Electrode Potential
Trending Questions
When do we use the formula delta G=-nFE. and when do we use E=E0-0.059/nlogKc?
Are there any specific conditions for using these two?
In the button cell widely used in watches and other devices the following reaction takes place :
Zn(s)+Ag2O(s)+H2O(l)→Zn2+(aq)+2Ag(s)+2OH−(aq)DetermineΔrG∘andE∘forthereaction.[Given, E∘ZnZn2+=−0.76V and E∘Ag+Ag+0.34V]
4.5 g of aluminium is deposited at cathode from Al3+ solution by a certain quantity of electric charge. The volume of hydrogen produced at STP from H+ ions is solution by the same quantity of electric charge will be?
Which one will be the powerful reducing agent?
- A
- B
- C
- D
(i) EMF of cell = (Oxidation potential of anode) - (Reduction potential of cathode )
(ii) EMF of cell = (Oxidation potential of anode) + (Reduction potential of cathode )
(iii) EMF of cell = (Reduction potential of anode ) + (Reduction potential of cathode )
(iv) EMF of cell = (Oxidation potential of anode) - (Oxidation potential of cathode)
Which of the above relations are correct?
- (iii) and (i)
- (i) and (ii)
- (iii) and (i)
- (ii) and (iv)
Given below are the half-cell reaction:
Mn2++2e⊖→Mn; E⊖=−1.18V
2(Mn3++e⊖→Mn2+; E⊖=+1.51V
The E⊖ for 3Mn2+→Mn+2Mn3+ will be:
(IIT-JEE 2014)
(a) -0.33 V; the reaction will not occur
(b) -0.33 V; the reaction will occur
(c) -2.69 V; the reaction will not occur
(d) -2.69 V; the reaction will occur
Zn→Zn2++2e−, Eo=0.76 VFe→Fe2++2e−, Eo=0.41 V
The EMF for the cell oxidation,
Fe2++Zn→Zn2++Fe(Give your answer upto two decimal only)
- increase
- decrease
- remain unchanged
- depends on the acid
Extraction of gold and silver involves leaching the metal with CN− ion. The metal is recovered by ________________.
calcination followed by roasting.
thermal decomposition of metal complex.
displacement of metal by some other metal from the complex ion.
- roasting of metal complex.
- Negative value of E∘ signifies that the redox couple is a stronger oxidising agent than the H+/H2 couple
- Negative value of E∘ signifies that the redox couple is a stronger reducing agent than the H+/H2 couple
- Polarity of the electrode system and the electrode reaction can be easily predicted.
- Spontaneity and feasibility of the cell can be easily predicted
Which of the following does not corrode when exposed to the atmosphere?
Iron
Copper
Gold
Silver
- Increases by 189 mV
- Decreases by 189 mV
- Increases by 19 mV
- Decreases by 19 mV
FPGAs are power efficient when compared to GPU
Ni(s)+2 Fe3+→ Ni2++2Fe2+ is Ni(s)|Ni2+(aq)||Fe3+(aq), Fe2+(aq)
- True
- False
- 3.01×10−12
- 2.98×10−15
- 3.98×10−20
- 2.89×10−18
Given the standard electrode potentials,
K+K=−2.93 V,
Ag+Ag=0.80 V
Hg2+Hg=0.79V,
Mg2+Mg=−2.37 V
Cr3+Cr=−0.74 V
Arrange these metals in their increasing order of reducing power.
The standard reduction electrode potentials of four elements are
A=-0.250 V B=-0.136 V
C=-0.126 D=-0.402 V
The element that displaces A from its compounds is
B
C
D
None of these
Arrange Ag, Cr, and Hg metals in the increasing order of reducing power.
Given:
Mn2++2e−→Mn;E∘=−1.18eV
2(Mn3++e−→Mn2+);E0=+1.51eV
The E∘for 3Mn2+→Mn+2Mn3+ will be
- -0.33 V; the reaction will not occur
- -0.33 V; the reaction will occur
- -2.69 V; the reaction will not occur
- -2.69 V; the reaction will occur
E∘Ni2+(aq)/Ni(s)=−0.24 V
E∘Fe2+(aq)/Fe(s)=−0.44 V
then calculate the EMF of the cell
- +0.68 V
- −0.68 V
- +0.2 V
- −0.2 V
On what factor does the amount of product formed at an electrode depend?
- It is the potential difference developed between the two ends of a salt bridge in a solution.
- It is the potential difference developed between an electrode and an electrolyte in a solution.
- It is the potential difference developed between electrolyte and salt bridge in a solution.
- None of the above
The tendency of a chemical species to acquire electrons from an electrode is measured by:
Oxidation potential
Reduction potential
- Cell potential
- None of these
Define electrochemical cell.