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In the following electrochemical cell : ...

In the following electrochemical cell `:`
`Zn|Zn^(2+)||H^(o+)|(H_(2))Pt`
`E_(cell)=E^(c-)._(cell).` This will be when
1) `[Zn^(2+)]=[H^(o+)]=1M` and `p_(H_(2))1 atm`
2) `[Zn^(2+)]=0.01 M,[H^(o+)]=0.1M, `and `p_(H_(2))=1atm`
3) `[Zn^(2+)]=1m,[H^(o+)]=0.1M,` and `p_(H_(2))=0.01atm`
4) `[Zn^(2+)]=[H^(o+)]=0.1M` and `p_(H_(2))=0.1 atm`

A

`[Zn^(2+)]=[H^(+)]=1M ` and `pH_(2)=1` atm.

B

`[Zn^(2+)]=0.01M.[H^(+)]=0.1M` and `pH_(2)=1` atm

C

`[Zn^(2+)]=1M. [H^(+)]=0.1M` and `pH_(2)=0.01`atm

D

All of the above.

Text Solution

Verified by Experts

The correct Answer is:
d
`Zn+2H^(+)rarrZn^(2+)+H_(2)`
`E_("cell")^(@)=E_("cell")^(@)-(RT)/(nF)InQ`
`=E_("cell")^(@)-(RT)/(nF)In((Zn^(2+))(pH_(2)))/([H^(+)]^(2))`
(A) When `[Zn^(2+)]=[H^(+)]`=1M and `pH_(2)` =1atm
`Q=1, "In" Q=0`
`therefore E_("cell")^(@)=E_("Cell")^(@)`
(C) When `[Zn^(2+)]=1M.[H^(+)]=0.1` and
`pH_(2)=0.01`
`Q=(0.01xx1)/((0.1)^(2))=1`
`therefore E_("cell")=E_("cell")^(@)`
Thus all the choices (A,B and C) are correct. Thus the correct answer is D.
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