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ChemistryElectrochemistryGalvanic cells/Nernst Equation/Concentration CellsMedium2 minQB

ChemistryMediumsingle choice

If in a standard hydrogen electrode half cell $2 H^{+} (aq) + 2 e ⟶ H_{2} (g); 1 atm$ ; on addition of base hydrogen ion concentration is reduced to $1 0^{- 1} (M)$ then the emf will be

Options:

Answer:

Solution:

$E = - \frac{RT}{2 F} ln \frac{1}{[ H ^{+} ] ^{2}}$

$E = - 0.0591 lo g_{10} \frac{1}{[ H ^{+} ] ^{2}}$

$E = - 0.0591 lo g_{10} 10 = - 0.0591 V o l$

Stream:JEESubject:ChemistryTopic:ElectrochemistrySubtopic:Galvanic cells/Nernst Equation/Concentration Cells

⏱ 2mℹ️ Source: QB

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