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Which of the following relations represe...

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)`

A

I,II and III

B

II and III

C

I, II and IV

D

I and IV

Text Solution

Verified by Experts

The correct Answer is:
C
`DeltaG =- 2.303 RT log K, -nFR^(@) =- 2.303 RT log K`
`log K = (nFE^(@))/(2.303RT) = 0.4342 (nFE^(@))/(RT)` ..(v)
ln `K = (nFE^(@))/(RT)`
`K = (-nFE^(@))/(e^(RT))` ...(ii)
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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).

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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

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 ]

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 If statndard heat of dissociation of PCl_(5) is 230 cal then slope of the graph of log vs (1)/(T) is :

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