Question

A silver wire placed in a saturated solution of AgI in water registers a voltage of + 0.358 V against the standard hydrogen electrode. E̊(Ag+/ Ag) = +0.80 V; Use the Nernst equation to determine the concentration of silver ions and use the result to calculate Ks for AgI at 298 K. using the equation Ecell = E cell - RT/ZF In Q

Answer 1

For SHE $E^o_{SHE}=0.0 \ V$

Hence,

$\\ \ \\ \ E_{cell} = E_{Ag|Ag^+} \\ \ \\ \ E^0_{cell} =E_{Ag|Ag^+}^o$

$Ag^+(aq)+e^- \to Ag(s)$

$\frac{1}{2}H_2(g) \to [H^+] +e^-$

$E_{cell}=E^0_{cell}-\frac{RT}{ZF}log_{10}\frac{[H^+]}{[Ag^+] \times \sqrt{P_{H2}}}$

At 298 K, $\frac{RT}{ZF}=0.0592$

For SHE, $[H^+\frac{RT}{ZF}=0.0592]=1 \ M, \ P_{H2}=1 \ atm$

$E_{cell}=E^0_{cell}-\frac{0.0592}{Z}log_{10}\frac{1}{[Ag^+] }$

$+0.358 \ V=+0.80 \ V -\frac{0.0592}{1}log_{10}\frac{1}{[Ag^+]}$

$\frac{0.0592}{1}log_{10}\frac{1}{[Ag^+]}=+0.80 \ V -0.358 \ V$

$0.0592log_{10}\frac{1}{[Ag^+]}=0.442$

$log_{10}\frac{1}{[Ag^+]}=\frac{0.442}{0.0592}$

$log_{10}\frac{1}{[Ag^+]}=7.466$

$\frac{1}{[Ag^+]}=2.93 \times 10^7$

$[Ag^+]=\frac{1}{ 2.93 \times 10^7 }$

$[Ag^+]=3.42 \times 10^{-8} \ M$

$AgI \rightleftharpoons Ag^++I^-$

The solubility product

$K_{sp}=[Ag^+] \times [I^-]$

But

$[Ag^+] = [I^-]$

Hence,

$K_{sp}=[Ag^+] ^2$

$K_{sp}=\left ( 3.42 \times 10^{-8} \right )^2$

$K_{sp}=1.17 \times 10^{-15}$