A) +0.44 V, yes
B) -0.44 V, No
C) + 0.44 V, No
D) - 0.44 V, yes
Correct Answer: A
Solution :
| \[F{{e}^{3+}}+{{e}^{-}}\xrightarrow[{}]{{}}F{{e}^{2+}}\] | (1) | ||
| \[{{E}^{o}}_{1}=0.77V\] \[\Delta {{G}_{1}}^{o}=-1\times F\times 0.77\] | |||
| \[F{{e}^{3+}}+3{{e}^{-}}\xrightarrow[{}]{{}}Fe\] | (2) | ||
| \[{{E}^{o}}_{2}=-0.036V\] \[\Delta {{G}_{2}}^{o}=3\times F\times 0.036\] | |||
| \[Fe\xrightarrow[{}]{{}}F{{e}^{2+}}+2{{e}^{-}}\] | (3) | ||
| \[{{E}_{3}}^{o}=\Delta {{G}_{3}}^{o}=-2\times F\times {{E}_{3}}^{o}\] |
| (1) - (2) = 3 |
| \[\Delta {{G}_{1}}^{o}-\Delta {{G}_{2}}^{o}=\Delta {{G}_{3}}^{o}\] |
| \[\Rightarrow -F\times 0.77-3F-0.036=-2\times F\times {{E}_{3}}^{o}\] |
| \[\Rightarrow (0.77\times 0.108)=2\times {{E}_{3}}^{o}\] |
| \[\Rightarrow \,{{E}_{3}}^{o}\simeq 0.44V\] Ans. |
| Disproportionation reaction. | ||
| \[3F{{e}^{2+}}\xrightarrow[{}]{{}}Fe+2F{{e}^{3+}}\] | (4) | |
| \[F{{e}^{2+}}+2{{e}^{-}}\xrightarrow[{}]{{}}Fe\] | ||
| \[{{E}^{o}}=-0.44V\] \[\Delta {{G}^{o}}=-2\times F\times (-0.44)\] | (5) | |
| \[F{{e}^{2+}}\xrightarrow[{}]{{}}F{{e}^{3+}}+{{e}^{-}}\] | ||
| \[{{E}^{o}}=-0.77V\] \[\Delta {{G}^{o}}=-1\times F\times (-0.77)\] | (6) | |
| \[(5)+2\times (6)=4\] | ||
| \[\Rightarrow \Delta {{G}^{o}}_{4}=-2\times F\times (-0.44)+2(-1\times F\times (0.77)){{=}^{+}}ve\]\[\because \,\,\Delta {{G}_{4}}^{o}>0,\Rightarrow \]Reaction is not spontaneous. | ||
You need to login to perform this action.
You will be redirected in
3 sec
