$\therefore \,\,\,\,0 = 0.591 - 0.0591\,\,\,\log \,{K_c}$
$\therefore \frac{{0.591}}{{0.0591}} = \log \,{K_c}\,\,\,\,\therefore \,\,\,10 = \log \,{K_c}\,$
$\,\therefore \,\,\,{K_c} = {10^{10}}$
${[Fe\,{(CN)_6}]^{4 - }}\, \to \,{[Fe{(CN)_6}]^{3 - }}\, + \,{e^ - }\,;\,$ ${E^o}\, = \, - \,0.35\,V$
$\,F{e^{2 + }}\, \to \,F{e^{3 + }}\, + \,{e^ - }\,;$ ${E^o}\, = \, - \,0.77\,V$