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

It's because 3 if it was 10 it wouldn't look right do it goes up to 4d