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In the following concentration cell Ag...

In the following concentration cell
`Ag(s) // AgCl "(saturated)"// // AgNO_3(aq) (0.1M)//Ag_((s)), K_(SP) " of AgCl"=1 xx 10^(-10)` The cell potential will be

A

`E_("cell") = 0.295V `

B

`E_("cell")=0.236 V `

C

`E_("cell")=(0.059)/(1)log""([Ag^(+)]_("cathode"))/(sqrt(K_(sp) " of " AgCl))`

D

`E_("cell")=E_("cell")^(@) + (0.059)/(1) log ""(sqrt(K_(SP) " of AgCl"))/([Ag^(+)]_("cathode"))`

Text Solution

Verified by Experts

The correct Answer is:
B, C, D
`E=0 - (0.0591)/(1) log""([Ag^(oplus)]_a)/([Ag^(oplus)]_c), E = (-0.0591)/(1) log""(10^(-5))/(10^(-1)) = - (0.0591)/(1) xx (-4) = 0.2364V `
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