What would the potential of a standard hydrogen electrode (s. h. e.) be under the given conditions? [h+]=0.46 mph2=1.9 atmt=298 k?

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