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At what concetration of Ag^(2+) ions, wi...

At what concetration of `Ag^(2+)` ions, will the electrode have a potential of 0.0 V?

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Verified by Experts

The correct Answer is:
`2.9xx10^(-14) M`
`Ag^(+)+e^(-) to Ag`.
`(E_(Ag)^(+))/(Ag) to (E_(Ag^(+))^(@))/(Ag)-(0.0591)/(n)"log"(1)/([Ag^(+)])`
`0=0.80-(0.0591)/(1)"log"(1)/([Ag^(+)])=0.80+0.0591 " log "[Ag^(+)]`
`[Ag^(+)]=((-)0.80)/(0.0591)=-13.5364=overset(-" ")14.4636`.
`[Ag^(+)]="Antilog "log[overset(-" ")14.4636]=2.9xx10^(-14)" M "`.
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