Redox Reactions And Electrochemistry Ques 559
Question: Consider the following four electrodes:
$ P=C{u^{2+}}(0.0001M)/Cu(s) $ $ Q=C{u^{2+}}( 0.1M)/Cu(s ) $ $ R=C{u^{2+}}(0.01M/)Cu(s) $ $ S=C{u^{2+}}(0.001M/)Cu(s) $ If the standard reduction potential of $ C{u^{2+}}/Cu $ is +0.34 V, the reduction potentials in volts of the above electrodes follow the order.
Options:
A) $ P>S>R>Q $
B) $ S>R>Q>P $
C) $ R>S>Q>P $
D) $ Q>R>S>P $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ E_{red}=E_red^{{}^\circ }+\frac{0.591}{n}\log [{M^{n+}}] $ Lower the concentration of $ {M^{n+}} $ , lower is the reduction potential. Hence order of reduction potential is: $ Q>R>S>P $
