Assume a cell with the following reaction Cu ( s ) + 2 Ag + 1 × 10 - 3 M → Cu 2 + ( 0 . 250 M ) + 2 Ag ( s ) E cell ∘ = 2 . 97 V E cell for the above reaction is V . (Nearest integer) [ Given : log 2 . 5 = 0 . 3979 , T = 298 K ]

E = E ° - 0 . 059 2 log Cu + 2 Ag + 2 = 2 . 97 - 0 . 059 2 log 0 . 25 10 - 3 2 = 2 . 81 V

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ChemistryElectrochemistryGalvanic cells/Nernst Equation/Concentration CellsEasy2 minPYQ_2021

ChemistryEasystatement

Assume a cell with the following reaction ${Cu}_{(s)} + 2 {Ag}^{+} (1 \times 10^{- 3} M) \to {Cu}^{2 +} (0.250 M) + 2 {Ag}_{(s)}$
$E_{cell}^{\circ} = 2.97 V$

$E_{cell}$ for the above reaction is $V .$

(Nearest integer)

$[$ Given : $\log 2.5 = 0.3979, T = 298 K]$

Answer:

Solution:

$E = E ° - \frac{0.059}{2} \log \frac{[{Cu}^{+ 2}]}{{[{Ag}^{+}]}^{2}}$
$= 2.97 - \frac{0.059}{2} \log \frac{0.25}{{(10^{- 3})}^{2}} = 2.81 V$

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

⏱ 2mℹ️ Source: PYQ_2021

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