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The voltage of the cell consisting of Li...

The voltage of the cell consisting of Li (s) and `F_(2)(g)` electrodes is 5.92 V at standard condition at 298K. What is the voltage if the electrolyte consists of 2M LiF. (ln 2=0.693, R `=8.314J K^(-1)` and F=96500 C `mol^(-1)`)

A

5.90V

B

5.937V

C

5.88V

D

4.9V

Text Solution

Verified by Experts

The correct Answer is:
A
Now, the cell reaction is `Li(s) + (1)/(2) F_(2)(g) rarr Li^(+) + F^(-)`
we know that, `E_("cell")= E_("cell")^(@)- (RT)/(nF) "ln" (["Product"])/(["Reactant"])`
`=E_("cell")^(@)- (2.303RT)/(nF) log [Li^(+)] [F^(-)]`
`=5.92 -(2.303 xx 8.314 xx 298)/(1 xx 96500) log (2 xx 2)`
`=5.92 -(0.059)/(1) xx 2 log 2`
`=5.92-0.035= 5.88V`
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