Question

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
0 + 0.1255 V​

Answer 1

we have to find the reduction potential of H - electrode at 298K containing 10^-5 M H⁺ where pressure of hydrogen atom, PH₂ = 1 atm.

solution : reduction reaction of hydrogen ions is ... 2H⁺ + 2e¯⇒H₂

now using Nernst equation,

$\quad E_{cell}=E_0-\frac{0.0591}{2}log\left(\frac{P_{H_2}}{[H^+]^2}\right)$

= 0 - 0.0591/2 log( 1/(10^-5)²)

= -0.0591/2 log(10¹⁰)

= -0.0591 × 5

= -0.2955 V

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

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