Question

The emf of a cell corresponding to the reaction,$Zn+2{{H}^{+}}(aq)\to Z{{n}^{+2}}(0.1M)+{{H}_{2}}(g)1atm$
is 0.28 volt at 25 degree Celsius. Write the half-cell reactions and calculate the pH of the solution at the hydrogen electrode. ${{E}^{0}}_{Z{{n}^{2+}}/Zn}$ =-0.76 volt, ${{E}^{0}}_{{{H}^{+}}/{{H}_{2}}}$=0.

Answer 1

Look at the given information in this question. As we can see, we’ve been given the values for electrode potential along with the concentration and number of electron exchange. Therefore, we need to apply Nernst equation.

Complete step by step answer:
The given reaction is mentioned below:
$Zn+2{{H}^{+}}\to Z{{n}^{+2}}+{{H}_{2}}$
The half cell reaction which occurs at anode will be:
$Zn\to Z{{n}^{+2}}+2{{e}^{-}}$ -- 1
The half-cell reaction which occurs at cathode will be:
$2{{H}^{+}}+2{{e}^{-}}\to {{H}_{2}}$-- 2
- From the above equations of half-cell, we can say that there is a net exchange of 2 electrons.
- From equation 1, we can say that zinc gets oxidized. Therefore, oxidation potential of Zn will be
${{E}^{0}}_{Z{{n}^{2+}}/Zn}$ = 0.76volt
From equation 2, we can say that hydrogen gets reduced but the reduction potential of hydrogen is always zero.
Therefore, the standard electrode potential of the cell will be –

& E_{cell}^{\circ } = E_{{{H}^{+}}/{{H}_{2}}}^{\circ }+E_{Zn/Z{{n}^{2+}}}^{\circ } \\\ & {{E}^{o}}_{cell} = 0.76\text{ }+\text{ }0\text{ }=\text{ }0.76volt \\\ \end{aligned}$$ \- Applying the Nernst equation, $${{E}_{cell}}=E_{cell}^{\circ }-\dfrac{0.0591}{n}\log \left[ \dfrac{[Z{{n}^{+2}}]}{{{\left[ {{H}^{+}} \right]}^{2}}} \right]$$ Substituting the value in the Nernst equation we get – $$0.28=0.76-\dfrac{0.0591}{n}\log \left[ \dfrac{0.1}{{{\left[ {{H}^{+}} \right]}^{2}}} \right]$$ From the reaction above, we can see, n = 2. $$\dfrac{0.48\text{x}2}{0.0591}=\log \left[ \dfrac{0.1}{{{\left[ {{H}^{+}} \right]}^{2}}} \right]$$ $$1.86\text{x}{{10}^{16}}=\dfrac{0.1}{{{\left[ {{H}^{+}} \right]}^{2}}}$$ $${{\left[ {{H}^{+}} \right]}^{2}}=5.37\text{x}{{10}^{-18}}$$ $$\left[ {{H}^{+}} \right]=2.31\text{x}{{10}^{-9}}$$ Since, $$\begin{aligned} & pH\text{ }=\text{ }-log\text{ }({{H}^{+}}) \\\ & pH=\text{ }-\text{ }log\text{ }(2.31\times {{10}^{-9}}) \\\ & pH\text{ }=\text{ }8.62 \\\ \end{aligned}$$ Therefore, the pH of the solution at the hydrogen electrode is 8.62. **Note:** pH is a scale for measuring the acidity or basicity of any solution. Lesser the pH, more acidic is the compound.