Standard electrode potentials — SHE reference, electrochemical series

Overview

To compare the "driving force" of different half-reactions, we need a standard electrode potential ($E°$) measured against a universal reference electrode. The Standard Hydrogen Electrode (SHE) serves as the zero-reference point, and all other electrode potentials are measured relative to it. The electrochemical series ranks these potentials from most negative (strongest reducing agents) to most positive (strongest oxidizing agents).

[!intuition] Why We Need a Reference Electrode

The Problem: A half-reaction like $Cu^{2+} + 2e^- \rightarrow Cu$ has an inherent "electron-pulling power," but we can't measure it in isolation—you need a complete circuit. Every measurement involves TWO half-cells.

The Solution: Pick ONE half-reaction as the universal reference and define its potential as exactly zero. Then measure all others relative to this reference. It's like setting sea level as the zero-point for altitude—all heights are measured relative to that standard.

Why SHE?: The hydrogen electrode is:

• Reproducible: Easy to set up with pure H₂ gas, H⁺ ions at unit activity
• Chemically simple: Involves only H⁺/H₂, no complex ions
• Historically established: Universal convention since the early 20th century

[!definition] Standard Hydrogen Electrode (SHE)

The Standard Hydrogen Electrode consists of:

• A platinum electrode (inert, catalyzes the reaction)
• Pure H₂ gas at 1 bar pressure bubling over the electrode
• Aqueous solution with $[\text{H}^+] = 1 \text{ M}$ (unit activity)
• Temperature: 25°C (298 K)

Half-reaction: $\text{2H}^+ \text{ (aq, 1M)} + 2e^- \rightleftharpoons \text{H}_2 \text{ (g, 1 bar)}$

Defined potential: $E°_{\text{H}^+/\text{H}_2} = 0.00 \text{ V (by convention)}$

This is an arbitrary zero-reference, not a physical measurement. All other electrode potentials are measured by coupling a half-cell to the SHE.

[!formula] Measuring Standard Electrode Potentials

The Setup

To find $E°$ for a half-reaction like $\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}$:

1. Left half-cell: SHE (reference, $E° = 0.00 \text{ V}$)
2. Right half-cell: Zn electrode in $[\text{Zn}^{2+}] = 1 \text{ M}$
3. Connect with a salt bridge and measure the cell potential $E°_{\text{cell}}$

Cell notation: $\text{Pt(s) | H}_2\text{(g, 1bar) | H}^+ \text{(1M) || Zn}^{2+} \text{(1M) | Zn(s)}$

Derivation from First Principles

The cell potential is the difference between the cathode (reduction) and anode (oxidation) potentials:

$E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}}$

For the Zn/SHE cell:

• Observation: Zn dissolves (oxidation), H₂ forms (reduction)
• So: Cathode = SHE, Anode = Zn

$E°_{\text{cell}} = E°_{\text{H}^+/\text{H}_2} - E°_{\text{Zn}^{2+}/\text{Zn}}$

Since $E°_{\text{H}^+/\text{H}_2} = 0.00 \text{ V}$:

$E°_{\text{cell}} = 0.00 - E°_{\text{Zn}^{2+}/\text{Zn}}$

Measurement gives $E°_{\text{cell}} = +0.76 \text{ V}$, so:

$E°_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \text{ V}$

Why negative?: The Zn half-cell acts as thenode (loses electrons). A negative $E°$ means the reduction is non-spontaneous relative to SHE—Zn prefers to oxidize.

General Formula

For any half-cell X coupled to SHE:

$E°_{\text{X}} = E°_{\text{cell}} \text{ (if X is cathode)}$ $E°_{\text{X}} = -E°_{\text{cell}} \text{ (if X is anode)}$

Key insight: The sign of $E°_{\text{cell}}$ tells you which electrode oxidizes. Positive $E°_{\text{cell}}$ → spontaneous as written; negative → reverse is spontaneous.

[!example] Worked Example 1: Measuring $E°$ for Copper

Setup: $\text{Pt | H}_2\text{(1 bar) | H}^+ \text{(1M) || Cu}^{2+} \text{(1M) | Cu}$

Observation: H₂ bubles disappear, Cu deposits on Cu electrode.

