JEE Advance - Chemistry (1998 - No. 6)
Calculate the equilibrium constant for the reaction:
2Fe3+ + 3I- $$\leftrightharpoons$$ 2Fe2+ + $$I_3^-$$. The standard reduction potentials in acidic conditions are 0.78 V and 0.54 V respectively for Fe3+ | Fe2+ and $$I_3^-$$ | I- couples.
2Fe3+ + 3I- $$\leftrightharpoons$$ 2Fe2+ + $$I_3^-$$. The standard reduction potentials in acidic conditions are 0.78 V and 0.54 V respectively for Fe3+ | Fe2+ and $$I_3^-$$ | I- couples.
104
106
108
1010
1012
Explanation
Given, $$E_{F{e^{3 + }}|F{e^{2 + }}}^0 = 0.78V$$ and $$E_{I_3^ - |{I^ - }}^0 = 0.54V$$
The half-cell reactions are as follows:
At anode : $$3{I^ - } \to I_3^ - + 2e$$, $${E^0} = 0.54$$
At cathode : $$2F{e^{3 + }} + 2e \to 2F{e^{2 + }}$$, $${E^0} = + 0.78V$$
Cell reaction : $$2F{e^{3 + }} + 3{I^ - }$$ $$\rightleftharpoons$$ $$2F{e^{2 + }} + I_3^ - $$
$$\therefore$$ $$E_{cell}^0 = (0.78 - 0.54) = 0.24V$$
We know, $$E_{cell}^0 = {{0.059} \over 2}\log {K_{eq}}$$ or, $$0.24 = 0.029\log {K_{eq}}$$
or, $$\log {K_{eq}} = {{0.24} \over {0.029}} = 8.27 \simeq 8.00$$ $$\therefore$$ $${K_{eq}} = {10^8}$$
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