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Given concentration cell Zn//Zn^(2+) (1M...

Given concentration cell `Zn//Zn^(2+) (1M) //Zn^(2+) (0.15M) //Z`
Calculate E cell. As the cell discharges, does the difference in concentrations between the two solutions become smaller or larger?

Text Solution

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At anode `Zn to Zn^(2+) (I M) +2e`
At cathode `Zn^(2+) (0.15M) +2e to Zn`
Overall reaction `Zn^(2+) (0.15M) to Zn^(2+) (I M)`
E cell `=E^(@)" cell "(0.059)/(2) log ""(1)//(0.15)`
For a concentration cell, `E_("cell")^(@)=0`
E cell `=-0.0295 xx 0.825 V=-0.0224`
As the cell discharges, the reaction proceeds to the left, that is, the IM zinc ions is used and 0.15M zinc is produced. Thus, the two solutions approach each other in concentration.
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