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Answer :
Sr(s)+Mg²+(aq)→Sr²+(aq)+Mg(s)
Number of e-'s transfered are, n=2. Equilibrium constant,
K=2.69×10∧12
ΔG=-2.303RT logK
R=gasconstant=8.314J/mol-k
T= temperature in K= 25 oC=25+273=298K
The value we get ΔG = -70922.3J. But ΔG = -nFE
n= number of e-'s transfered in the reaction =2
F= farady = 96500C
E=potential of the cell is what?
∴E = ΔG.nF
=-(-70922.3)/2×96500)
=0.367v.
Number of e-'s transfered are, n=2. Equilibrium constant,
K=2.69×10∧12
ΔG=-2.303RT logK
R=gasconstant=8.314J/mol-k
T= temperature in K= 25 oC=25+273=298K
The value we get ΔG = -70922.3J. But ΔG = -nFE
n= number of e-'s transfered in the reaction =2
F= farady = 96500C
E=potential of the cell is what?
∴E = ΔG.nF
=-(-70922.3)/2×96500)
=0.367v.
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Final answer:
The given equilibrium constant can be used, along with the Nernst equation, to solve for the standard cell potential. This involves substituting the known values into the equation and solving for E⁰.
Explanation:
The cell reaction's equilibrium constant (K) gives a numerical measure of how far the response goes at equilibrium. Using the Nernst Equation, we can relate this to the standard cell potential (E⁰). According to this equation, we have E⁰ = - (RT/nF)ln(K), where R is the gas constant (8.314 J/mol.K), T is the temperature in Kelvin (here, T=25°C=298K), n is the number of moles of electrons transferred in the cell reaction (which is 2 when the Sr and Mg reactions are balanced), and F is the Faraday's constant (96485 C/mol).
By substituting the known values into this equation, we can solve for E⁰. Using the given value of K (2.69 × 10¹²), R, T, n, and F, we find E⁰ = - (8.314 J/mol.K * 298K / (2 * 96485 C/mol))*ln(2.69 × 10¹²). You can solve this equation on your calculator to find the value of E⁰ for the cell.
Learn more about Cell Potential here:
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