Advertisements
Advertisements
Question
The cell constant of a conductivity cell is 0.146 cm-1. What is the conductivity of 0.01 M solution of an electrolyte at 298 K, if the resistance of the cell is 1000 ohm?
Solution
cell constant G* = k × R
k = `("G""*")/"R" = 0.146/ 1000 = 1.46 xx 10-4 "Scm"^-1.`
APPEARS IN
RELATED QUESTIONS
How many electrons flow through a metllic wire if a current of 0·5 A is passed for 2 hours? (Given : 1 F = 96,500 C mol−1)
Why cannot we store AgNO3 solution in copper vessel?
A gas X at 1 atm is bubbled through a solution containing a mixture of 1MY− and 1MZ− at 25°C. If the reduction potential of Z > Y > X, then ____________.
Two metals M1 and M2 have reduction potential values of −xV and +yV respectively. Which will liberate H2 and H2SO4.
Can absolute electrode potential of an electrode be measured?
A galvanic cell has electrical potential of 1.1V. If an opposing potential of 1.1V is applied to this cell, what will happen to the cell reaction and current flowing through the cell?
Consider a cell given below:
\[\ce{Cu | Cu^{2+} || Cl^{-} | Cl_{2},Pt}\]
Write the reactions that occur at anode and cathode
Which of the following statements about galvanic cell is incorrect
Which of the following is incorrect?
The two half cell reaction of an electrochemical cell is given as
\[\ce{Ag+ + e- -> Ag}\], `"E"_("Ag"^+//"Ag")^circ` = - 0.3995 V
\[\ce{Fe^{2+} -> Fe^{3+} + e-}\], `"E"_("Fe"^{3+}//"Fe")^{2+}` = - 0.7120 V
The value of EMF will be ______.