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 standard emf `(E^(c-))` of a half cell `Y^(2)|Y^(o+)` is `0.15V`, the standard emf of the half cell `Y^(o+)|Y` will be

Options

A. `0.19V`
B. `0.53V`
C. `0.49V`
D. `0.64V`

Answer

B

Explanation

When connected, electron flow from `X` to `Y` means,
`E_(Y^(2+)//Y)^(@) gt E_(X^(2+)//X)^(@)`
and when `X` is connected to `SHF`, e flow from `X` to `SHE` means.
`E_(X^(2+)//X)^(@)`

So, `0.34 xx2 = 0.15 xx 1 +a xx 1`
`rArr a = 0.68 - 0.15 = 0.53 V`