A graph is plotted between `E_(cell)` and `log .([Zn^(2+)])/([Cu^(2+)])` . The curve is linear with intercept on `E_(cell)` axis equals to `1.10V`. Calculate `E_(cell)` for the cell. `Zn(s)||Zn^(2+)(0.1M)||Cu^(2+)(0.01M)|Cu`

A graph is plotted between `E_(cell)` and `log .([Zn^(2+)])/([Cu^(2+)]

1.	A graph is plotted between `E_(cell)` and `log .([Zn^(2+)])/([Cu^(2+)])` . The curve is linear with intercept on `E_(cell)` axis equals to `1.10V`. Calculate `E_(cell)` for the cell. `Zn(s)\|\|Zn^(2+)(0.1M)\|\|Cu^(2+)(0.01M)\|Cu`
Answer» Cell reaction is `: Zn+Cu^(2+) rarr Zn^(2+)+Cu` `E_(cell)=E^(c-)._(cell)-(0.059)/(2) log[(Zn^(2+))/(Cu^(2+))]" "(i)` `(y=c+mx)" "(ii)` Equation `(i)` represent a straight line like Eq. `(ii)` . `:` Intercept `(c)=E^(c-)._(cell)=1.10V` `:. E_(cell)=1.10V-(0.059)/(2)log .(0.1)/(0.01)` `=1.10V-(0.059)/(2)log 10` `=1.10-0.03" "(` tAKE `0.059~~0.06)` `=1.07V`

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A graph is plotted between `E_(cell)` and `log .([Zn^(2+)])/([Cu^(2+)])` . The curve is linear with intercept on `E_(cell)` axis equals to `1.10V`. Calculate `E_(cell)` for the cell. `Zn(s)||Zn^(2+)(0.1M)||Cu^(2+)(0.01M)|Cu`

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