Nernst Equation
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Pt, 12H2(g)|H+(10−8M)|| M+(0.001M)|12H2(g)Pt
- 0.295V
- 0.0295V
- 0.0590V
- Zero
- Precipitation does not occur
- Can't be determined from given data
- Precipitation occurs
- None of these
M|Mn+(0.02 M)||H+(1 M)|H2(g)(1 atm), Pt at 25∘C is 0.81 V. Calculate the valency of the metal if the standard oxidation potential of the metal is 0.76 V
[ E0Cu+2/Cu= 0.34 V, E0Zn+2/Zn= -0.76 V ]
In a separate experiment, 50 ml of 1.4 M NH3 is added to CuSO4 solution. EMF of the cell is:
Kf([Cu(NH3)4])=6×1013
- 0.993 V
- 1.327 V
- 1.467 V
- 0.720 V
Pt|H2(1 atm)|CH3COONa(0.1M)+CH3COOH(0.01M)||NH4Cl(0.2M)+NH4OH(0.1M)|H2(1 atm)|Pt
Given: pKa of CH3COOH and pKb of NH4OH=4.74
- −0.04 V
- 0.04 V
- −0.189 V
- 0.189 V
( <!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> AIPMT Priliminary Examination 2011)
Electra has trouble finding the correct Nernst equation for the cell Zn(s)|Zn2+||Cu2+|Cu(s) at 25∘C
Can you help her?
Ecell=E∘−0.0296 log([Zn2+][Cu2+])
Ecell=E∘−0.0296 log([Cu2+][Zn2+])
Ecell=E∘−0.0296 log(ZnCu)
Ecell=E∘−0.0296 log(CuZn)
<!--td {border: 1px solid #cccccc;}br {mso-data-placement:same-cell;}--> JEE MAIN 2021
- Reduces by 0.03 V
- Increases by 1 V
- Reduces by 1 V
- Increases by 0.03 V
Pb2+aq+2e−→Pb(s) E0=−0.121VPbO2(s)+4H+(aq)+SO2−4(aq)+2e−→PbSO4(s)+2H2O E0=1.71VPb4+(aq)+2e−→Pb2+(aq) E0=1.689V
For the cell PbSO4(s)+2e−→Pb(s)+SO2−4(aq),
The E0 is missing but it is found that in an aqueous solution of H2SO4 having a pH=1.7, when PbSO4(s) is put, then at saturated condition
Solubility of PbSO4=5×10−5Mol/L
(Take 2.303RTF=0.06 and log2=0.3)
List-I contains questions and List-II contains their answers.
List-IList-IIWhat is the E0(in V) for the half-cell(I) PbSO4(s)+2e−→Pb(s)+SO2−4(aq)?(P) -0.168What is the E0cell(in V) of a lead storage battery? It has(II) the cell reaction asPb(s)+PbO2(s)+2H2SO4(aq)→2PbSO4(s)+2H2O(l)(Q) 0.7What is the E0(in V) for the half cell(III)PbO2(s)+4H+(aq)+4e−→Pb(s)+2H2O(I)(R) 2.0Ksp is the equilibrium constant for the reaction(IV) PbO2(s)+4H+(aq)⇌Pb4+(aq)+2H2O(l)(S) -0.068What is the value of Eo (in V)?(T) -0.31(U) 2.02
Which of the following options has the correct combination considering the reactions in List-I and their answers in List-II?
