Embibe Experts Solutions for Chapter: Electrochemistry, Exercise 1: JEE Main - 15th April 2018

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Embibe Experts Chemistry Solutions for Exercise - Embibe Experts Solutions for Chapter: Electrochemistry, Exercise 1: JEE Main - 15th April 2018

Attempt the free practice questions on Chapter 9: Electrochemistry, Exercise 1: JEE Main - 15th April 2018 with hints and solutions to strengthen your understanding. EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR CHEMISTRY solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Electrochemistry, Exercise 1: JEE Main - 15th April 2018 with Hints & Solutions

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