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Galvanic cells generate electrical energ...

Galvanic cells generate electrical energy at the expense of a spontaneous redox reaction. In an electrode concentration cell two like electrodes having different concentrations, either because they are gas electrodes operating at different pressures or because they are amalgams (solutions in mercury) with different concentrations are dipped into the same solution.
Eg. An example is a cell composed of two chlorine electrodes with different pressure of `Cl_(2)`:
`Pt_(L)|Cl_(2)(P_(L))|HCl(aq)|Cl_(2)(P_(R))|Pt_(R)`
Where `P_(L)` and `P_(R)` are the `Cl_(2)` pressure at the left and right electrodes.
Calculate the EMF of the electrode concentration cell represented by :
`Hg-Zn(c_(1))|Zn^(+2)(aq)|Hg-Zn(c_(2))`
At `25^(@)C.c_(1)=2g` of `Zn` per `100g` of `Hg` and `c_(2)=1g` of `Zn` per `50g` of `Hg`

A

`0.059V`

B

`3`

C

`0`

D

data insufficient

Text Solution

Verified by Experts

The correct Answer is:
C

`E_("cell")=E_("cell")^(@)-(RT)/(nF)InQ`
`E_("cell")=0-(RT)/(nF) In[(C_(2))/(C_(1))]`
`=-(RT)/(nF) In[(1/50)/(2/100)]`
`=0V`
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Galvanic cells generate electrical energy at te expense of a spontaneous redox reaction. In an electrode concentration cell two like electrodes having different concentrations, either because they are gas electrodes operating at different pressures or because they are amalgams (solutions in mercury) with different concentratiions are dipped into the same solution. Eg. An example is a cell composed of two chlorine electrodes with different pressure of Cl_(2) : Pt_(L)|Cl_(2)(P_(L))|HCl(aq)|Cl_(2)(P_(R))|Pt_(R) Where P_(L) and P_(R) are the Cl_(2) pressure at the left and right electrodes. Calculate the EMF of the electrode concentration cell represented by : Hg-Zn(c_(1))|Zn^(+2)(aq)|Hg-Zn(c_(2)) A cell represented by the following cell diagram is generated to measure the solubility product of AgCl . Ag|Ag^(+)(aq)|Cl^(-)(aq)|AgCl(s)|Agg K_(sp)^(0)(AgCl) at 298K and 1 bar pressure is: Given: AgCl+e^(-)toAg+Cl^(-) E^(0)=0.22V Ag^(+)+e^(-)toAg E^(0)=0.79V (Take sqrt(10)=3.16 & (RT.(2.303))/F=0.06 )

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