Question

The reduction potential of H-electrode in a solution of 0.1MH^(+)(aq) is (p_(2)=1atm)​

Answer 1

Answer:

Correct option is

A

0.00V

2H

+

+2e

−

→H

2

E

cell

=E

0

−

2

0.059

log[

(H

+

)

2

P

H

2

]

E

cell

=0−0.059log[

(0.1)

1

]

E

cell

=0.05glog(8.1)

=−0.05gV

Answer 2

Answer:

The reduction potential of H-electrode is -0.118 V.

Concept:

Nernst equation

Given:

Concentration of hydrogen, [H+] = 0.1 M

Pressure, $P_2$ = 1 atm

Find:

Reduction potential of H-electrode.

Solution:

At hydrogen electrode, $H^+ + e^- = \frac{1}{2} H_2$

According to Nernst equation, we have

$E_{cell} = E_{cell}^0 - 0.059\hspace{1mm} log [\frac{P_2}{[H^+]} ]$

$= 0 - 0.059\hspace{1mm}log[\frac{1}{(0.1)^2} ]\\= -0.059\hspace{1mm} log 100\\= -0.059\times 2\\= -0.118 V$

Hence, the reduction potential of H-electrode is -0.118 V.

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