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ChemistryElectrochemistryMiscellaneous/MixedEasy2 minPYQ_2020

ChemistryEasysingle choice

${Cu}^{2+} + {2e}^{-} \to Cu; log [{Cu}^{2+}] vs . E_{red}$ graph is of the type as shown in figure where $OA = 0.34 V$ then electrode potential of the half cell of $C u | C u^{2 +} (0.1 M)$ will be

Options:

Answer:

Solution:

$E_{Cu / {Cu}^{2 +}} = E_{Cu / {Cu}^{2 +}}^{0} - \frac{0.059}{2} \log [{Cu}^{2 +}]$

If $\log [{Cu}^{2 +}] = 0$ i.e., $[{Cu}^{2 +}] = 1$ , then $E_{Cu / {Cu}^{2 +}} = E_{Cu / {Cu}^{2 +}}^{o}$

or $E_{Cu / {Cu}^{2 +}}^{o} = - E_{{Cu}^{2 +} / Cu}^{o} = - 0.34$

Now, $E_{Cu / {Cu}^{2 +}} = - 0.34 - \frac{0.059}{2} \log 0.1$

$= - 0.34 + \frac{0.059}{2}$

Stream:NTA_ABHYASSubject:ChemistryTopic:ElectrochemistrySubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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