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17 October, 19:47

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?

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  1. 17 October, 20:11
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    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
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