What is the approximate standard free energy change per mole of Zn (in kJ mol -1 ) for a Daniel cell at 298 K ?

Question

What is the approximate standard free energy change per mole of Zn (in kJ mol -1 ) for a Daniel cell at 298 K ? (A) -212.3 (B) 230.0 (C) 0.0 (D) -1.10

Tardigrade Admin · Accepted Answer

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 ∴ Gibbs free energy=-2 × 96500 × 1.1 J =-212.3 kJ

Q. What is the approximate standard free energy change per mole of $Z n$ (in $k J m o l^{- 1}$ ) for a Daniel cell at $298 K$ ?

-212.3

230.0

0.0

-1.10

For Daniell cell we have

$Z n (s) + C u^{2 +} (a q) ⟶ Z n^{2 +} (a q) + C u (s)$

Gibbs free energy $= - n FE$

$n =$ number of electrons $= 2$

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

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

$∴$ Gibbs free energy $= - 2 \times 96500 \times 1.1 J = - 212.3 k J$

Q. What is the approximate standard free energy change per mole of Zn (in kJmol−1 ) for a Daniel cell at 298K ?

-212.3

230.0

0.0

-1.10

For Daniell cell we have Zn(s)+Cu2+(aq)⟶Zn2+(aq)+Cu(s) Gibbs free energy =−nFE n= number of electrons =2 F= Faraday's constant =96500C E= emf of cell =1.1V ∴ Gibbs free energy=−2×96500×1.1J=−212.3kJ

Q. What is the approximate standard free energy change per mole of $Z n$ (in $k J m o l^{- 1}$ ) for a Daniel cell at $298 K$ ?

For Daniell cell we have

$Z n (s) + C u^{2 +} (a q) ⟶ Z n^{2 +} (a q) + C u (s)$

Gibbs free energy $= - n FE$

$n =$ number of electrons $= 2$

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

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

$∴$ Gibbs free energy $= - 2 \times 96500 \times 1.1 J = - 212.3 k J$