| The electrode potentials for [AIPMT (S) 2011] |
| \[C{{u}^{2+}}(aq)+{{e}^{-}}\xrightarrow[{}]{{}}C{{u}^{+}}(aq)\] and\[C{{u}^{+}}(aq)+{{e}^{-}}\xrightarrow[{}]{{}}Cu(s)\] |
| Are + 0.15 V and + 0.50 v respectively. The value of \[{{E}^{o}}_{C{{u}^{2+}}/Cu}\] will be |
A) 0.325 V
B) 0.650 V
C) 0.150 V
D) 0.500 V
Correct Answer: A
Solution :
| [a] \[C{{u}^{2+}}+{{e}^{-}}\xrightarrow[{}]{{}}C{{u}^{+}};\] |
| \[E_{1}^{o}=015V;\Delta G_{1}^{o}=-{{n}_{1}}E_{1}^{o}F\] |
| \[C{{u}^{+}}+{{e}^{-}}\xrightarrow[{}]{{}}Cu;\] |
| \[\frac{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,E_{2}^{o}=0.50V;\Delta G_{2}^{o}=-{{n}_{2}}E_{2}^{o}F}{C{{u}^{2+}}+2{{e}^{-}}\xrightarrow[{}]{{}}Cu;{{E}^{o}}=?;\Delta {{G}^{o}}=-n{{E}^{o}}F}\] |
| \[\Delta {{G}^{o}}=\Delta G_{1}^{o}+\Delta G_{2}^{o}\] |
| or\[-2{{E}^{o}}F=-1F\times 0.15+(-1F\times 0.50)\] |
| or\[-2{{E}^{o}}F=-0.15F-0.50F\] |
| or\[-2F{{E}^{o}}=-F(0.15+0.50)\] |
| \[\therefore \]\[{{E}^{o}}=\frac{0.65}{2}=0.325V\] |
You need to login to perform this action.
You will be redirected in
3 sec
