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In a cell reaction, Cu((s))+ 2Ag((aq))^(...

In a cell reaction, `Cu_((s))+ 2Ag_((aq))^(+) to Cu_((aq))^(2+) + 2Ag_((s)) E_"cell"^@`=+0.46 V . If the concentration of `Cu^(2+)` ions is doubled then `E_"cell"^@` will be

A

Doubled

B

Halved

C

Unchanged

D

Decreases by small fraction.

Text Solution

Verified by Experts

The correct Answer is:
E
`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.
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