Question

The reduction potential of H-electrode at 298 K containing 10–5 M H+ is (pH2 = 1 atm)

Answer 1

Answer:

The reduction potential of H-electrode at 298 K containing 10-5 M H+ is (PH2 = 1 atm) + 0.2955 V. 0 – 0.2955 V. 0 – 0.1255 V.

Answer 2

Answer:

Reduction potential of H - electrode is $-0.2955$ Volts.

Explanation:

Given: H-electrode at 298 K containing 10–5 M H+ is (pH2 = 1 atm)

To find: Reduction potential of H-electrode

Solution:

we have to find the reduction potential of $\mathrm{H}$- electrode at $298 \mathrm{~K} containing 10^{\wedge}-5 \mathrm{M} \mathrm{H}^{+} where pressure of hydrogen atom, \mathrm{PH}_{2}=1 \mathrm{~atm} .$

The reduction reaction of hydrogen ions is $\ldots 2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \Rightarrow \mathrm{H}_{2}$

now using Nernst equation,

$\begin{aligned}&E_{\text {cell }}=E_{0}-\frac{0.0591}{2} \log \left(\frac{P_{H_{2}}}{\left[H^{+}\right]^{2}}\right) \\&=0-0.0591 / 2 \log \left(1 /\left(10^{\wedge}-5\right)^{2}\right) \\&=-0.0591 / 2 \log \left(10^{10}\right) \\&=-0.0591 \times 5 \\&=-0.2955 \mathrm{~V}\end{aligned}$

Therefore reduction potential of H - electrode is $-0.2955$ Volts.