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A volataic cell consists of an electode ...

A volataic cell consists of an electode of solide silver immerse in a `0.10M AgNO_(3)` solution and an electrode of unknown metal `'X'` immersed in a `0.10M` solution `X(NO_(3))_(2)`. A porous barrier separates the two half of the cell. Also given `:`
`E^(c-)._((Ag^(o+)|Ag))=0.80V` and `E^(c-)._(cell)=1.05V` at `25^(@)C`
`X(s)|X^(2+)(1.0M)||Ag^(o+)(0.1M)|Ag(s)`
If `Ag^(o+)|Ag` half cell in the above voltaic cell is replaced by `Zn^(2+)|Zn` half cell `(E^(c-)._(Zn^(2+)|Zn)=-0.76V)`

A

The direction of current flow will remain same.

B

Polarity of the electrodes will be reversed.

C

Cell will stop working.

D

`EMF` of the cell will increase.

Text Solution

Verified by Experts

The correct Answer is:
b
As cell operates, `[X^(2+)]` will increase and `[Ag^(o+)]` will decrease.
`[` cell reaction `:X+2Ag^(o+) rarr 2Ag+X^(2+)]`
If `Zn^(2+)|Zn` replaces `Ag^(o+)|Ag`, cell polarities will change.
`E^(c-)._(cell)=-0.25-(-0.76)=0.51V`
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