Respuesta :
The equation is
2H⁺ 2e⁻ → H₂
The number of electrons transferred in reaction are2.
The potential of standard hydrogen(S.H.E) electrode can be calculated using Nervast equation
Ecell = E₀(cell)- RT/nF In [cathode]/[anode]
Where E₀cell is the standard potential of the cell with a value of 0.00V, where R is the gas
Constant = 8.34JK-1 mol-1 T is temperature in Kelvin
= 298k and F is the Faraday constant whose value is 96500 C/mol, n being the number of electrons transferred to the reaction.
substitute given
Ecell = E₀(cell) - RT/nF In [cathode]/[anode]
Ecell = E₀(cell) -RT/nF In [PH₂]/[H⁺]²
Ecell = 0.00 - 8.314 JK⁻1 mol⁻1 × 298k/2× 96500 C/mol In (1.9 atm]/[0.46]²
0.00 - 0.0129 In(8.98)
Ecell = 0.00 - 0.0129 × 6.951
Ecell is = 0.0897V
2H⁺ 2e⁻ → H₂
The number of electrons transferred in reaction are2.
The potential of standard hydrogen(S.H.E) electrode can be calculated using Nervast equation
Ecell = E₀(cell)- RT/nF In [cathode]/[anode]
Where E₀cell is the standard potential of the cell with a value of 0.00V, where R is the gas
Constant = 8.34JK-1 mol-1 T is temperature in Kelvin
= 298k and F is the Faraday constant whose value is 96500 C/mol, n being the number of electrons transferred to the reaction.
substitute given
Ecell = E₀(cell) - RT/nF In [cathode]/[anode]
Ecell = E₀(cell) -RT/nF In [PH₂]/[H⁺]²
Ecell = 0.00 - 8.314 JK⁻1 mol⁻1 × 298k/2× 96500 C/mol In (1.9 atm]/[0.46]²
0.00 - 0.0129 In(8.98)
Ecell = 0.00 - 0.0129 × 6.951
Ecell is = 0.0897V
The potential of the standard hydrogen electrode at the given conditions will be [tex]\boxed{0.0897\text{ V}}[/tex].
Explanation:
Given:
The concentration of the aqueous [tex]H^+[/tex] ions is [tex]0.46\text{ M}[/tex].
The pressure of the gas formed is [tex]1.9\text{ atm}[/tex].
The temperature of the surroundings is [tex]298\text{ K}[/tex].
Concept:
The equation of the Redox-half reaction of the standard platinum electrode is given as:
[tex]\boxed{2H^+(aq)\ +2e^-\ \rightarrow\ H_2(g)}[/tex]
The potential of the standard hydrogen electrode is given by the expression:
[tex]\boxed{E_{cell}=E_0-\dfrac{RT}{nF}\ ln\dfrac{[P_{H_2}]}{[H^+]^2}}[/tex]
Here, [tex]E_{cell}[/tex] is the potential of the cell, [tex]E_0[/tex] is the standard potential of cell, [tex]R[/tex] is the gas constant and [tex]F[/tex] is the Faraday's constant.
The value of the gas constant is [tex]8.34\dfrac{J}{K\cdot mol}[/tex] and the value of [tex]F[/tex] is [tex]96500\text{ C/mol}[/tex].
Substitute the value in above expression.
[tex]\begin{aligned}E_{cell}&=0.0-\dfrac{8.34\times298}{2\times96500}\ ln\dfrac{1.9}{(0.46)^2}\\&=0.0-0.0129\ ln(8.98)\\&=0.0897\text{ V}\end{aligned}[/tex]
Thus, The potential of the standard hydrogen electrode at the given conditions will be [tex]\boxed{0.0897\text{ V}}[/tex].
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Answer Details:
Grade: Senior school
Subject: Chemistry
Topic: Standard Hydrogen Electrode
Keywords:
Electrode, hydrogen, electric, potential, standard, aqueous, temperature, platinum, Standard hydrogen electrode, S.H.E., concentration, pressure.
