Question

calculate delta G° and log Kc for the following reaction at 298 K. 2Al(s)+3Cu2+(aq)-------->2Al3+(Aq)+3cu(s) and given:E°cell=2.02v​

Answer 1

Del G°=- nFE°

Where,n=6, E°=2.02,F=96500

Del G°=-6*2.02*96500=-1169580

Answer 2

The value of $\Delta G^o$ is -389860 J/mol and $\log K_c$ is 68.33

Explanation:

For the given chemical equation:

$2Al(s)+3Cu^{2+}(aq.)\rightarrow 2Al^{3+}(aq.)+3Cu(s)$

The half reaction follows:

Oxidation half reaction:  $Al(s)\rightarrow Al^{3+}(aq.)+3e^-$      ( × 2 )

Reduction half reaction:  $Cu^{2+}(aq.)+2e^-\rightarrow Cu(s)$      ( × 3 )

To calculate standard Gibbs free energy, we use the equation:

$\Delta G^o=-nFE^o_{cell}$

Where,

n = number of electrons transferred = 6

F = Faraday's constant = 96500 C

$E^o_{cell}$ = standard cell potential = 2.02 V

Putting values in above equation, we get:

$\Delta G^o=-2\times 96500\times 2.02=-389860J/mol$

To calculate the $\log K_c$ for given value of Gibbs free energy, we use the relation:

$\Delta G^o=-2.303RT\log K_c$

where,

$\Delta G^o$ = standard Gibbs free energy = -389860 J/mol

R = Gas constant = $8.314J/K mol$

T = temperature = 298 K

$K_c$ = equilibrium constant in terms of concentration = ?

Putting values in above equation, we get:

$-389860J/mol=-(2.303\times 8.314J/Kmol)\times 298K\times \log K_c\\\\\log K_c=68.33$

Learn more about Gibbs free energy:

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