# 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?

Jun 2, 2018

pH = 2.99

#### Explanation:

The cell diagram for your cell is

$\text{Pt(s)|H"_2"(1 atm)|H"^"+""("xcolor(white)(l) "mol/L)||H"^"+""(1 mol/L)|H"_2"(1 atm)|Pt(s)}$

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))
$\text{Overall:"color(white)(m) "H"_2"(1atm)" + "2H"^"+"("1 mol/L") ⇌ "2H"^"+"(xcolor(white)(l)"mol/L") + "H"_2"(1 atm)}$

Because the two half-reactions are identical,

${E}_{\textrm{c e l l}}^{\circ} = {E}_{\textrm{red}}^{\circ} - {E}_{\textrm{\otimes}} = 0$

Cell potential as a function of pH

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

${E}_{\textrm{c e l l}} = {E}_{\textrm{c e l l}}^{\circ} - \frac{R T}{z F} \ln Q$

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 ${E}_{\textrm{c e l l}}^{\circ} = 0$ and $\ln Q = 2.303 \log Q$, ao

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} {E}_{\textrm{c e l l}} = - \frac{2.303 R T}{z F} \log Q \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

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"

${E}_{\textrm{c e l l}} = - \frac{2.303 R T}{z F} \log Q$

= ("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

${E}_{\textrm{c e l l}} = \text{0.0592 V × pH}$

"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