calculate equilibrium constant for reaction at 298 k.zn+cu2+gives zn2+ +cu given that Eo zn2+/zn=0.76v Eo cu2+/cu + 0.34v.

∆G

^{0}= -nFE

^{0}--------------(1)

Also from kinetic perspective we have,

∆G

^{0}

^{ }= -RT ln K --------------(2)

Thus, from these 2 equations we get

$\mathrm{ln}K=\frac{nF{E}^{0}}{RT}\phantom{\rule{0ex}{0ex}}Thus,\phantom{\rule{0ex}{0ex}}\mathrm{log}K=\frac{n{E}^{0}}{0.0592}\phantom{\rule{0ex}{0ex}}$

E

^{0}= E cathode - E anode

Here Zn which has higher oxidation potential will act as anode where as Cu electrode will act as cathode

E

^{0}= + 0.34 -(-0.76)

= 1.1 V

Substituting values we get,$\mathrm{log}K=\frac{2x1.1}{0.0592}=37.162\phantom{\rule{0ex}{0ex}}K=anti\mathrm{log}37.162\phantom{\rule{0ex}{0ex}}K=1.44x{10}^{37}$

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