Question #72818

1 Answer
Apr 8, 2017

Answer:

#K_text(sp) = 7.4 × 10^"-17"#

Explanation:

You have a concentration cell:

#"Ag(s)"∣"Ag"^"+" (xcolor(white)(l) "mol/L")"∣∣""Ag"^"+"("0.135 mol/L")∣∣"Ag(s)"#

The reactions are:

Anode: #color(white)(m)"Ag(s)" → "Ag"^"+"(xcolor(white)(l) "mol/L") + e^"-"#
Cathode: #"Ag"^"+"("0.135 mol/L") + e^"-" → "Ag(s)"#
Cell: #color(white)(mml)"Ag(s)" + "Ag"^"+"("0.135 mol/L") → "Ag"^"+"(xcolor(white)(l) "mol/L") "Ag(s)"#

#E_text(cell) = E_text(cell)^° - (RT)/(nF)lnQ = 0 - (RT)/(nF)lnQ = -(RT)/(nF)lnQ = "0.180 V"#

#-(8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1"))) × 298 color(red)(cancel(color(black)("K"))))/(1 × "96 485" color(red)(cancel(color(black)("J·V"^"-1"))))lnQ = 0.180 color(red)(cancel(color(black)("V")))#

#lnQ = "-7.010"#

#Q = e^"-7.010" = 9.030 × 10^"-4"#

#Q =( ["Ag"^"+"]("products"))/( ["Ag"^"+"]("reactants")) = x/0.135 = 9.030 × 10^"-4"#

#x = 0.135 × 9.030 × 10^"-4" = 1.219 × 10^"-4"#

∴ In the anode half-cell,

#["Ag"^"+"] = x color(white)(l)"mol/L" = 1.219 × 10^"-4"color(white)(l) "mol/L"#

#["PO"_4^"3-"] = ⅓["Ag"^"+"] "mol/L" = ⅓xcolor(white)(l) "mol/L"#

#"Ag"_3"PO"_4"(s)" ⇌ "3Ag"^"+""(aq)" + "PO"_4^"3-""(aq)"#

#K_text(sp) = ["Ag"^"+"]^3["PO"_4^"3-"] = x^3 × ⅓x = ⅓x^4 = ⅓ × (1.219 × 10^"-4")^4#

#= 7.4 × 10^"-17"#