What is the pH of the unknown solution?

A voltaic cell is constructed in which the cathode is a standard hydrogen electrode and the anode is a hydrogen electrode (P(H2)= 1atm) immersed in a solution of unknown [H+]. If the cell potential is 0.177 V, what is the pH of the unknown solution at 298 K?

1 Answer
Jun 2, 2018

Answer:

pH = 2.99

Explanation:

The cell diagram for your cell is

#"Pt(s)|H"_2"(1 atm)|H"^"+""("xcolor(white)(l) "mol/L)||H"^"+""(1 mol/L)|H"_2"(1 atm)|Pt(s)"#

You have a hydrogen concentration cell.

The half-cell reactions are

#"Anode:"color(white)(ml) "H"_2"(1 atm)"color(white)(mmmmml) ⇌ "2H"^"+"(xcolor(white)(l)"mol/L") + color(red)(cancel(color(black)("2e"^"-")))#
#"Cathode:"color(white)(l) ul("2H"^"+"("1 mol/L") + color(red)(cancel(color(black)("2e"^"-"))) ⇌ "H"_2"(1 atm)"color(white)(mmmmmmmmmmm))#
#"Overall:"color(white)(m) "H"_2"(1atm)" + "2H"^"+"("1 mol/L") ⇌ "2H"^"+"(xcolor(white)(l)"mol/L") + "H"_2"(1 atm)"#

Because the two half-reactions are identical,

#E_text(cell)^@ = E_text(red)^@ - E_text(ox) = 0#

Cell potential as a function of pH

The cell is not at standard conditions, so we must use the Nernst Equation:

#E_text(cell) = E_text(cell)^@ - (RT)/(zF)lnQ#

where

#R# is the Universal Gas Constant
#T# is the temperature
#z# is the moles of electrons transferred per mole of hydrogen
#F# is the Faraday constant
#Q# is the reaction quotient

But #E_text(cell)^@ =0# and #lnQ = 2.303logQ#, ao

#color(blue)(bar(ul(|color(white)(a/a)E_text(cell) = - (2.303RT)/(zF)logQcolor(white)(a/a)|)))" "#

#Q = (["H"^"+"]_text(prod)^2p_text(H₂, product))/(["H"^"+"]_text(react)^2p_text(H₂, reactant)) = (["H"^"+"]^2 ×1)/(1 × 1) = ["H"^"+"]^2#

#logQ = log["H"^"+"]^2 = 2log["H"^"+"] = "-2pH"#

#E_text(cell) = -(2.303RT)/(zF)logQ#

#= ("2.303 × 8.314 V"·color(red)(cancel(color(black)("C·K"^"-1""mol"^"-1"))) × 298.15 color(red)(cancel(color(black)("K"))))/("2 × 96 485" color(red)(cancel(color(black)("C·mol"^"-1")))) × "2pH"#

#color(blue)(bar(ul(|color(white)(a/a)E_text(cell) = "0.0592 V × pH"color(white)(a/a)|)))" "#

Calculate the pH

#E_text(cell) = "0.0592 V × pH"#

#"pH" = E_text(cell)/"0.0592 V" = (0.177 color(red)(cancel(color(black)("V"))))/(0.0592 color(red)(cancel(color(black)("V")))) = 2.99#