JEE MAIN - Chemistry (2003 - No. 33)
For a cell reaction involving a two-electron change, the standard e.m.f. of the cell is found to be 0.295 V at 25oC. The equilibrium constant of the reaction at 25oC will be
29.5 $$\times$$ 10-2
10
1 $$\times$$ 1010
1 $$\times$$ 10-10
Explanation
The equlibrium constant is released to the standard emf of cell by the expression
$$\log \,K = {E^ \circ }_{cell} \times {n \over {0.059}}$$
$$ = 0.295 \times {2 \over {0.059}}$$
$$\log \,K = {{590} \over {59}} = 10$$
or $$\,\,\,\,\,K = 1 \times {10^{10}}$$
$$\log \,K = {E^ \circ }_{cell} \times {n \over {0.059}}$$
$$ = 0.295 \times {2 \over {0.059}}$$
$$\log \,K = {{590} \over {59}} = 10$$
or $$\,\,\,\,\,K = 1 \times {10^{10}}$$
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