EMF Calculations (Standard Conditions)

CHEGalvanic Cells & Electrode Potential

These problems compute the standard cell EMF from tabulated standard electrode potentials, identify anode and cathode, and judge spontaneity. They often extend to standard Gibbs energy and the equilibrium constant, so combine the EMF formula with $\Delta G^\circ = -nFE^\circ$ .

Standard cell EMF

E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}

use standard REDUCTION potentials for both; do not flip the sign when a half-reaction runs as oxidation.

Gibbs energy from EMF

\Delta G^\circ = -nFE^\circ_{cell}

n

= electrons exchanged,

F = 96500\,\text{C}

; result in joules (

\div1000

for

\text{kJ}

Equilibrium constant

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

298\,\text{K}

; large positive

E^\circ_{cell}

gives

K \gg 1

$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$ , using standard reduction potentials for both half-cells (do not change the sign of $E^\circ$ when reversing a half-reaction to oxidation).
The electrode with the higher (more positive) reduction potential is the cathode (reduction); the one with the lower potential is the anode (oxidation).
A positive $E^\circ_{cell}$ means $\Delta G^\circ < 0$ and the reaction is spontaneous as written; a negative $E^\circ_{cell}$ means it is non-spontaneous (runs in reverse).
Example logic: for $Cr\,|\,Cr^{3+}\,||\,Fe^{2+}\,|\,Fe$ with $E^\circ(Cr^{3+}/Cr) = -0.74\,\text{V}$ and $E^\circ(Fe^{2+}/Fe) = -0.44\,\text{V}$ , $E^\circ_{cell} = (-0.44) - (-0.74) = +0.30\,\text{V}$ (Cr is anode, Fe is cathode).
EMF is intensive: it does NOT change when the balanced equation is multiplied by a factor, but $n$ (and hence $\Delta G^\circ$ ) does.
Balancing electrons: to combine $Cr^{3+}/Cr$ ( $3e^-$ ) with $Fe^{2+}/Fe$ ( $2e^-$ ) you scale to $n=6$ for the overall reaction, but $E^\circ_{cell}$ stays the same; only $\Delta G^\circ$ scales with $n$ .
Link to energy: $\Delta G^\circ = -nFE^\circ_{cell}$ (in joules with $F = 96500\,\text{C}$ ); a more positive $E^\circ_{cell}$ gives a more negative $\Delta G^\circ$ .
Equilibrium constant from EMF: $\log K = \dfrac{nE^\circ_{cell}}{0.0591}$ at $298\,\text{K}$ , so a large positive $E^\circ_{cell}$ implies $K \gg 1$ .
Worked check (Mg/Ag, $n=2$ ): $E^\circ_{cell} = 0.80 - (-2.37) = +3.17\,\text{V}$ ; $\Delta G^\circ = -2\times96500\times3.17 \approx -612\,\text{kJ mol}^{-1}$ .
When a metal is paired with the SHE, treat $E^\circ(H^+/H_2) = 0$ ; for Zn/SHE, $E^\circ_{cell} = 0 - (-0.76) = +0.76\,\text{V}$ with Zn as anode (spontaneous).
A cell with $E^\circ_{cell}$ very close to zero is near equilibrium ( $K \approx 1$ ) and gives little useful work.

Where the marks go

Double-counting signs: $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$ uses reduction potentials directly — do NOT also reverse the sign of the anode value.
Multiplying $E^\circ_{cell}$ when scaling the equation — EMF is intensive and unchanged, but $n$ in $\Delta G^\circ = -nFE^\circ$ does change.
Picking the cathode wrongly — the higher (more positive) reduction potential is the cathode; choosing the lower one flips the sign.
Leaving $\Delta G^\circ$ in joules and reporting it as $\text{kJ}$ (or vice versa) — divide by $1000$ to convert.
Forgetting that $E^\circ(H^+/H_2)=0$ when one electrode is the SHE.

How the board asks it

Numerical $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$
For the cell $Cr\,|\,Cr^{3+}\,||\,Fe^{2+}\,|\,Fe$ , given $E^\circ(Cr^{3+}/Cr) = -0.74\,\text{V}$ and $E^\circ(Fe^{2+}/Fe) = -0.44\,\text{V}$ , calculate the standard EMF of the cell and identify which electrode is the anode and which is the cathode.
Numerical $\Delta G^\circ = -nFE^\circ_{cell}$
For the cell $Mg\,|\,Mg^{2+}\,||\,Ag^+\,|\,Ag$ with $E^\circ_{cell} = +3.17\,\text{V}$ and $n = 2$ , calculate the standard Gibbs energy change $\Delta G^\circ$ in $\text{kJ mol}^{-1}$ (take $F = 96500\,\text{C mol}^{-1}$ ).
Numerical $\log K = \dfrac{nE^\circ_{cell}}{0.0591}$
Calculate the equilibrium constant $K$ at $298\,\text{K}$ for a cell reaction with $E^\circ_{cell} = +0.30\,\text{V}$ and $n = 6$ .
Give reasonspositive $E^\circ_{cell}$ means $\Delta G^\circ < 0$
Given $E^\circ(Zn^{2+}/Zn) = -0.76\,\text{V}$ and $E^\circ(Cu^{2+}/Cu) = +0.34\,\text{V}$ , predict whether the reaction $Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu$ is feasible under standard conditions. Give reasons for your answer.
Give reasonsemf is intensive; only $n$ scales $\Delta G^\circ$
When a balanced cell reaction is multiplied by $2$ , state whether $E^\circ_{cell}$ and $\Delta G^\circ$ change, and justify your answer using the relation $\Delta G^\circ = -nFE^\circ_{cell}$ .
Define / state $E^\circ(H^+/H_2) = 0$ for the SHE
State the value of the standard electrode potential of the standard hydrogen electrode, and write the expression used to calculate the standard EMF of a cell from the standard reduction potentials of its two electrodes.

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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.