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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

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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

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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}$

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

$E=+0.794V$