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ChemistryElectrochemistryMiscellaneous/MixedEasy2 minPYQ_2020

ChemistryEasysingle choice

The voltage of the cell consisting of $L i_{(s)}$ and $F_{2 (g)}$ electrodes is 5.92 V at standard condition at 298 K. What is the voltage if the electrolyte consists of 2 M LiF.

(ln2 = 0.693, R = 8.314 J $K^{- 1} m o l^{- 1}$ and F = 96500 C $m o l^{- 1}$ )

Options:

Answer:

Solution:

$2 L i_{(s)} + F_{2 (g)} \to 2 L i F_{(a q)}$

$E_{c e l l}^{o} = 5.92 V$

$E_{c e l l} = E_{c e l l}^{o} - \frac{0.0591}{2} l o g [L i F]^{2}$

$E_{c e l l} = 5.92 - \frac{0.059}{2} l o g 2^{2}$

$= 5.92 - 0.059 \times 0.3010 = 5.90$ V

Stream:NTA_ABHYASSubject:ChemistryTopic:ElectrochemistrySubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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