Calculate the equilibrium constant and free energy change of given following reaction for Daniell cell at 298 K temperature. $Z n_{(s)} + C u_{(a q)}^{2 +} ⇌ Z n_{(a q)}^{2 +} + C u_{(s)}$ Cell potential $= 1.1 v o l t$ ( $F = 96500$ coulomb)

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Solution

$E_{c e l l} = E_{c e l l}^{0} - \frac{R T}{n F} l n Q$
At equilibrium, $Q = K$ and $E_{c e l l} = 0$
So, $E_{c e l l}^{0} = \frac{R T}{n F} l n K$

$E_{c e l l}^{0} = \frac{2.303 R T}{n F} l o g K$
In the given reaction, exchange of two electrons is taking place. Therefore, $n = 2$ .
$1.1 = \frac{2.303 \times 8.314 \times 298}{2 \times 96500} l o g K$
$l o g K = 37.207$
$K = 1.62 \times 10^{37}$
Gibb's free energy change $Δ G = - n F E = - 2 \times 96500 \times 1.1$
$Δ G = 212, 300 J = 212.3 k J$

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Calculate the equilibrium constant and free energy change of given following reaction for Daniell cell at 298 K temperature. Zn(s)+Cu2+(aq)⇌Zn2+(aq)+Cu(s) Cell potential =1.1 volt (F=96500 coulomb)

Calculate the equilibrium constant and free energy change of given following reaction for Daniell cell at 298 K temperature. $Z n_{(s)} + C u_{(a q)}^{2 +} ⇌ Z n_{(a q)}^{2 +} + C u_{(s)}$ Cell potential $= 1.1 v o l t$ ( $F = 96500$ coulomb)