| \[P=C{{u}^{2+}}(0.0001M)/Cu(s)\] |
| \[Q=C{{u}^{2+}}\left( 0.1\text{ }M)/Cu(s \right)\] |
| \[R=C{{u}^{2+}}(0.01M/)Cu(s)\] |
| \[S=C{{u}^{2+}}(0.001M/)Cu(s)\] |
| 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. |
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] \[{{E}_{red}}=E_{red}^{{}^\circ }+\frac{0.591}{n}\log [{{M}^{n+}}]\] Lower the concentration of\[{{M}^{n+}}\], lower is the reduction potential. Hence order of reduction potential is: \[Q>R>S>P\]
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