Consider the following cell reaction
`Cu(s)+2Ag^(+)(aq) rarr Cu^(2+) (aq) +2Ag(s)`
`E_("cell")^(@)=0.46 V` By boubling the concentration of `Cu^(2+)`, `E_("cell")` is

Options

A. Doubled
B. Halved
C. Unchanged
D. Decreases by small fraction.

Answer

E

Explanation

`E_("cell")=E_("cell")^(@)+(0.059)/(2)"log"([Ag^(2+)]^(2))/([Cu^(2+)])`
Doubling the conc. Of `Cu^( 2)+` will make log `(Ag^(+))^(2)//Cu^(+)` negative and `E_("cell")` will marginally decrease.