1.

Consider the following four electrodes: P=Cu^(2+)(0.0001M)//Cu(s)""Q=Cu^(2+)(0.1M)//Cu(s) R=Cu^(2+)(0.01M)//Cu(s)""S=Cu^(2+)(0.001M)//Cu(s) If the standard electrode potential of Cu^(2+)//Cu is +0.34V, the reduction potentials in volts of the above electrodes follow the order:

Answer»

PgtSgtRgtQ
SgtRgtQgtP
RgtSgtQgtP
QgtRgtSgtP

Solution :`Cu^(2+)+2E^(-)TOCU,E_(red)=E_(red)^(@)+(0.0591)/(2)log[Cu^(2+)]`
Thus, greater is `Cu^(2+)` ION concentration, greater will be the REDUCTION potential.


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