ElectrochemistryMedium
Question
The solution of CuSO4, in which copper rod is immersed, is diluted to 10 times. The reduction electrode potential
Options
A.increases by 0.0295 V
B.decreases by 0.0295 V
C.increases by 0.059 V
D.decreases by 0.059 V
Solution
$Cu^{2 +} + 2e^{-} \rightleftharpoons Cu$
$E = E^{o} - \frac{0.059}{2}.\log\frac{1}{\left\lbrack Cu^{+ 2} \right\rbrack} = E^{o} - \frac{0.059}{2}\log\frac{1}{0.1} $$$= E^{o} - 0.0295\text{ V}$$
Create a free account to view solution
View Solution FreeMore Electrochemistry Questions
F2 gas cannot be obtained by the electrolysis of any aqueous fluoride salt because...A galvanic cell is composed of two hydrogen electrodes, one of which is a standard one. In which of the following soluti...Both oxidation and reduction takes place in :...During electrolysis of water the volume of O2 Liberated is 2.24 dm3. The volume of hydrogen liberated, under same condit...Using the information in the preceding problem, calculate the solubility product of AgI in water at 25oC [Eo(Ag+, Ag) = ...