A) \[-0.072V\]
B) \[0.385V\]
C) \[0.770V\]
D) \[-0.270V\]
Correct Answer: C
Solution :
| Given,\[F{{e}^{3}}+3{{e}^{-}}\xrightarrow{{}}Fe;\] | |
| \[{{E}_{1}}{}^\circ =-0.336V\]?? (i) | |
| \[F{{e}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Fe;\,\,\,\,\,\,\,\,{{E}_{2}}{}^\circ =-0.439V\] | ? (ii) |
| We need to calculate | |
| \[F{{e}^{3+}}+{{e}^{-}}\xrightarrow{{}}F{{e}^{2+}},{{E}_{3}}{}^\circ =?\] | ?. (iii) |
| We can obtain the Eq. (iii) by subtracting Eq. (ii) from Eq. (i), but \[{{E}_{3}}{}^\circ \], cannot be obtained by this method because electrode potential is intensive property. Thus we can it by determine |
| \[\Delta {{G}_{3}}=\Delta {{G}_{1}}-\Delta {{G}_{2}}\] |
| (\[\Delta G\] is an extensise property) |
| \[\Delta {{G}_{3}}=3\times 0.336V-2\times 0.439V\] |
| \[\Delta {{G}_{3}}=0.108V-0.878V\]\[-1\times V\times E_{3}^{{}^\circ }\]\[=-0.770\,\,V\]\[\Rightarrow E_{3}^{{}^\circ }=0.770V\] |
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