JEE Advance - Chemistry (2004 - No. 2)
Find the equilibrium constant for the reaction,
In2+ + Cu2+ $$\to$$ In3+ + Cu+ at 298 K
given
$$E_{C{u^{2 + }}/C{u^ + }}^o$$ = 0.15 V; $$E_{l{n^{2 + }}/l{n^ + }}^o$$ = -0.40 V; $$E_{l{n^{3 + }}/l{n^ + }}^o$$ = -0.42 V;
In2+ + Cu2+ $$\to$$ In3+ + Cu+ at 298 K
given
$$E_{C{u^{2 + }}/C{u^ + }}^o$$ = 0.15 V; $$E_{l{n^{2 + }}/l{n^ + }}^o$$ = -0.40 V; $$E_{l{n^{3 + }}/l{n^ + }}^o$$ = -0.42 V;
10
105
1010
10-10
10-5
Explanation
We know, $$\Delta {G^0} = - nF{E^0}$$
(1) $$C{u^{2 + }} + e \to C{u^ + }$$ ; $$\Delta G_1^0 = - 0.15F$$
(2) $$I{n^{2 + }} + e \to I{n^ + }$$ ; $$\Delta G_2^0 = + 0.4F$$
(3) $$I{n^{3 + }} + 2e \to I{n^ + }$$ ; $$\Delta G_3^0 = + 0.84F$$
Adding equation (1) and (2) and subtracting equation (3) we get, $$C{u^{2 + }} + I{n^{2 + }} \to C{u^ + } + I{n^{3 + }}$$
$$\Delta G_{}^0 = \Delta G_1^0 + \Delta G_2^0 - \Delta G_3^0 = - 0.59F$$
$$\Delta G_{}^0 = - 2.303RT\log {K_{eq}}$$ $$\therefore$$ $$ - 0.59F = - 2.303RT\log {K_{eq}}$$
or, $$\log {K_{eq}} = {{0.59 \times 96500} \over {2.303 \times 8.314 \times 298}} \approx 10$$ $$\therefore$$ $${K_{eq}} = {10^{10}}$$
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