For the redox reaction: `Zn(s) +Cu^(2+) (0.1M) rarr Zn^(2+) (1M)+Cu(s)` taking place in a cell, `E_(cell)^(@) is 1.10` volt. `E_(cell)` for the cell will be `(2.303 (RT)/(F) = 0l.0591)`A. `1.07` voltB. `0.82` voltC. `2.14` voltD. `1.80` volt

For the redox reaction: `Zn(s) +Cu^(2+) (0.1M) rarr Zn^(2+) (1M)+Cu(s)

1.	For the redox reaction: `Zn(s) +Cu^(2+) (0.1M) rarr Zn^(2+) (1M)+Cu(s)` taking place in a cell, `E_(cell)^(@) is 1.10` volt. `E_(cell)` for the cell will be `(2.303 (RT)/(F) = 0l.0591)`A. `1.07` voltB. `0.82` voltC. `2.14` voltD. `1.80` volt
Answer» Correct Answer - A Applying Nernst equation `E_(cell) = E_(cell)^(@) =- (0.0591)/(2)log. \|(Sn^(2+))/(Cu^(2+))\|` `=1.1 -(0.0591)/(2)log.(1)/(0.1)` `1.1 - 0.02955` `=1.07`

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For the redox reaction: `Zn(s) +Cu^(2+) (0.1M) rarr Zn^(2+) (1M)+Cu(s)` taking place in a cell, `E_(cell)^(@) is 1.10` volt. `E_(cell)` for the cell will be `(2.303 (RT)/(F) = 0l.0591)`A. `1.07` voltB. `0.82` voltC. `2.14` voltD. `1.80` volt

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