For the reaction
2Fe3+(aq) + 2I$-$(aq) $\to$ 2Fe2+(aq) + I2(s)
the magnitude of the standard molar Gibbs free energy change, $\Delta$rG$_m^o$ = $-$ ___________ kJ (Round off to the Nearest Integer).
$$\left[ {\matrix{
{E_{F{e^{2 + }}/Fe(s)}^o = - 0.440V;} & {E_{F{e^{3 + }}/Fe(s)}^o = - 0.036V} \cr
{E_{{I_2}/2{I^ - }}^o = 0.539V;} & {F = 96500C} \cr
} } \right]$$
Answer (integer)
46
Solution
$E_{F{e^{3 + }}/Fe}^o = 0.036V$<br><br>$F{e^{3 + }} + 3{e^ - } \to Fe$<br><br>$\therefore$ $\Delta G_1^o = - nF{E^o}$<br><br>$= - 3F( - 0.036)$<br><br>$E_{F{e^{2 + }}/Fe}^o = 0.440V$<br><br>$F{e^{2 + }} + 2{e^ - } \to Fe;{E^o} = - 0.440V$<br><br>$Fe \to F{e^{2 + }} + 2{e^ - };{E^o} = 0.440V$<br><br>$\therefore$ $\Delta G_2^o = - 2F(0.440)$<br><br>$E_{{I_2}/2{I^ - }}^o = 0.539V$<br><br>${I_2} + 2{e^ - } \to 2{I^ - };{E^o} = 0.539V$<br><br>$2{I^ - } \to {I_2} + 2{e^ - };{E^o} = - 0.539V$<br><br>$\Delta G_3^o = - 2F( - 0.539)$<br><br>$\therefore$ $\Delta {G^o} = 2\left[ {\Delta G_1^o + \Delta G_2^o} \right] + \Delta G_3^o$<br><br>$= 2\left[ {3F(0.036) - 2F(0.440)} \right] + 2F(0.539)$<br><br>$= - 45934 = - 45.9KJ \simeq - 46KJ$
About this question
Subject: Chemistry · Chapter: Electrochemistry · Topic: Electrochemical Cells
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