| Consider the following four electrodes: |
| \[P=C{{u}^{2+}}(0.0001M)/C{{u}_{(s)}}\] |
| \[Q=C{{u}^{2+}}(0.1M)/C{{u}_{(s)}}\] |
| \[R=C{{u}^{2+}}(0.01M)/C{{u}_{(s)}}\] |
| \[S=C{{u}^{2+}}(0.001M)/C{{u}_{(s)}}\] |
| 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 |
A) \[P>S>R>Q\]
B) \[S>R>Q>P\]
C) \[R>S>Q>P\]
D) \[Q>R>S>P\]
Correct Answer: D
Solution :
[d]: \[E={{E}^{o}}+\frac{0.591}{n}\log [{{M}^{n+}}]\] Lower the concentration of \[{{M}^{n+}}\], lower is the E.
You need to login to perform this action.
You will be redirected in
3 sec