• Oxidation (anode): $\text{H}_2 \rightarrow 2\text{H}^+ + 2e^-$ at Pt
• Reduction (cathode): $Cu^{2+} + 2e^- \rightarrow Cu$ at Cu

Measured cell potential: $E°_{\text{cell}} = +0.34 \text{ V}$

Calculation: $E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}}$ $+0.34 = E°_{\text{Cu}^{2+}/\text{Cu}} - 0.00$ $E°_{\text{Cu}^{2+}/\text{Cu}} = +0.34 \text{ V}$

Why this step? The Cu electrode is the cathode (reduction occurs there), so $E°_{\text{cathode}} = E°_{\text{Cu}^{2+}/\text{Cu}}$. The positive value means Cu²⁺ reduction is spontaneous relative to H⁺ reduction.

[!example] Worked Example 2: Predicting Spontaneous Direction

Question: Will $\text{Fe}^{3+}$ oxidize $\text{I}^-$ to $\text{I}_2$ under standard conditions?

Given: $E°_{\text{Fe}^{3+}/\text{Fe}^{2+}} = +0.77 \text{ V}$, $E°_{\text{I}_2/\text{I}^-} = +0.54 \text{ V}$

Proposed reaction: $2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2$

Step 1: Identify half-reactions

• Reduction: $\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}$ ($E° = +0.77 \text{ V}$)
• Oxidation: $2\text{I}^- \rightarrow \text{I}_2 + 2\text{e}^-$ (reverse of $E° = +0.54 \text{ V}$)

Step 2: Calculate $E°_{\text{cell}}$ $E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}}$ $E°_{\text{cell}} = 0.77 - 0.54 = +0.23 \text{ V}$

Why this step? The cathode is where reduction occurs ($\text{Fe}^{3+}$), the anode is where oxidation occurs ($\text{I}^-$). We subtract the anode potential from the cathode potential.

Step 3: Determine spontaneity $\Delta G° = -nFE°_{\text{cell}} = -2 \times 96485 \times 0.23 < 0$

Conclusion: $E°_{\text{cell}} > 0$ → Spontaneous. $\mathrm{Fe}^{3+}$ will oxidize $\text{I}^-$.

Physical interpretation: Fe³⁺ is a stronger oxidizing agent (higher $E°$) than I₂, so it "wins" and gets reduced, forcing I⁻ to oxidize.

[!definition] The Electrochemical Series

The electrochemical series is a ranked list of standard electrode potentials, ordered from most negative (top) to most positive (bottom).

Standard Reduction Potentials (excerpt):

Reading the Series

Reducing agents (electron donors, get oxidized):

• Top of the series (negative $E°$): Strong reducers (Li, K, Zn)
• These species resist reduction but easily undergo oxidation

Oxidizing agents (electron acceptors, get reduced):

• Bottom of the series (positive $E°$): Strong oxidizers (F₂, Cl Ag⁺)
• These species easily undergo reduction but resist oxidation

Rule of thumb:

• Higher $E°$ (more positive) → stronger oxidizing agent (oxidized form)
• Lower $E°$ (more negative) → stronger reducing agent (reduced form)

[!formula] Predicting Redox Reactions from the Series

Principle: A spontaneous redox reaction occurs when the stronger oxidizing agent (higher $E°$) oxidizes the stronger reducing agent (lower $E°$).

Algorithmic approach:

1. Write two half-reactions from the series
2. The one with higher $E°$ proceeds as reduction (cathode)
3. The one with lower $E°$ proceeds as oxidation (anode, reverse the half-reaction)
4. Calculate $E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}}$
5. If $E°_{\text{cell}} > 0$ → spontaneous; if $E°_{\text{cell}} < 0$ → non-spontaneous

Derivation of the rule:

From thermodynamics: $\Delta G° = -nFE°_{\text{cell}}$

For spontaneity, $\Delta G° < 0$, so: $E°_{\text{cell}} > 0$

Expanding: $E°_{\text{cathode}} - E°_{\text{anode}} > 0$ $E°_{\text{cathode}} > E°_{\text{anode}}$

Why this works: The species with higher $E°$ "pulls" electrons more strongly, so it acts as the cathode. The one with lower $E°$ releases electrons more readily, so it acts as the anode.

[!example] Worked Example 3: Will Zinc Reduce Copper?

