The standard $$EMF$$  of a galvanic cell involving cell reaction with $$n = 2$$  is found to be $$0.295$$ $$V$$ at $${25^ \circ }C.$$  The equilibrium constant of the reaction would be
( Given $${F = 96500\,C\,mo{l^{ - 1}},}$$    $${R = 8.314\,J{K^{ - 1}}\,mo{l^{ - 1}}}$$    )

Solution :
$$\eqalign{ & {\text{By Nernst equation,}} \cr & {E_{cell}} = E_{cell}^ \circ - \frac{{2.303\,RT}}{{nF}}{\text{lo}}{{\text{g}}_{10}}\,K \cr & {\text{At equilibrium, }}{E_{cell}} = 0 \cr & {\text{Given that,}} \cr & R = 8.314\,J{K^{ - 1}}\,mo{l^{ - 1}} \cr & T = {25^ \circ }C + 273 = 298\,K \cr & F = 96500\,C\,\,{\text{and}}\,\,n = 2 \cr & \therefore \,\,E_{cell}^ \circ = \frac{{2.303 \times 8.314 \times 298}}{{2 \times 96500}}{\text{lo}}{{\text{g}}_{10}}\,K \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{0.0591}}{2}{\text{lo}}{{\text{g}}_{10}}\,K \cr & {\text{Given that}}\,E_{cell}^ \circ = 0.295\,V \cr & \therefore \,\,0.295 = \frac{{0.0591}}{2}{\text{lo}}{{\text{g}}_{10}}\,K \cr & \,\,{\text{lo}}{{\text{g}}_{10}}\,K = \frac{{0.295 \times 2}}{{0.0591}} = 10 \cr & {\text{antilog}}\,\,{\text{lo}}{{\text{g}}_{10}}\,K = {\text{antilog}}\,10 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,K = 1 \times {10^{10}} \cr} $$