# Question ef321

Aug 17, 2016

$\textsf{p H = 3}$

#### Explanation:

In a cell reduction occurs at the cathode. In this case this will be the hydrogen/Pt half cell. We can find the $\textsf{{H}^{+}}$ concentration and hence the pH.

At $\textsf{{25}^{\circ} C}$ The Nernst Equation can be written:

$\textsf{{E}_{c e l l} = {E}_{c e l l}^{\circ} - \frac{0.05916}{z} \log Q}$

$\textsf{z}$ is the no. of moles of electrons transferred

$\textsf{Q}$ is the reaction quotient

The cell reaction when it is working is:

$\textsf{Z n + 2 {H}^{+} \rightarrow Z {n}^{2 +} + {H}_{2}}$

We know it proceeds in this direction since $\textsf{{E}_{c e l l}}$ is +ve.

The reaction quotient is given by:

$\textsf{Q = \frac{\left[Z {n}^{2 +}\right] \times {p}_{{H}_{2}}}{{\left[{H}^{+}\right]}^{2}}}$

We can use partial pressures and concentrations together in this expression as they are normalised against the standard conditions of 1 atmosphere and unit concentration.

This becomes:

$\textsf{Q = \frac{1}{{H}^{+}} ^ 2}$

So The Nernst Equation becomes:

sf(E_(cell)=E_(cell)^(@)-0.05916/(z)log[1/[H^+]^2]

The standard electrode potentials are:

$\textsf{\text{ "E^@} \left(V\right)}$

$\textsf{Z {n}^{2 +} + 2 e r i g h t \le f t h a r p \infty n s Z n \text{ } - 0.76}$

$\textsf{2 {H}^{+} + 2 e r i g h t \le f t h a r p \infty n s {H}_{2} \text{ "0.00}}$

To find sf(E_(cell)^@ you subtract the least positive potential from the most positive $\textsf{\Rightarrow}$

$\textsf{{E}_{c e l l}^{\circ} = 0.00 - \left(- 0.76\right) = + 0.76 \textcolor{w h i t e}{x} V}$

We can now put the numbers in to solve for $\textsf{\left[{H}^{+}\right]}$ and hence find the pH.

sf(0.58=0.76-0.05916/(2)log[1/[H^+]^2]#

$\textsf{- 0.18 = - 0.02958 \log \left[\frac{1}{{H}^{+}} ^ 2\right]}$

$\therefore$$\textsf{\log \left[\frac{1}{{H}^{+}} ^ 2\right] = 6.085}$

From which:

$\textsf{\left[{H}^{+}\right] = 9.314 \times {10}^{- 4} \textcolor{w h i t e}{x} \text{mol/l}}$

$\textsf{p H = - \log \left[{H}^{+}\right] = - \log \left[9.314 \times {10}^{- 4}\right]}$

$\textsf{p H = 3}$