ElectrochemistryHard
Question
A resistance of 50W is registered when two electrodes are suspended into a beaker containing a dilute solution of a strong electrolyte such that exactly half of the them are submerged into solution as shown in figure. If the solution is diluted by adding pure water (negligible conductivity) so as to just completely submerge the electrodes, the new resistance offered by the solution would be
Options
A.50 Ω
B.100 Ω
C.25 Ω
D.200 Ω
Solution
R = 
The k is halved while the A is doubled. Hence R remains 50 Ω.
The k is halved while the A is doubled. Hence R remains 50 Ω.Create a free account to view solution
View Solution FreeMore Electrochemistry Questions
CH2 = CH - CH = CH - CH3 is more stable than CH3 - CH = C = CH - CH3 because (I) (II)...For Na+, the value of symbol λm° is 50.0 Ω−1 cm2 mol−1. The speed of Na+ ion in the solution, if in the cell, electrodes...The correct order of EoM2+/M values with negative sign for the four successive elements Cr, Mn, Fe and Co is...A galvanic cell is composed of two hydrogen electrodes, one of which is a standard one. In which of the following soluti...The preparation of LiOH by the electrolysis of a 35% solution of LiCl using a platinum anode led to a current efficiency...