MEDIUM
JEE Main/Advance
IMPORTANT
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When 9.65 ampere current was passed for 1.0 hour into nitrobenzene in acidic medium, the amount of p-aminophenol produced is:

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Important Questions on Electrochemistry

MEDIUM
JEE Main/Advance
IMPORTANT
The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 Faraday. The amount of PbSO4  electrolyzed in g during the process is: (Molar mass of PbSO4=303g mol-1)
MEDIUM
JEE Main/Advance
IMPORTANT
If the standard electrode potential for a cell is 2 V at 300 K, the equilibrium constant K for the reaction.

Zns+Cu2+aqZn2+aq+Cus 

at 300 K is approximately:

R=8 JK-1mol-1, F=96000 C mol-1
EASY
JEE Main/Advance
IMPORTANT
Consider the following reduction processes
Zn2++2e-Zns; Eo=-0.76 V
Ca2++2e-Cas; Eo=-2.87 V
Mg2++2e-Mgs; Eo=-2.36 V
Ni2++2e-Nis; Eo=-0.25 V

The reducing power of the metals increases in the order
HARD
JEE Main/Advance
IMPORTANT
In the cell, PtsH2g, 1bar HCl aqAgClsAgsPts,the cell potential is 0.92 V when a 10-6 molar HCl solution is used. The standard electrode potential of Ag|AgCl|Cl- electrode is:

(Given, 2.303RTF=0.06 V at 298 K)
MEDIUM
JEE Main/Advance
IMPORTANT
For the cell Zn(s)|Zn2+(aq)||Mx+(aq)|M(s), different half cells and their standard electrode potential are given bellow:
Mx+aq/Ms Au3+aq/Aus Ag+aq/Ags Fe3+aq/Fe2+aq Fe2+aq/Fes
EoMx+/Ms in Volt 1.40 0.80 0.77 -0.44
If EZn2+/Zno=-0.76, Which cathode will give a maximum value of E°cell per electron transferred?
EASY
JEE Main/Advance
IMPORTANT
Given the equilibrium constant: Kc of the reaction:

Cus+2Ag+aqCu2+aq+2Ags is 10×1015 , calculate the Ecello of this reaction at 298 K

2.303RTFat298K=0.059V
HARD
JEE Main/Advance
IMPORTANT

The standard electrode potential Eo and its temperature coefficient dEdT for a cell are 2V and -5×10-4 V K-1 at 300 K, respectively. The reaction is Zn s+Cu2+ aqZn2+ aq+Cu s. The standard reaction enthalpy ΔrH- at 300K in mol-1 is 

[Use R=8 J K-1 mol-1 and F=96,500 Cmol-1]

HARD
JEE Main/Advance
IMPORTANT
mo for NaCl,HCl and NaA are 126.4,425.9 and 100.5Scm2mol-1 respectively. If the conductivity of 0.001MHA is 5×10-5 S cm-1, degree of dissociation of HA is