If it is desired to construct the following galvanic cell: to have V, what should be the value of to — Electrochemistry Chemistry Question
Question
If it is desired to construct the following galvanic cell: $Ag(s) \| Ag^+ (\text{saturated } AgI) \|\| Ag^+ (\text{saturated } AgCl, x \times 10^{-4} \text{ M } Cl^-) \| Ag(s)$ to have $E_{cell} = 0.102$ V, what should be the value of $x$ to get $[Cl^-]$, which must be present in the cathodic half-cell to achieve the desired EMF. Given $K_{sp}$ of AgCl and AgI are $1.8 \times 10^{-10}$ and $8.1 \times 10^{-17}$, respectively. ($2.303RT/F = 0.06$)
💡 Solution & Explanation
Anode $Ag^+$ concentration from saturated AgI: $[Ag^+]_a = \sqrt{8.1 \times 10^{-17}} = 9 \times 10^{-9}$ M. For the concentration cell, $E = 0.06 \log([Ag^+]_c / [Ag^+]_a)$. Substituting values: $0.102 = 0.06 \log([Ag^+]_c / (9 \times 10^{-9})) \Rightarrow \log([Ag^+]_c / (9 \times 10^{-9})) = 1.7$. $[Ag^+]_c / (9 \times 10^{-9}) = 10^{1.7} = 10^{2 - 0.3} = 100 / 2 = 50$. $[Ag^+]_c = 50 \times 9 \times 10^{-9} = 4.5 \times 10^{-7}$ M. The cathode solution is also saturated with AgCl, so $K_{sp} = [Ag^+]_c [Cl^-] \Rightarrow 1.8 \times 10^{-10} = (4.5 \times 10^{-7}) \times [Cl^-]$. $[Cl^-] = (1.8 \times 10^{-10}) / (4.5 \times 10^{-7}) = 0.4 \times 10^{-3} = 4 \times 10^{-4}$ M. Thus $x = 4$. Therefore, correct answer is 4.