A) \[3b-a\]
B) \[3b-4a\]
C) \[3b-2a\]
D) \[2a+3b\]
Correct Answer: C
Solution :
\[F{{e}^{2+}}+2e\,\,\,\to \,\,Fe\,\,\,\,{{E}^{0}}_{F{{e}^{2+}},\,\,Fe}=a\,\,\Delta G{{{}^\circ }_{1}}\,\,\,\,\,\,\,\,\,\,\,\,......(1)\] \[F{{e}^{3+}}-3e\,\,\to \,\,Fe\,\,\,\,\,{{E}^{0}}_{F{{e}^{3+}}\,,\,\,Fe}\,=\,\,b,\,\,\,\Delta G{{{}^\circ }_{2}}\,\,\,\,\,\,\,....(2)\] \[F{{e}^{3+}}+e\,\,\to \,\,F{{e}^{2+}}\,\,\,{{E}^{0}}_{F{{e}^{3+}}\,,\,\,F{{e}^{2+}}}\,=\,\,?\,,\,\,\,\,\Delta G{{{}^\circ }_{3}}\,\,\,\,\,\,\,\,\,\,\,\,\,....(3)\] Clearly: \[\left( 3 \right)=\left( 2 \right)-\left( 1 \right)\] \[\Delta G{{{}^\circ }_{3}}=\Delta G{{{}^\circ }_{2}}-\Delta G{{{}^\circ }_{1}}\] \[-E{{{}^\circ }_{3}}F=-3Fb+2Fa\] Or \[-E{{{}^\circ }_{3}}=-3b+2a\] Or \[E{{{}^\circ }_{3}}=3b-2a\]
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