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Which of the following relation represen...

Which of the following relation represents correct relation between standard electrode potential and equilibrium constant?
I. `logK = (nFE^(@))/(2.303RT)`
II. `K = e^((nFE)/(RT))`
III. `log K = (-nFE^(@))/(2.303 RT)`
IV. `log K = 0.4342 (-nFE^(@))/(RT)`
Choose the correct statement(s).

A

I, II and III are correct

B

II and III are correct

C

I, II and IV are correct

D

I and IV are correct

Text Solution

Verified by Experts

The correct Answer is:
C
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Which of the following relations represent correct relation between standard electrode potential and equilibrium constant ? I. logK = (nFE^(@))/(2303 RT) II. K = e^(-(nFE^(@))/(RT)) III. log K = (-nFE^(@))/(2303 RT) IV. log K = 0.4342 (nFE^(@))/(RT)

Show that 2.303 log K=-(triangleH^(0))/(RT)+(triangles^(0))/(R )

For a cell involving one electron E_(cell)^(0)=0.59V and 298K, the equilibrium constant for the cell reaction is: [Given that (2.303RT)/(F)=0.059V at T=298K ]

Variation of equilibrium constan K with temperature is given by van't Hoff equation InK=(Delta_(r)S^(@))/R-(Delta_(r)H^(@))/(RT) for this equation, (Delta_(r)H^(@)) can be evaluated if equilibrium constans K_(1) and K_(2) at two temperature T_(1) and T_(2) are known. log(K_(2)/K_(1))=(Delta_(r)H^(@))/(2.303R)[1/T_(1)-1/T_(2)] Select the correct statement :

Effect of temperature on the equilibrium process analysed by using the thermodynamics From the thermodynamics reaction DeltaG^(@)=-2.30RTlogk DeltaG^(@): Standing free energy change DeltaG^(@)=DeltaH^(@)-TDeltaS^(@) …(ii) DeltaH^(@) : Standard heat of the reaction gt From eqns.(i) and(ii) -2RTlogk=DeltaH^(@)=TDeltaS^(@) DeltaS^(@) : standard entropy change implies" "logK=-(DeltaH^(@))/(2.3RT)+(DeltaS^(@))/(2.3R) Clearly, if a plot of k vs 1/T is made then it is a straight lone having slope =(-DeltaH^(@))/(2.3R) amd y intercept =(DeltaS^(@))/(2.3R) If at temperature T_(1) equilibrium constant be k_(1) and at temperature T_(2) equilibrium constant be k_(2) then : implies" "logK_(1)=-(DeltaH^(@))/(2.3RT_(1))+(DeltaS^(@))/(2.3R) ..(iv) implies" "logK_(2)=-(DeltaH^(@))/(2.3RT_(2))+(DeltaS^(@))/(2.3R) ...(v) Substracting e.q (iv) from (v), we get from the relation we can conclude that the of equilibrium constant increase in temperature for endothermic reaction eith but value of equilibrium constant decrease with the increase in temperature for exothermic reaction For exothermic reaction if DeltaS^(@)lt0 then the sketch of log k vs (1)/(T) may be

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