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Chemistry - Electrochemistry Question with Solution | TestHub

ChemistryElectrochemistryGalvanic cells/Nernst Equation/Concentration CellsEasy2 minPYQ_2020

ChemistryEasysingle choice

How much will the reduction potential of a hydrogen electrode change when its solution initially at $p H = 0$ is neutralised to $p H = 7$

Options:

Answer:

Solution:

$E = E^{o} - \frac{0.0591}{' n'} \log \frac{1}{[H^{+}]}$
$E = 0 - \frac{0.0591}{1p} H$
If $p H = 0$ $E = 0$
If $p H = 7$ $E = \frac{- 0.0591}{1} \times 7 = - 0.4137$

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

⏱ 2mℹ️ Source: PYQ_2020

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