1.

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

Answer»

`E_("CELL") = 0.295V `
`E_("cell")=0.236 V `
`E_("cell")=(0.059)/(1)log""([Ag^(+)]_("cathode"))/(sqrt(K_(SP) " of " AgCl))`
`E_("cell")=E_("cell")^(@) + (0.059)/(1)log ""(sqrt(K_(SP) " of AgCl"))/([Ag^(+)]_("cathode"))`

Solution :`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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