Question
For a cell reaction involving two electron change, the standard $$EMF$$ of the cell is $$0.295\,V$$ at $${2^ \circ }C.$$ The equilibrium constant of the reaction at $${25^ \circ }C$$ will be:
A.
$$29.5 \times {10^{ - 2}}$$
B.
$$10$$
C.
$$1 \times {10^{10}}$$
D.
$$2.95 \times {10^{ - 10}}$$
Answer :
$$1 \times {10^{10}}$$
Solution :
$$\eqalign{
& {\text{Using the relation,}} \cr
& {E^ \circ }_{cell} = \frac{{2.303\,RT}}{{nF}}\log \,{K_{eq}} = \frac{{0.0591}}{n}\log \,{K_{eq}} \cr
& \therefore \,\,0.295\,V = \frac{{0.0591}}{2}\log \,{K_{eq}} \cr
& {\text{or}}\,\,\log \,{K_{eq}} = \frac{{2 \times 0.295}}{{0.0591}} = 10 \cr
& {\text{or}}\,\,{K_{eq}} = 1 \times {10^{10}} \cr} $$