The pKsp of AgI is 16-07- If the E- value for Ag-Ag is 0-7991 V- find the E- for the half cell recation AgI-s - e- Ag - I

AgI_(s)⟶ Ag^++I^ K_sp=[Ag^+][I^ ] At cathode: AgI+e^ ⟶ Ag+I^ #160; #160; #160; #160; E^o_Red=? Anode: Ag⟶ Ag^++e^ #16 ...

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Standard Electrode Potentials

The pKsp of...

Question

The $p K_{s p}$ of $A g I$ is 16.07. If the $E^{\circ}$ value for $A g^{+} | A g$ is 0.7991 V. find the $E^{\circ}$ for the half cell recation $A g I / (s) + e^{-} \to A g + I$

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p K_{s p}

E^{\circ}

A g^{+} / A g

E^{\circ}

A g I + e^{-} \to A g + I^{-}

149 \times 10^{- x}

Given the data $25^{0} C$

A g + I^{-} \to A g I + e^{-}, E^{0} = 0.152 V

A g \to A g^{+} e^{-}, E^{0} = 0.800 V

What is the value of log $K_{s p}$ for AgI?

25^{0} C

A g + I^{-} \to A g I + e^{-}; E^{0} = 0.152 V

A g \to A g^{+} + e^{-}; E^{0} = - 0.800 V

K_{s p}

A g I

(G i v e n : 2.303 R T / F = 0.059 V)

25^{o}

A g_{(s)} + I_{(a q)} \to A g I_{(s)} + e^{-}, E^{o} = 0.152

A g_{(s)} \to A g_{(a q)}^{+} + e^{-}, E^{o} = - 0.800

K_{s p}

(W h e r e K_{s p} = solubility product, 2.303 \frac{R T}{F} = 0.059 V)

A g I

E_{A g^{+} / A g}^{\circ} = 0.80 v o l t

E_{I / A g I / A g}^{\circ} = - 0.15 v o l t

25^{\circ} C

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