Question

The EMF of a cell corresponding to the reaction:
$Zn(s)+2H^{+}(aq)\to {\left(Zn\right)}^{2+}\left(aq\right)(0.1M)+H_2(g)(1atm.)is0.28voltat25^{\circ }C.$
The pH of the solution at the hydrogen electrode?
${\text{E}}_{\text{Z}{\text{n}}^{2+}/\text{Z}\text{n}}^{\text{o}}=-0.76\text{v}\text{o}\text{l}\text{t};{\text{E}}_{{\text{H}}^{+}/{\text{H}}_2}^{\text{o}}=0$

• A. 2.30
• B. 7.8
• C. 9.2
• D. 8.30

Answer 1

Answer: (D)

The cell may be represented as $\text{Zn}(\text{s})|{\left(\text{Zn}\right)}^{+2}(\text{aq})||{\left(\text{H}\right)}^{+}(\text{aq})|{\left(\text{H}\right)}_2(\text{g})1\text{atm |Pt}(\text{s})$
$$\begin{gathered} \text{Zn}(\text{s})\to {\left(\text{Zn}\right)}^{2+}(\text{a}\text{q})+2\bar{\text{e}}(\text{ Anode }) \\ 2{\left(\text{H}\right)}^{+}(\text{a}\text{q})+2\bar{\text{e}}\to {\left(\text{H}\right)}_2(\text{g})1\text{a}\text{t}\text{m}(\text{ Cathode }) \end{gathered}$$
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$\text{Zn}(\text{s}){\left(\text{+2H}\right)}^{\text{+}}(\text{aq})\stackrel{\text{2}\stackrel{\text{-}}{e}}{\to }{\left(\text{Zn}\right)}^{\text{2+}}(\text{aq}){\left(\text{+H}\right)}_{\text{2}}\text{(g)}\text{1atm}\text{(CellReaction)}$
${\text{E}}_{\text{cell}}=0\text{.}28={\text{E}}_{\text{cell}}^{\text{o}}-\frac{0\text{.}059}{2}\text{log}\frac{\left[{\text{Zn}}_{\text{aq}}^{2+}\right]}{{\left[{\text{H}}^{+}\right]}^2}$
$0\text{.}28=0-\left(-0\text{.}76\right)-\frac{0\text{.}059}{2}\text{ log }\frac{0\text{.}1}{{\left(\left[{\left(\text{H}\right)}^{+}\right]\right)}^2}$
$0\text{.}28-0\text{.}76=-0\text{.}48=\frac{-0\text{.}059}{2}\text{ log }\frac{0\text{.}1}{{\left[{\text{H}}^{+}\right]}^2}$
$\frac{0\text{.}48\times 2}{0\text{.}059}=16\text{.}27=\text{ log }\frac{0\text{.}1}{{\left[{\text{H}}^{+}\right]}^2}$
$\frac{0\text{.}1}{{\left[{\text{H}}^{+}\right]}^2}=1\text{.}862\times 10^{+16}$
${\left[{\text{H}}^{+}\right]}^2=\frac{0\text{.}1}{1\text{.}862\times 10^{+16}}$
$=0.0537\times 10^{-16/2}$
$\left[H^{+}\right]=5.37\times 10^{-9}M$
$pH=-log\left(5.37\times {\left(10\right)}^{-9}\right)pH=9-log\left(5.37\right)pH=9-0.73pH=8.3$