- (II), (R) and (IV), (S)
- (II), (R) and (IV), (P)
- (II), (U) and (IV), (S)
- (II), (U) and (IV), (P)
Pt (s) | Cl−(C1) | Cl2(g)(P1 atm) || Cl2(g)(P1 atm)| Cl−(C2)| Pt (s)
Identify in which of the following condition working of cell takes place:
- C1 > C2 and P1 = P2
- P2 > P1 and C1 = C2
- C1 < C2 and P1 = P2
- P2 < P1 and C1 = C2
- 0 V
- 0.136 V
- 0.04135 V
- Data insufficient
The solubility product (Ksp:mol3dm−9) of MX2 at 298 K based on the information available the given concentration cell is (take 2.303×R×298/F=0.059V)
- 1×10−15
- 4×10−15
- 1×10−12
- 4×10−12
- Zn(s)|Zn2+(aq)(0.02M)||H3O+(aq)(0.1M)|H2(g)(0.5atm), Pt
- Cu(s)|Cu2+(aq)(0.25M)||Ag+(aq)(0.5M)|Ag
- Cd(s)|Cd2+(aq)(0.01M)||H+(pH=1)|H2(g)(1atm), Pt
- Zn(s)|Zn2+(aq)(0.1M)||H+(pH=1)|H2(g)(1atm), Pt
M|Mn+(0.02 M)||H+(1 M)|H2(g)(1 atm), Pt at 25∘C is 0.81 V. Calculate the valency of the metal if the standard oxidation potential of the metal is 0.76 V
Pt(s)|H2(g, 1bar)|H+(aq, 1M)||M4+(aq), M2+(aq)|Pt(s)
Ecell=0.092V when[M2++(aq)][M4+(aq)]=10x
Given:E∘M4+/M2+=0.151V;2.303RTF=0.059V
The value of x is
- -2
- -1
- 1
- 2
- 0.177 V
- −0.177 V
- 0.087 V
- 0.059 V
- 0.136 V
- Data insufficient
- 0.04135 V
- 0 V
The e.m.f. of the cell at 25∘C
CuCu2+(0.01M)||Ag+(0.1M)Ag
is [E0(Cu2+Cu)=0.34V and E0(Ag+Ag)=0.80V]
0.46V
1.14V
0.43V
1.11V
- Zn(s)|Zn2+(aq)(0.02M)||H3O+(aq)(0.1M)|H2(g)(0.5atm), Pt
- Cu(s)|Cu2+(aq)(0.25M)||Ag+(aq)(0.5M)|Ag
- Cd(s)|Cd2+(aq)(0.01M)||H+(pH=1)|H2(g)(1atm), Pt
- Zn(s)|Zn2+(aq)(0.1M)||H+(pH=1)|H2(g)(1atm), Pt
Calculate the emf of the cell
Pt, H2(1.0atm) | CH3COOH (0.1M) || NH3(aq, 0.001M) | H2(1.0 atm), Pt Ka(CH3COOH) = 10−5, Kb(NH3) = 10−5
+0.368 V
-0.413 V
-0.233 V
+0.567 V
One day Mr.Nernst told her to find the EMF of the following cell using experiments at 25∘C
Mg(s)+2Ag+(0.0001M)→Mg2+(0.130M)+2Ag(s)
Electra reported the value to be +2.6V. Mr.Nernst told her to recheck the value using his theory, knowing Standard EMF = 3.17 V.
What was the right value? Round off the answer to nearest integer.
Type your answer here.
[ E0Cu2+/Cu= 0.34 V, E0Zn2+/Zn= -0.76 V ]
The emf of the cell increases when small amount of concentrated NH3 is added to:
- ZnSO4 solution
- CuSO4 solution
- Both ZnSO4 and CuSO4
- Can't say
The Edison storage cell is represented as :
Fe(s)/FeO(s)/KOH(aq)/Ni2O3 (s)/Ni(s)
The half-cell reactions are:
Ni2O3(s)+H2O(l)+2e−→2NiO(s)+2OH−;E∘+0.40
FeO(s)+H2O(l)+2e−→Fe(s)+2OH−;E∘=−0.87,
What is the maximum amount of electrical energy that can be obtained from one mole of Ni2O3 ?
344.6 KJ
87.9 KJ
2478.8 KJ
245.11 KJ
- True
- False
Zn|ZnSO4 (0.01M)|| CuSO4(1.0M)Cu, the emf of this Daniell cell is E1. When the concentration of ZnSO4 is changed to 1.0 M and that of CuSO4 changed to 0.01 M, the emf changes to E2. From the followings, which one is the relationship between E1 and E2? (Given, RT/F = 0.059)
- E1<E2
- E1>E2
- E2=−E1
- E1=E2
- −(E1+E2)2×0.059
- (E1+E2)2×0.059
- (E2−E1)2×0.059
- −(E1+E2)0.059