Question: Will Zn metal reduce $\text{Cu}^{2+}$ ions?

Half-reactions:

• $Zn^{2+} + 2e^- \rightarrow Zn$, $E° = -0.76 \text{ V}$
• $Cu^{2+} + 2e^- \rightarrow Cu$, $E° = +0.34 \text{ V}$

Identify cathode/anode:

• Higher $E°$: Cu²⁺/Cu → cathode (reduction)
• Lower $E°$: Zn²⁺/Zn → anode (oxidation, reverse the reaction)

Overall reaction: $Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu$

Calculate $E°_{\text{cell}}$: $E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}} = 0.34 - (-0.76) = +1.10 \text{ V}$

Why this step? We subtract the anode potential (Zn) from the cathode potential (Cu). The double negative becomes a positive.

Conclusion: $E°_{\text{cell}} = +1.10 \text{ V} > 0$ → Spontaneous. Zn will reduce Cu²⁺.

Physical observation: Drop a Zn strip into $\text{CuSO}_4$ solution → Cu deposits on Zn, blue color fades.

[!mistake] Common Mistake 1: Flipping Signs When Reversing Half-Reactions

Wrong idea: "If I reverse $Cu^{2+} + 2e^- \rightarrow Cu$ ($E° = +0.34 \text{ V}$) to $\text{Cu} \rightarrow \text{Cu}^{2+} + 2e^-$, the potential becomes $E° = -0.34 \text{ V}$."

Why it feels right: Reversing a reaction reverses $\Delta G°$, and since $\Delta G° = -nFE°$, shouldn't $E°$ flip sign?

The truth: $E°$ values in the electrochemical series are always for the reduction half-reaction. When you reverse a half-reaction for use as the anode, you don't change the $E°$ value—you use it in the $E°_{\text{cell}}$ formula as $E°_{\text{anode}}$ with a minus sign already built into the formula.

Correct approach:

• Find $E°$ for both half-reactions as reductions
• Use $E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}}$
• The subtraction accounts for the reversal

Example: For $\text{Cu} \rightarrow \text{Cu}^{2+} + 2e^-$ used as anode: $E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{Cu}^{2+}/\text{Cu}} = E°_{\text{cathode}} - 0.34$

The $-0.34$ effect comes from the subtraction, not from flipping the sign of $E°$.

[!mistake] Common Mistake 2: Forgetting Standard Conditions

Wrong idea: "$E° = +0.34 \text{ V}$ for Cu²⁺/Cu, so the reaction is always spontaneous."

Why it feels right: $E° > 0$ suggests spontaneity.

The truth: $E°$ applies only under standard conditions:

• 25°C
• 1 M concentrations (unit activity)
• 1 bar gas pressures

Under non-standard conditions, use the Nernst equation: $E = E° - \frac{RT}{nF} \ln Q = E° - \frac{0.0592}{n} \log Q \text{ (at 25°C)}$

Example: For $Cu^{2+} + 2e^- \rightarrow Cu$, if $[\text{Cu}^{2+}] = 0.001 \text{ M}$: $E = 0.34 - \frac{0.0592}{2} \log \frac{1}{0.001} = 0.34 - 0.089 = +0.25 \text{ V}$

Still positive, but reduced driving force. At very low $[\text{Cu}^{2+}]$, $E$ could even become negative.

Fix: Always check if conditions are standard. If not, apply the Nernst equation (covered in 2.7.03-Nernst-equation-and-concentration-effects).

[!mistake] Common Mistake 3: Multiplying $E°$ by Stoichiometric Coefficients

Wrong idea: "The reaction $\text{2Ag}^+ + \text{2e}^- \rightarrow \text{2Ag}$ has twice the electrons, so $E° = 2 \times 0.80 = 1.60 \text{ V}$."

Why it feels right: $\Delta G° = -nFE°$ depends on $n$, so shouldn't $E°$ scale?

