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# Nernst Equation Calculator

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Nernst Equation determines the half cell reduction potential in non-standard state condition. If the cell potential is zero, this means the reaction is at equilibrium state.

Nernst equation at no-standard state is given by,
$E=E^{^{\circ}}-$$\frac{RT}{nF}$$lnQ$

The Nernst Equation calculator determines the reduction potential of a reaction at 25$^{\circ}$ C. Therefore, the Nernst equation at 25$^{\circ}$ C is given by,
$E=E^{\circ}-$$\frac{0.05916}{n}$$log_{10}$$\frac{a_{Red}}{a_{Ox}} Where, E = reduction potential in V, E^{\circ} = standard cell potential in V, R = gas constant = 8.314 J/mol-K, T = temperature in K, n = number of electron moles transferred in mol, F = Faraday's constant = 96500 coulombs/mol, Q = reaction quotient. ## Steps Back to Top Step 1 : Read the problem and put down the given values. Step 2 : Substitute the values into the Nernst equation and get the cell potential of the reaction. ## Problems Back to Top Given below are some of the problems based on nernst equation. ### Solved Examples Question 1: Determine the reduction potential of the reaction Sn(s)|Sn^{2+}(0.15 M)||Ag^{+}(1.7 M)|Ag(s), if the cell potential is given as +0.94V at 25^{\circ}. Solution: Step 1 : Given parameter values : [Sn^{2+}] = 0.15 M, [Ag^{+}] = 1.7 M, E^{\circ} = +0.94V, n = 2. Step 2 : Reduction potential of the reaction is, E=E^{\circ}-$$\frac{0.05916}{n}$$log_{10}$$\frac{a_{Red}}{a_{Ox}}$

$E=+0.94-$$\frac{0.05916}{2}$$log_{10}$$\frac{[0.15]}{[1.7]^2} E=+0.94 - 0.02958 \times log_{10}[0.0519] E=+0.978V Question 2: Determine the reduction potential of the reaction Fe(s)|Cu^{2+}(aq)(0.3 M)||Fe^{2+}(aq)(0.1 M)|Cu(s), if the cell potential is given as +0.78V at 25^{\circ}. Solution: Step 1 : Given parameter values : [Cu^{2+}] = 0.3 M, [Fe^{2+}] = 0.1 M, E^{\circ} = +0.78V, n = 2. Step 2 : Reduction potential of the reaction is, E=E^{\circ}-$$\frac{0.05916}{n}$$log_{10}$$\frac{a_{Red}}{a_{Ox}}$

$E=+0.78-$$\frac{0.05916}{2}$$log_{10}$$\frac{[0.3]}{[0.1]^2}$

$E=+0.78 - 0.02958 \times log_{10}[0.33]$

$E=+0.794V$