Given the equilibrium constant : $$K_{c}$$ of the reaction : $$Cu(s)+2Ag^{+}(aq)rightarrow Cu^{2+}(aq)+2Ag(s)$$ is 10 X10 15 ,calculate the $$mathrm{E_{cell}^{varnothing }textrm{ in V}}$$ of this reaction at 298 K $$left [ 2.303frac{RT}{F}at: 298 K=0.059 V right ]$$ (Response should be upto only twodecimal digit like 78.76 or 0.78 or 0.03).

Given the equilibrium constant :

$K_{c}$ of the reaction :

$Cu(s)+2Ag^{+}(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)$ is 10 X10¹⁵calculate the $E_{cell}^{\varnothing }$ of this reaction at 298 K

$\left [ 2.303\frac{RT}{F}at\: 298 K=0.059 V \right ]$
Option 1)
0.4736 mV
Option 2)
0.04736 V
Option 3)
0.4736 V
Option 4)
0.04736 mV

Answers (1)

A admin

Equilibrium constant and Enthalpy Change -

The equilibrium constant for an endothermic reaction ( positive $\bigtriangleup H$ ) increases as the temperature increases.

The equilibrium constant for the reverse reaction -

The equilibrium constant for the reverse reaction is the inverse of the equilibrium constant for the reaction in the forward direction.

- wherein

$K'_{c}=\frac{1}{K_{c}}$

${K'_{c}}$ is equilibrium constant for reverse direction.

$E_{cell}^{0}=\frac{0.0591}{N}log\: k_{c}=\frac{0.0591}{2}log(1\times 10^{16})=0.4728 V$

Option 1)

0.4736 mV

Option 2)

0.04736 V

Option 3)
0.4736 V
Option 4)

0.04736 mV

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Given the equilibrium constant : of the reaction : is 10 X1015 calculate the of this reaction at 298 K Option 1)0.4736 mVOption 2)0.04736 VOption 3)0.4736 VOption 4)0.04736 mV

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