JEE MAIN - Chemistry (2009 - No. 20)
Given : $$E_{F{e^{3 + }}/Fe}^o$$ = -0.036V; $$E_{F{e^{2 + }}/Fe}^o$$ = -0.439 V
The value of standard electrode potential for the change,
Fe3+ (aq) + e- $$\to$$ Fe2+ (aq) will be
The value of standard electrode potential for the change,
Fe3+ (aq) + e- $$\to$$ Fe2+ (aq) will be
-0.072 V
0.385 V
0.770 V
0.270
Explanation
Given
$$F{e^{3 + }} + 3{e^ - } \to Fe,\,\,{E^ \circ }_{F{e^{3 + }}/Fe}$$
$$\,\,\,\,\,\,\,\,\,\,\, = - 0.036V\,\,\,...\left( i \right)$$
$$F{e^{2 + }} + 2{e^ - } \to Fe,\,\,{E^ \circ }_{F{e^{2 + }}/Fe}$$
$$\,\,\,\,\,\,\,\,\,\,\, = - 0.439V\,\,\,...\left( {ii} \right)$$
we have to calculate
$$F{e^{3 + }} + {e^ - } \to F{e^{2 + }},\,\,\Delta G = ?$$
To obtain this equation subtract equ $$(ii)$$ from $$(i)$$ we get
$$F{e^{3 + }} + {e^ - } \to F{e^{2 + }}\,\,\,...\left( {iii} \right)$$
As we know that $$\Delta G = - nFE$$
Thus for reaction $$(iii)$$
$$\Delta G = \Delta {G_1} - \Delta G;\,$$
$$\, - nF{E^ \circ } = - nF{E_1} - \left( { - nF{E_2}} \right)$$
$$\, - nF{E^ \circ } = nF{E_2} - nF{E_1}$$
$$ - 1F{E^ \circ } = 2 \times 0.439F - 3 \times 0.036\,F$$
$$ - 1F{E^ \circ } = 0.770\,F$$
$$\therefore$$ $$\,\,\,\,{E^ \circ } = - 0.770V$$
$${O^{ - - }} > {F^ - } > N{a^ + } > M{g^{ + + }} > A{l^{3 + }}$$
$$F{e^{3 + }} + 3{e^ - } \to Fe,\,\,{E^ \circ }_{F{e^{3 + }}/Fe}$$
$$\,\,\,\,\,\,\,\,\,\,\, = - 0.036V\,\,\,...\left( i \right)$$
$$F{e^{2 + }} + 2{e^ - } \to Fe,\,\,{E^ \circ }_{F{e^{2 + }}/Fe}$$
$$\,\,\,\,\,\,\,\,\,\,\, = - 0.439V\,\,\,...\left( {ii} \right)$$
we have to calculate
$$F{e^{3 + }} + {e^ - } \to F{e^{2 + }},\,\,\Delta G = ?$$
To obtain this equation subtract equ $$(ii)$$ from $$(i)$$ we get
$$F{e^{3 + }} + {e^ - } \to F{e^{2 + }}\,\,\,...\left( {iii} \right)$$
As we know that $$\Delta G = - nFE$$
Thus for reaction $$(iii)$$
$$\Delta G = \Delta {G_1} - \Delta G;\,$$
$$\, - nF{E^ \circ } = - nF{E_1} - \left( { - nF{E_2}} \right)$$
$$\, - nF{E^ \circ } = nF{E_2} - nF{E_1}$$
$$ - 1F{E^ \circ } = 2 \times 0.439F - 3 \times 0.036\,F$$
$$ - 1F{E^ \circ } = 0.770\,F$$
$$\therefore$$ $$\,\,\,\,{E^ \circ } = - 0.770V$$
$${O^{ - - }} > {F^ - } > N{a^ + } > M{g^{ + + }} > A{l^{3 + }}$$
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