A hydrogen electrode is made by dipping platinum wire in a solution of nitric acid of $\mathrm{pH}=9$ and…

A hydrogen electrode is made by dipping platinum wire in a solution of nitric acid of $\mathrm{pH}=9$ and passing hydrogen gas around the platinum wire at $1.2 \mathrm{~atm}$ pressure. The oxidation potential of such an electrode equals .V.
  1. +0.059
  2. -0.0531
  3. -0.059
  4. +0.0531

Solution

For hydrogen electrode, oxidation half reaction is, $ \underset{1.2 \mathrm{~atm}}{\mathrm{H}_2} \longrightarrow \underset{\text { (at pH 9) }}{2 \mathrm{H}^{+}}+2 e^{-} $ If $\mathrm{pH}=9$ then $\mathrm{H}^{+}$ion $=1 \times 10^{-\mathrm{pH}}=1 \times 10^{-9}$ Using Nernst equation, $ E_{\text {cell }}=E_{\text {cell }}^0-\frac{0.0591}{n} \log \frac{\left[\mathrm{H}^{+}\right]^2}{\mathrm{p}_{\mathrm{H}_2}} $ For hydrogen electrode, $E_{\text {cell }}^0=0, n=2$ $ \begin{aligned} & E_{\text {Cell }}=0-\frac{0.0591}{2} \log \frac{\left(10^{-9}\right)^2}{1.2} \\ & E_{\text {Cell }}=-\frac{0.0591}{2}\left[\log 10^{-18}-\log 1.2\right] \Rightarrow E_{\text {Cell }}=+0.0531 \end{aligned} $

Asked in: AP EAMCET 2021 (25 Aug Shift 1)

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