The reduction potential of a hydrogen electro | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The reduction potential of the hydrogen electrode is calculated using the Nernst equation for the reaction: $2H^{+} + 2e^{-} \rightarrow H_2$. $E_

Vedclass · Accepted Answer

$-0.413$

Vedclass · Answer

(C) The reduction potential of the hydrogen electrode is calculated using the Nernst equation for the reaction: $2H^{+} + 2e^{-} ightarrow H_2$. $E_{red} = E_{red}^{0} - \frac{0.0591}{n} \log \frac{P_{H_2}}{[H^{+}]^2}$. For a neutral solution,the $H^{+}$ ion concentration is $10^{-7} \ M$ because the $pH = 7$. Substituting the values: $E_{H^{+}|H_2} = 0 - \frac{0.0591}{2} \log \frac{1}{(10^{-7})^2}$. $E_{H^{+}|H_2} = -\frac{0.0591}{2} \log 10^{14}$. $E_{H^{+}|H_2} = -\frac{0.0591 	imes 14}{2} \log 10$. $E_{H^{+}|H_2} = -0.4137 \ V \approx -0.413 \ V$.

The reduction potential of a hydrogen electrode at $25^{\circ} C$ in a neutral solution is $(P_{H_2} = 1 \ atm)$. (in $V$)

Similar Questions

In the following reaction,what is the value of equilibrium constant?
$Cu_{(s)} + 2 Ag_{(aq)}^{+} \rightarrow Cu_{(aq)}^{2+} + 2 Ag_{(s)}$
$E_{cell}^0 = 0.46 \ V$

For the cell reaction $Zn_{(s)} + 2H^+_{(aq)} \to Zn^{2+}_{(aq)} + H_{2(g)}$,what happens when $H_2SO_4$ is added to the cathode compartment?

In which of the following conditions will the reduction potential of a hydrogen half-cell be negative?

Calculate the cell potential for $Zn_{(s)} | Zn^{2+} (0.6 \ M) || Cd^{2+} (0.2 \ M) | Cd_{(s)}$ at $298 \ K$. (in $V$)

The potential of a hydrogen electrode with $pH = 10$ with respect to a standard hydrogen electrode is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module