Consider the following four electrodes:
`A=Cu^(2+)(0.0001M)//Cu_((s))`
`B=Cu^(2+)(0.1M)//Cu_((s))`
`C= Cu^(2+)(0.01M)//Cu_((s))`
`D=Cu^(2+)(0.001M)//Cu_((s))`
If the standard reduction potential of `Cu^(+2)//Cu` is `+0.34V`, the reduction potentials (in volts) of the above electrodes follow the order

Options

A. `PgtSgtRgtQ`
B. `SgtRgtQgtP`
C. `RgtgtRgtP`
D. `QgtRgtSgtP`

Answer

D

Explanation

`Cu^(2+)+2e ^(-)rarrCu`
`E_("red")=E_("red")^(@)+(0.0591)/(2)"log"[Cu^(2+)]`
`therefore` greater is `Cu^(2+)` ion concentration, greater will be the reduction potential.