JEE MAIN - Chemistry (2004 - No. 37)
The standard e.m.f of a cell, involving one electron change is found to be 0.591 V at 25oC.
The equilibrium constant of the reaction is (F = 96,500 C mol-1: R = 8.314 JK-1 mol-1)
1.0 $$\times$$ 101
1.0 $$\times$$ 1030
1.0 $$\times$$ 1010
1.0 $$\times$$ 105
Explanation
$$E_{cell}^o = E_{cell}^o - {{0.059} \over n}\log \,{K_c}$$
or $$\,\,\,\,\,\,$$ $$0 = 0.591 - {{0.0591} \over 1}\log {K_c}$$
or $$\,\,\,\,\,\,$$ $$\log \,{K_c} = {{0.591} \over {0.0591}} = 10$$
or $$\,\,\,\,\,\,$$ $${K_c} = 1 \times {10^{10}}$$
or $$\,\,\,\,\,\,$$ $$0 = 0.591 - {{0.0591} \over 1}\log {K_c}$$
or $$\,\,\,\,\,\,$$ $$\log \,{K_c} = {{0.591} \over {0.0591}} = 10$$
or $$\,\,\,\,\,\,$$ $${K_c} = 1 \times {10^{10}}$$
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