Question

At 25 C temperature, for the reaction, Sn + Cu2+ -----> Sn2+ + Cu the value of E cell is

0.48 V then what will be the equilibrium constant of the reaction?​

Answer 1

Answer:

o.222 degree ÷55 09 answer is 44

Answer 2

Final answer: Equilibrium constant = $1.64\times10^{16}$

Given that: We are given,

$Sn + Cu^{2+}$ → $Sn^{2+} + Cu$

$E^{0}_{cell} = 0.48$ $V$

To find: We have to find equilibrium constant of the reaction.

Explanation:

• $\triangle G^{o}$ for a reaction is proportional to both the potential and the number of electrons transferred.

                         $\triangle G^{0} = -nFE^{0}_{cell}$

• Relationship between $\triangle G^{o}$ and the equilibrium constant $K$ is given by:

                       $\triangle G^{0}=-2.030RT\log K$

• We can write,

                 $-nFE^{0}_{cell}=-2.030RT\log K$

• $E^{0}_{cell}=\frac{2.303RT}{nF} \log K$

$T=25\°C=273+25=298K$ (given)

R = Gas constant= $8.314 Jmol^{-1}K^{-1}$

F = Faraday constant = $96500$ $C$

• At $298$ $K$, $\frac{2.303RT}{nF} = \frac{0.0592}{n}$

• Substitute this value into $E^{0}_{cell}$.

               $E^{0}_{cell}=\frac{0.0592}{n}\log K$

• $E^{0}_{cell} = 0.48$ $V$ (given)

• n = Number of electron involved

Reaction at anode: $Sn$ → $Sn^{2+}+2e^{-}$

Reaction at cathode: $Cu^{2+}+2e^{-}$ → $Cu$

Reaction: $Sn + Cu^{2+}$ → $Sn^{2+} + Cu$

So $n=2$

• Substitute the values of  $n$ and $E^{0}_{cell}$.

  $0.48 =\frac{0.0592}{2} \log K$

         $=0.0296 \times \log K$

$\log K=\frac{0.48}{0.0296}=16.216$

• Taking anti-logarithm

$K=1.64\times10^{16}$

• Hence, equilibrium constant = $1.64\times10^{16}$

To know more about the concept please go through the links

https://brainly.in/question/10308374

https://brainly.in/question/9980535

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