The truth: $E°$ is an intensive property (like temperature or pressure)—it doesn't change with amount. When you double the stoichiometry:

• $\Delta G°$ doubles ($n$ doubles)
• $E°$ stays the same (the $n$ in $\Delta G° = -nFE°$ compensates)

Proof: $\Delta G° = -nFE°$

For $\text{Ag}^+ + e^- \rightarrow \text{Ag}$: $\Delta G°_1 = -(1)FE°$

For $\text{2Ag}^+ + \text{2e}^- \rightarrow \text{2Ag}$: $\Delta G°_2 = -(2)FE°$

Since $\Delta G°_2 = 2\Delta G°_1$ (extensive property): $-(2)FE° = 2[-(1)FE°]$ $E° = E°$

Physical analogy: The "voltage" of a battery doesn't change if you connect two identical batteries in parallel (same voltage, double current). $E°$ is the "voltage per electron."

Fix: Always use the tabulated $E°$ value regardless of how you balance the equation.

[!recall]- Explain to a 12-Year-Old

Imagine you have a bunch of kids in a playground, and some REALLY want to play with a toy (electrons). Let's say:

• Strong oxidizers (like F₂, Cl₂) are the pushy kids who REALLY want the toy and will grab it from anyone.
• Strong reducers (like Li, Zn) are the generous kids who are happy to give away the toy.

Now, if you put a pushy kid (F₂) next to a generous kid (Li), the pushy one will take the toy—that's a spontaneous reaction!

But how do we measure "pushiness"? We can't ask each kid individually—we need them to compete! So we pick one kid as the reference (the Standard Hydrogen Electrode, SHE) and say, "This kid has zero pushiness by definition." Then we measure everyone else by how hard they fight against the reference kid.

• If someone takes the toy from the reference kid, they're pushier (positive $E°$, good oxidizer).
• If someone gives the toy to the reference kid, they're more generous (negative $E°$, good reducer).

We write all the "pushiness scores" in a list—that's the electrochemical series! The pushiest kid (F₂, $E° = +2.87 \text{ V}$) is at the bottom, and the most generous kid (Li, $E° = -3.04 \text{ V}$) is at the top.

When you mix two kids (two half-cells), the pushier one (higher $E°$) will take the toy (get reduced), and the more generous one (lower $E°$) will give it away (get oxidized). If the difference in pushiness ($E°_{\text{cell}}$) is positive, the "trade" happens on its own—spontaneous!

[!mnemonic] Remembering the Electrochemical Series

"Lucky Lions Can Zoom High, Courageously Fighting Bears After Fierce Competition"

• Li ($-3.04$ V)
• Li (wait, that's K for Potassium!) → K ($-2.93$ V)
• Ca ($-2.87$ V)
• Zn ($-0.76$ V)
• H₂ (0.00 V, reference)
• Cu ($+0.34$ V)
• Fe³⁺ ($+0.77$ V)
• Ag ($+0.80$ V) (Bears? close enough!)
• Br₂ ($+1.07$ V)
• Cl₂ ($+1.36$ V) (After)
• F₂ ($+2.87$ V) (Fierce Competition)

Alternate mnemonic for oxidizer strength: "Fine Chefs Bake Awesome Cookies Hot Zesty Creations Keep Lads" (bottom to top, strongest to weakest oxidizers).

Connections

• Previous: 2.7.01-Oxidation-states-and-balancing-redox-equations — How to assign oxidation states and balance redox reactions
• Next: 2.7.03-Nernst-equation-and-concentration-effects — How $E$ changes with concentration
• Relates to: 2.7.04-Electrochemical-cellsand-cell-potential — Applying $E°$ to galvanic cells
• Relates to: 2.7.05-Gibs-free-energy-and-equilibrium-constant — Connecting $E°$ to $\Delta G°$ and $K$
• Foundation: 1.5.02-Thermodynamics-spontaneity-and-Gibbs-energy — Why $\Delta G° < 0$ means spontaneous

#flashcards/chemistry

What is the Standard Hydrogen Electrode (SHE), and what is its defined potential? :: The SHE consists of a Pt electrode in 1 M H⁺ solution with H₂ gas at 1 bar and 25°C. The half-reaction is $2H^+ + 2e^- \rightleftharpoons H_2$. Its potential is defined as exactly 0.00 V by convention, serving as the universal reference for measuring all other electrode potentials.

What does a negative standard electrode potential ($E° < 0$) indicate? :: A negative $E°$ means the reduction half-reaction is non-spontaneous relative to SHE—the species prefers to oxidize (act as a reducing agent). Example: $E°_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \text{ V}$, so Zn readily loses electrons.

Concept Map

Hinglish (regional understanding)