daniel Cell. She noticed that when the reaction reaches equilibrium, the cell potential goes to 0.
She asked Nernst why this happened. His reply was
- Because all the ions have been depleted.
- Because there is change in temperature
- Because there is no displacement reaction
- Because there is no change in concentration of Cu2+ and Zn2+
Q. E∘ for the cell is Zn |Zn2+ (aq)||Cu2+ (aq)| Cu at 25∘C, the equilibrium constant for the reaction Zn+Cu2+(aq)⇌Cu+Zn2+ (aq) is of the order of
- 10-37
- 10-28
- 10+18
- 10+17
Q. In a cell that utilizes the reaction Zn(s)+2H+(aq)→Zn2+(aq)+H2 (g) addition of H2SO4 to cathode compartment, will
- Increase the E and shift equilibrium to the right
- Lower the E and shift equilibrium to the right
- Lower the E and shift equilibrium to the left
- Increase the E and shift equilibrium to the left
Q. If the E0cell for a given reaction has a negative value, which of the following gives the correct relationships for the values of ΔG0 and Keq ?
- ΔG∘>0;Keq<1
- ΔG∘>0;Keq>1
- ΔG∘<0;Keq>1
- ΔG∘<0;Keq<1
Q. In the acid-base titration
[H3PO4(0.1M)+NaOH(0.1M)] , the e.m.f of the solution is measured by coupling this electrode with a suitable reference electrode. When an alkali is added, the pH of the solution is in accordance with the equation:
Ecell=E∘cell+0.059 pH
For H3PO4:Ka1=10−3, Ka2=10−8, Ka3=10−13
What is the e.m.f of the cell at the IInd end point of the titration if E∘cell at this stage is 1.3805 V
[H3PO4(0.1M)+NaOH(0.1M)] , the e.m.f of the solution is measured by coupling this electrode with a suitable reference electrode. When an alkali is added, the pH of the solution is in accordance with the equation:
Ecell=E∘cell+0.059 pH
For H3PO4:Ka1=10−3, Ka2=10−8, Ka3=10−13
What is the e.m.f of the cell at the IInd end point of the titration if E∘cell at this stage is 1.3805 V
Q. Maxium work that can be done by a chemical cell is equal to the change in gibbs free energy.
- True
- False
Q. EMF of the cell, Ag|AgNO3(0.1M)|KBr(1N), AgBr(s)|Ag is −0.6 V at 298 K
Calculate the Ksp of AgBr at 298 K.
Calculate the Ksp of AgBr at 298 K.
- Ksp=4.0×10−12
- Ksp=3.5×10−12
- Ksp=4.8×10−12
- Ksp=3.8×10−12
Q. The equilibrium constant for the following reaction is 1030. Calculate E∘ (in volts) for the cell at 298 K.
2X2(s)+3Y2+(aq)→2X3+2(aq)+3Y(s)
2X2(s)+3Y2+(aq)→2X3+2(aq)+3Y(s)
- +0.295 V
- -0.295 V
- +0.105 V
- +0.0985 V