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For the cell reaction , |Cu| Cu^(2+)(aq)...

For the cell reaction , `|Cu| Cu^(2+)(aq)|Ag^(+)|Ag` is `E^(@) Cu^(2)|Cu=+0.34 V` adb `nE^(@) Ag^(+) |Ag=0.80 V `and `[Cu^(+2)]=0.01` and `[Ag^(+)] = 1.0 "mol dm"^(-3)` the emf of the cell is

A

0.34 V

B

0.46 V

C

0.52 V

D

1.14 V

Text Solution

Verified by Experts

The correct Answer is:
C
`E_("cell")=E_("cell")^(@)+(0.059)/(2)"log"([Ag^(+)])/([Cu^(+2)])`
`=0.80-0.34+(0.059)/(2)"log"((1.0)^(2))/(0.01)`
`=0.46 +(0.059)/(2) xx 2=0.46+0.059`
=0.52 V
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