Question
If the \[E_{cell}^{\circ }\] for a given reaction has a negative value, which of the following gives correct relationships for the values of $$\Delta {G^ \circ }$$ and $${K_{eq}}?$$
A.
$$\Delta {G^ \circ } > 0;\,{K_{eq}} < 1$$
B.
$$\Delta {G^ \circ } > 0;\,{K_{eq}} > 1$$
C.
$$\Delta {G^ \circ } < 0;\,{K_{eq}} > 1$$
D.
$$\Delta {G^ \circ } < 0;\,{K_{eq}} < 1$$
Answer :
$$\Delta {G^ \circ } > 0;\,{K_{eq}} < 1$$
Solution :
$${\text{Given,}}\,E_{cell}^ \circ = - ve$$
The relation between $$\Delta {G^ \circ }$$ and $$E_{cell}^ \circ $$ is given as
$$\Delta {G^ \circ } = - nF\,E_{cell}^ \circ \,\,\,...{\text{(i)}}$$
If $$E_{cell}^ \circ $$ is negative, so $$\Delta {G^ \circ }$$ comes out to be positive. Again, relation between $$\Delta {G^ \circ }$$ and $${K_{eq}}$$ is given as
$$\Delta {G^ \circ } = - 2.303nRT\,{\text{log}}\,{K_{eq}}\,\,\,...{\text{(ii)}}$$
From Eq. (i) we get that $$\Delta {G^ \circ }$$ is positive. Now, if $$\Delta {G^ \circ }$$ is positive then $${K_{eq}}$$ comes out to be negative from eq (ii).
$${\text{i}}{\text{.e}}.\,\,\,\Delta {G^ \circ } > 1\,\,\,{\text{and}}\,\,\,{K_{eq}} < 1$$
Short trick As $$E_{cell}^ \circ $$ is negative so reaction is non-spontaneous or you can say reaction is moving in backward direction. For non-spontaneous reaction, $$\Delta {G^ \circ }$$ is always positive and $${K_{eq}}$$ is always less than 1.