We are asked to** find the Ksp for AgSCN.**

$\overline{){\mathbf{E}}{{\mathbf{\xb0}}}_{{\mathbf{cell}}}{\mathbf{}}{\mathbf{=}}\frac{\mathbf{0}\mathbf{.}\mathbf{0592}\mathbf{}\mathbf{V}}{\mathbf{n}}{\mathbf{logK}}}$

E°_{cell} = standard cell potential

n = number of e^{-} transferred

Q= reaction quotient = [products]/[reactants]

We have to determine the E°_{cell} first and as well as the **anode (oxidation)** and **cathode (reduction)** in the concentration cell and the **number of electrons transferred (n)**.

The standard potential for the reduction of AgSCN(s) is +0.09 V.

AgSCN(s) + e^{-} → Ag(s) + SCN^{-} (aq).

Using this value and the electrode potential for Ag^{+} (aq), calculate the K_{sp} for AgSCN.

Ag^{+}(aq) + e^{-} → Ag(s) E^{o}= +0.80 V

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