Chemistry - Electrochemistry Question with Solution | TestHub

The cell may be represented as$\text{Zn}(\text{s})\left|{\text{Zp}}^2\left(\text{aq}\right)\right|\left|\text{H}\left(\text{aq}\right)\right|{\text{H}}_2(\text{g})1\text{atm}\text{Pt}(\text{s})$

$\begin{array}{l}\text{Zn}\left(\text{s}\right)\to {\text{Zn}}^{2+}\left(\text{a}\text{q}\right)+2\overline{\text{e}}(\text{ Anode })\\ 2{\text{H}}^{+}\left(\text{a}\text{q}\right)+2\overline{\text{e}}\to {\text{H}}_2(\text{g})1\text{a}\text{t}\text{m}(\text{ Cathode })\end{array}$
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$\text{Zn}\left(\text{s}\right){\text{+2H}}^{\text{+}}\left(\text{aq}\right)\stackrel{\text{2}\stackrel{\text{-}}{\text{e}}}{\to }{\text{Zn}}^{\text{2+}}\left(\text{aq}\right){\text{+H}}_{\text{2}}\text{(g)}\text{1atm}\text{(CellReaction)}$

${\text{E}}_{\text{cell}}=0\text{.}28={\text{E}}_{\text{cell}}^{\text{o}}-\frac{0\text{.}059}{2}\text{log}\frac{\left[{\text{Zn}}_{\text{aq}}^{2+}\right]}{{\left[{\text{H}}^{+}\right]}^2}$

$0\text{.}28=0-\left(-0\text{.}76\right)-\frac{0\text{.}059}{2}\text{ log }\frac{0\text{.}1}{{\left[{\text{H}}^{+}\right]}^2}$

$0\text{.}28-0\text{.}76=-0\text{.}48=\frac{-0\text{.}059}{2}\text{ log }\frac{0\text{.}1}{{\left[{\text{H}}^{+}\right]}^2}$

$\frac{0\text{.}48\times 2}{0\text{.}059}=16\text{.}27=\text{ log }\frac{0\text{.}1}{{\left[{\text{H}}^{+}\right]}^2}$

$\frac{0\text{.}1}{{\left[{\text{H}}^{+}\right]}^2}=1\text{.}862\times 10^{+16}$

${\left[{\text{H}}^{+}\right]}^2=\frac{0\text{.}1}{1\text{.}862\times 10^{+16}}$

$=0.0537\times {10}^{-16/2}$

$\left[{\text{H}}^{+}\right]=5\text{.}37\times 10^{-9}$

pH = - log (5.37 x 10 -9)

= 9 - log 5,37

= 9 - 0.73 = 8.27

$\approx 8\text{.}3$