Question

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

• A. 2.30
• B. 7.8
• C. 9.2
• D. 8.30

Answer 1

Answer: (D)

The cell may be represented as $\text{Zn}\left(\right.\text{s}\left.\right)\left|\right. \left(\text{Zn}\right)^{+ 2} \left(\right. \text{aq} \left.\right) \left|\right.\left|\right. \left(\text{H}\right)^{+} \left(\right. \text{aq} \left.\right) \left|\right.\left(\text{H}\right)_{2}\left(\right.\text{g}\left.\right)1\text{atm |Pt}\left(\right.\text{s}\left.\right)$
$ \, \text{Zn}\left(\right. \text{s} \left.\right) \, \, \, \rightarrow \, \, \left(\text{Zn}\right)^{2 +}\left(\right. \text{a} \text{q} \left.\right)+2\bar{\text{e}} \, \left(\right.\text{ Anode }\left.\right) \\ 2\left(\text{H}\right)^{+}\left(\right. \text{a} \text{q} \left.\right)+2\bar{\text{e}} \, \, \rightarrow \, \, \, \left(\text{H}\right)_{2}\left(\right.\text{g}\left.\right) \, \, 1\text{a}\text{t}\text{m} \, \left(\right.\text{ Cathode }\left.\right)$
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$\text{Zn} \left(\right. \text{s} \left.\right) \left(\text{+2H}\right)^{\text{+}} \left(\right. \text{aq} \left.\right) \overset{\text{2} \overset{\text{-}}{\text{e}}}{ \rightarrow } \left(\text{Zn}\right)^{\text{2+}} \left(\right. \text{aq} \left.\right) \left(\text{+H}\right)_{\text{2}} \text{(g)} \, \text{1atm} \, \text{(CellReaction)}$
$\text{E}_{\text{cell}} = 0 \text{.} 2 8 = \text{E}_{\text{cell}}^{\text{o}} - \frac{0 \text{.} 0 5 9}{2} \text{log} \frac{\left[\text{Zn}_{\text{aq}}^{2 +}\right]}{\left[\text{H}^{+}\right]^{2}}$
$0 \text{.} 2 8 = 0 - \left(- 0 \text{.} 7 6\right) - \frac{0 \text{.} 0 5 9}{2} \text{ log } \frac{0 \text{.} 1}{\left(\left[\left(\text{H}\right)^{+}\right]\right)^{2}}$
$0 \text{.} 2 8 - 0 \text{.} 7 6 = - 0 \text{.} 4 8 = \frac{- 0 \text{.} 0 5 9}{2} \text{ log } \frac{0 \text{.} 1}{\left[\text{H}^{+}\right]^{2}}$
$\frac{0 \text{.} 4 8 \times 2}{0 \text{.} 0 5 9} = 1 6 \text{.} 2 7 = \text{ log } \frac{0 \text{.} 1}{\left[\text{H}^{+}\right]^{2}}$
$\frac{0 \text{.} 1}{\left[\text{H}^{+}\right]^{2}} = 1 \text{.} 8 6 2 \times 1 0^{+ 1 6}$
$\left[\text{H}^{+}\right]^{2} = \frac{0 \text{.} 1}{1 \text{.} 8 6 2 \times 1 0^{+ 1 6}}$
$= 0.0537 \times 10^{- 16 / 2}$
$\left[H^{+}\right]=5.37\times 10^{- 9}M$
$pH=-log\left(5 . 37 \times \left(10\right)^{- 9}\right)pH=9-log\left(5 . 37\right)pH=9-0.73pH=8.3$