The magnitude `(` but not the sign `)` of the standard reduction potentials of two metals `X` and `Y` are `:`
`Y^(2)+2e^(-) rarr Y |E_(1)^(c-)|=0.34V`
`X^(2)+2e^(-) rarr X |E_(2)^(c-)|=0.25V`
When the two half cells of `X` and `Y `are connected to construct a cell, eletrons flow from `X` to `Y`. When `X` is connected to a standard hydrogen electrode `(SHE)`,electrons flow from `X` to `SHE`.
If a half call `X|X^(2)(0.1M)` is connected to another half cell `Y|Y^(2+)(1.0M)` by means of a salt bridge and an external circuit at `25^(@)C`, the cell voltage would be

Options

A. `0.06V`
B. `0.12V`
C. `0.62V`
D. `0.72V`

Answer

c

Explanation

When `X` is connected to `SHE`, electons flow from `X` to `SHE`. This means that `X` is acting as anode and `SHE` as cathode and its oxidation potential is positive. Also, the reduction potential of `Y` is greater than the reduction potential of `X(` as electrons flow from `X` to `Y).`
`implies Y^(2)+2e^(-) rarr Y(s)`
`E^(c-)._(Y^(2+)|Y)=0.34V,`
`X^(2+)+2e^(-) rarr X(s)`
`E^(c-)._(X^(2+)|X)=-0.25V`
Consider`: X(s)|X^(2+)(0.1M)||Y^(2+)(1.0M)|Y(s)`
`E^(c-)._(cell)=E^(c-)._(cell)-(0.059)/(2) log .([X^(2+)])/([Y^(2+)])`
`=[0.34-(-0.25)]-(0.059)/(2)log_(10).(0.1)/(1)=0.62V`