Nernst Equation Numericals

CHENernst Equation & Thermodynamics

The Nernst equation gives the cell or electrode potential under non-standard (any) concentration, correcting $E^\circ$ for the reaction quotient. It is used for Daniell-cell EMF at given ion concentrations, single-electrode potentials, concentration cells, and to extract the equilibrium constant from EMF.

Nernst equation (298 K)

E_{cell} = E^\circ_{cell} - \dfrac{0.0591}{n}\log Q

Q = \dfrac{[\text{products}]}{[\text{reactants}]}

for the cell reaction; solids and pure metals taken as

1

General (any T)

E_{cell} = E^\circ_{cell} - \dfrac{RT}{nF}\ln Q

R = 8.314\,\text{J K}^{-1}\text{mol}^{-1}

T

in K,

F = 96500\,\text{C}

; the

0.0591

comes from

\dfrac{2.303RT}{F}

298\,\text{K}

Concentration cell

E_{cell} = \dfrac{0.0591}{n}\log\dfrac{[\text{ion}]_{cathode}}{[\text{ion}]_{anode}}

E^\circ_{cell} = 0

; EMF positive only if the cathode compartment is more concentrated.

EMF and equilibrium constant

E^\circ_{cell} = \dfrac{0.0591}{n}\log K

at equilibrium

E_{cell} = 0

and

Q = K

At $298\,\text{K}$ : $E_{cell} = E^\circ_{cell} - \dfrac{0.0591}{n}\log Q$ , where $Q$ is written as $\dfrac{[\text{products}]}{[\text{reactants}]}$ for the cell reaction (pure solids and the metal are taken as $1$ ).
Daniell cell: $E_{cell} = E^\circ_{cell} - \dfrac{0.0591}{2}\log\dfrac{[Zn^{2+}]}{[Cu^{2+}]}$ . When $[Zn^{2+}] = [Cu^{2+}]$ the log term is $0$ , so $E_{cell} = E^\circ_{cell}$ .
Single electrode (reduction form) e.g. $Zn^{2+}/Zn$ : $E = E^\circ - \dfrac{0.0591}{2}\log\dfrac{1}{[Zn^{2+}]}$ , i.e. $E = E^\circ + \dfrac{0.0591}{2}\log[Zn^{2+}]$ .
Concentration cell: $E^\circ_{cell} = 0$ , so $E_{cell} = \dfrac{0.0591}{n}\log\dfrac{[\text{ion}]_{cathode}}{[\text{ion}]_{anode}}$ ; EMF is positive only if the cathode (right) compartment is more concentrated.
At equilibrium $E_{cell} = 0$ and $Q = K$ , giving $E^\circ_{cell} = \dfrac{0.0591}{n}\log K$ — used to find $K$ from a measured EMF.
Sign care: increasing reactant-ion concentration (or lowering product-ion concentration) raises $E_{cell}$ ; remember $\log(10) = 1$ , $\log(10^{-3}) = -3$ for quick mental checks.
Always identify $n$ from the balanced cell reaction (electrons exchanged), not from the formula of one ion.
The full constant is $\dfrac{2.303RT}{nF}$ ; the value $0.0591\,\text{V}$ is valid only at $298\,\text{K}$ — at other temperatures recompute $\dfrac{2.303RT}{F}$ .
Worked check (Daniell, $[Cu^{2+}]=10^{-3}$ , $[Zn^{2+}]=1$ ): $E = 1.10 - \dfrac{0.0591}{2}\log\dfrac{1}{10^{-3}} = 1.10 - 0.0296\times3 \approx 1.011\,\text{V}$ .
Worked check (concentration cell $0.001\to0.1$ , $n=2$ ): $E = \dfrac{0.0591}{2}\log\dfrac{0.1}{0.001} = 0.0296\times2 \approx 0.059\,\text{V}$ .
Worked check ( $E_{cell}=0.295\,\text{V}$ , $n=2$ ): $\log K = \dfrac{2\times0.295}{0.0591} \approx 10$ , so $K \approx 10^{10}$ .
For a half-cell, write the reduction form $\text{ox} + ne^- \rightarrow \text{red}$ so $Q = \dfrac{[\text{red}]}{[\text{ox}]}$ ; getting $Q$ upside down flips the sign of the correction.

Where the marks go

Inverting $Q$ : it is products over reactants for the cell as written; for Daniell that is $\dfrac{[Zn^{2+}]}{[Cu^{2+}]}$ (anode ion over cathode ion), not the reverse.
Using $0.0591$ when the temperature is not $298\,\text{K}$ — the constant is temperature dependent.
Taking $n$ from one ion (e.g. $n=1$ for $Cu^{2+}$ ) instead of the balanced equation ( $n=2$ for the Daniell cell).
Including the activity of pure solids or the metal electrode in $Q$ — these are taken as $1$ and omitted.
Sign error in a single-electrode calculation: $\log\dfrac{1}{[Zn^{2+}]} = -\log[Zn^{2+}]$ , so the two rearrangements differ by a sign that must be tracked carefully.

How the board asks it

Numericaldaniell-cell emf at given ion concentrations
Calculate the EMF of the cell $Zn\,|\,Zn^{2+}(0.1\,M)\,||\,Cu^{2+}(0.01\,M)\,|\,Cu$ at $298\,K$ , given $E^\circ_{cell} = 1.10\,V$ .
Numericalsingle-electrode potential from standard potential
Calculate the electrode potential of a zinc electrode dipped in $0.001\,M$ $ZnSO_4$ solution at $298\,K$ , given $E^\circ_{Zn^{2+}/Zn} = -0.76\,V$ .
Numericalemf and equilibrium constant
For a cell reaction with $E^\circ_{cell} = 0.295\,V$ and $n = 2$ , calculate the equilibrium constant $K$ at $298\,K$ .
Numericalconcentration cell with two like half-cells
Calculate the EMF of the concentration cell $Ag\,|\,Ag^{+}(0.001\,M)\,||\,Ag^{+}(0.1\,M)\,|\,Ag$ at $298\,K$ .
Numericalback-calculating an ion concentration from emf
For the Daniell cell at $298\,K$ , calculate $[Cu^{2+}]$ when $E_{cell} = 1.04\,V$ , $[Zn^{2+}] = 0.1\,M$ and $E^\circ_{cell} = 1.10\,V$ .
Give reasonssign of the log term in the nernst equation
Account for the fact that the EMF of a Daniell cell increases when $[Cu^{2+}]$ is increased while $[Zn^{2+}]$ is kept constant.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.