EASY
JEE Main
IMPORTANT
Earn 100

Following figure shows dependence of molar conductance of two electrolytes on concentration. Λm0 is the limiting molar conductivity.

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The number of Incorrect statement(s) from the following is _____

(A) Λm0 for electrolyte A is obtained by extrapolation
(B) For electrolyte B,   Λm Vs c  graph is a straight line with intercept equal to Λm0
(C) At infinite dilution, the value of degree of dissociation approach zero for electrolyte B.
(D) Λm for any electrolyte A or B can be calculated using λ° for individual ions.

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Important Questions on Electrochemistry

MEDIUM
JEE Main
IMPORTANT

The equilibrium constant for the reaction Zns+Sn2+aqZn2+aq+Sns is 1×1020 at 298 K. The magnitude of standard electrode potential of Sn/Sn2+ if EZn2*/Zn0=-0.76 V is _____ ×10-2 V. (Nearest integer)

Given : 2.303RTF=0.059 V

MEDIUM
JEE Main
IMPORTANT

Consider the cell

PtsH2g,1atmH+aq,1M|Fe3+aq,Fe2+aqPts

When the potential of the cell is 0.712 V at 298 K, the ratio Fe2+/Fe3+ is
(Nearest integer)
Given: Fe3++e-=Fe2+,E°Fe3+,Fe2+Pt=0.771 2.303RTF=0.06 V

HARD
JEE Main
IMPORTANT

The electrode potential of the following half cell at 298 K

XX2+0.001MY2+0.01MY is _____ ×10-2 V (Nearest integer)

Given : EX2+X0=-2.36 V

EY2+Y0=+0.36 V

2.303RTF=0.06 V

EASY
JEE Main
IMPORTANT

Which one of the following statements is correct for electrolysis of brine solution?

MEDIUM
JEE Main
IMPORTANT

The logarithm of equilibrium constant for the reaction Pd2++4Cl-PdCl42-   is
(Nearest integer)

Given: 2.303RTF=0.06 V

Pd(aq)2++2e-Pd(s)  Eo=0.83 V

PdCl42-(aq)+2e-Pd(s)+4Cl-(aq)

Eo=0.65 V

MEDIUM
JEE Main
IMPORTANT
The resistivity of a 0.8M solution of an electrolyte is 5×10-3Ωcm. Its molar conductivity is 104Ω-1 cm2 mol-1. (Nearest integer)
HARD
JEE Main
IMPORTANT

At what pH, given half cell MnO4-(0.1M)Mn2+ (0.001 M) will have electrode potential of 1.282 V ? (Nearest Integer)

Given EMnO4-/Mn2+o=1.54 V,2.303RTF=0.059 V

HARD
JEE Main
IMPORTANT

1×10-5 M AgNO3 is added to 1 L of saturated solution of AgBr. The conductivity of this solution at 298 K is _______×10-8 S m-1.

[Given : Ksp(AgBr)=4.9×10-13 at 298 K

λAg+0=6×10-3Sm2 mol-1

λBr-0=8×10-3Sm2 mol-1

λNO3-0=7×10-3Sm2 mol-1