Verified Solution • 2025–26 Session moderate

QJEEM09Ch82ec: JEE Main 2009 Chemistry - Class 12 JEEM09Ch82ec Solution: Given : = -0.036V; = -0.439 V The value of standard electrode…

Given : EFe3+/FeoE_{F{e^{3 + }}/Fe}^o = -0.036V; EFe2+/FeoE_{F{e^{2 + }}/Fe}^o = -0.439 V
The value of standard electrode potential for the change,
Fe3+ (aq) + e- →\to Fe2+ (aq) will be

Multiple Choice Options

  • A -0.072 V
  • B 0.385 V
  • C 0.770 V
  • D 0.27

Step-by-Step Verified Solution

Given

Fe3++3e−→Fe,  E∘Fe3+/FeF{e^{3 + }} + 3{e^ - } \to Fe,\,\,{E^ \circ }_{F{e^{3 + }}/Fe}

           =−0.036V   ...(i)\,\,\,\,\,\,\,\,\,\,\, = - 0.036V\,\,\,...\left( i \right)

Fe2++2e−→Fe,  E∘Fe2+/FeF{e^{2 + }} + 2{e^ - } \to Fe,\,\,{E^ \circ }_{F{e^{2 + }}/Fe}

           =−0.439V   ...(ii)\,\,\,\,\,\,\,\,\,\,\, = - 0.439V\,\,\,...\left( {ii} \right)

we have to calculate

Fe3++e−→Fe2+,  ΔG=?F{e^{3 + }} + {e^ - } \to F{e^{2 + }},\,\,\Delta G = ?

To obtain this equation subtract equ (ii)(ii) from (i)(i) we get

Fe3++e−→Fe2+   ...(iii)F{e^{3 + }} + {e^ - } \to F{e^{2 + }}\,\,\,...\left( {iii} \right)

As we know that ΔG=−nFE\Delta G = - nFE

Thus for reaction (iii)(iii)

ΔG=ΔG1−ΔG; \Delta G = \Delta {G_1} - \Delta G;\,

 −nFE∘=−nFE1−(−nFE2)\, - nF{E^ \circ } = - nF{E_1} - \left( { - nF{E_2}} \right)

 −nFE∘=nFE2−nFE1\, - nF{E^ \circ } = nF{E_2} - nF{E_1}

−1FE∘=2×0.439F−3×0.036 F- 1F{E^ \circ } = 2 \times 0.439F - 3 \times 0.036\,F

−1FE∘=0.770 F- 1F{E^ \circ } = 0.770\,F

∴\therefore     E∘=−0.770V\,\,\,\,{E^ \circ } = - 0.770V

{O^{ - - }} > {F^ - } > N{a^ + } > M{g^{ + + }} > A{l^{3 + }}

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