At p H = 2 , E Q u i n h y d r o n e 0 = 1 · 30 V , E Q u i n h y d r o n e will be

E = E 0 - 0.059 2 l o g H + 2 = 1 · 30 - 0.059 2 l o g 10 - 2 2 = 1 · 30 + 0.236 2 = 1 · 418 V

Chemistry - Electrochemistry Question with Solution | TestHub

ChemistryElectrochemistryMiscellaneous/MixedEasy2 minPYQ_2020

ChemistryEasysingle choice

At $p H = 2, E_{Q u i n h y d r o n e}^{0} = 1 \cdot 30 V, E_{Q u i n h y d r o n e}$ will be

Options:

Answer:

Solution:

$E = E^{0} - \frac{0.059}{2} l o g {[H^{+}]}^{2} = 1 \cdot 30 - \frac{0.059}{2} l o g {(10^{- 2})}^{2}$
$= 1 \cdot 30 + \frac{0.236}{2} = 1 \cdot 418 V$

Stream:NTA_ABHYASSubject:ChemistryTopic:ElectrochemistrySubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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