1.

The standard oxidation potential of Zn referred to SHE is 0.76V and that of Cu is -0.34V at 25^(@)C. When excess of Zn is added to CuSO_(4),Zn diplaces Cu^(2+) till equilibrium is reached. What is the ratio of Zn^(2+) to Cu^(2+) ions at equilibrium?

Answer»


Solution :`E_(CELL)^(@) = (0.0591)/(2)log.([ZN^(2+)])/([Cu^(2+)])`
`1.1 =(0.0591)/(2)log.([Zn^(2+)])/([Cu^(2+)])`
`([Zn^(2+)])/([Cu^(2+)]) =1.941 xx 10^(37)`


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