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Q. What is the approximate standard free energy change per mole of $Zn$ (in $kJ\, mol ^{-1}$ ) for a Daniel cell at $298\, K$ ?

TS EAMCET 2018

Solution:

For Daniell cell we have

$Zn (s)+ Cu ^{2+}(a q) \longrightarrow Zn ^{2+}(a q)+ Cu (s)$

Gibbs free energy $=-n F E$

$n=$ number of electrons $=2$

$F=$ Faraday's constant $=96500\, C$

$E=$ emf of cell $=1.1\, V$

$\therefore $ Gibbs free energy$=-2 \times 96500 \times 1.1\, J =-212.3\, kJ$