Electrode Potential, SHE & Electrochemical Series

CHEGalvanic Cells & Electrode Potential

This subtopic introduces the standard hydrogen electrode (SHE) as the universal reference, defines standard electrode potential, and uses the electrochemical series to predict displacement reactions, reducing/oxidising strength, and reactivity. Reasoning questions on why zinc displaces copper but not the reverse are typical.

SHE half-reaction (reference)

2H^+(1\,M) + 2e^- \rightleftharpoons H_2(1\,\text{bar}) \qquad E^\circ = 0.00\ \text{V}

assigned zero by convention at

298\,\text{K}

; all other

E^\circ

are measured against it.

Feasibility of a displacement reaction

E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} > 0

a positive

E^\circ_{cell}

means the reaction (e.g. metal displacing a less reactive metal ion) is spontaneous.

SHE: platinised Pt dipped in $1\,M\,H^+$ with $H_2$ gas at $1\,\text{bar}$ and $298\,\text{K}$ ; the half-reaction $2H^+ + 2e^- \rightleftharpoons H_2$ is assigned $E^\circ = 0.00\,\text{V}$ by convention as the reference against which all other potentials are measured.
Standard reduction potential $E^\circ$ is measured relative to SHE under standard conditions ( $1\,M$ ions, $1\,\text{bar}$ gas, $298\,\text{K}$ ); the SHE value is a chosen reference (zero), not a true absolute potential.
Electrode potential arises from the tendency of a metal to lose electrons (oxidation) or gain electrons (reduction) at the metal-solution interface, setting up an electrical double layer.
Electrochemical series arranges electrodes by increasing $E^\circ$ (reduction). The more negative $E^\circ$ , the stronger the reducing agent (more easily oxidised); the more positive $E^\circ$ , the stronger the oxidising agent.
A metal with lower (more negative) $E^\circ$ displaces the ion of a metal with higher $E^\circ$ from solution: $Zn$ ( $-0.76\,\text{V}$ ) displaces $Cu$ ( $+0.34\,\text{V}$ ) from $CuSO_4$ , but $Cu$ cannot displace $Zn$ because $E^\circ_{cell}$ would be negative (non-spontaneous).
Metals above hydrogen in the series (negative $E^\circ$ ) can displace $H_2$ from dilute acids; those below hydrogen (e.g. $Cu$ , $Ag$ ) cannot.
Reactivity (activity series) of metals follows the order of their reduction potentials: the more negative the $E^\circ$ , the more reactive the metal (e.g. $Li, K, Ca, Na$ at the top).
Applications of the electrochemical series: (i) compare reducing/oxidising power, (ii) predict whether a redox/displacement reaction is feasible, (iii) calculate $E^\circ_{cell}$ , (iv) predict which metal displaces $H_2$ from acids and the order of reactivity of metals.
Higher cathode $E^\circ$ and lower anode $E^\circ$ give a larger positive $E^\circ_{cell}$ , indicating a more spontaneous (and feasible) cell reaction.
In practice the SHE is awkward to set up, so secondary reference electrodes (calomel electrode) are used and their potentials are known relative to SHE.
$F_2$ ( $E^\circ = +2.87\,\text{V}$ ) is the strongest common oxidising agent and $Li$ ( $E^\circ = -3.04\,\text{V}$ ) is the strongest reducing agent at the extremes of the series.

Where the marks go

Confusing 'most negative $E^\circ$ ' with weakest reducing agent — most negative $E^\circ$ means the BEST reducing agent (most easily oxidised).
Predicting that Cu displaces Zn (or any less reactive metal displacing a more reactive one) — the resulting $E^\circ_{cell}$ is negative, so it does not occur.
Forgetting that metals below hydrogen (Cu, Ag, Au) cannot liberate $H_2$ from dilute acids.
Treating SHE's $0.00\,\text{V}$ as an absolute potential rather than a conventional reference point.
Reversing the sign of $E^\circ$ when reversing the half-reaction in the $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$ formula — the subtraction already accounts for the anode being oxidation.

How the board asks it

Give reasonsdisplacement feasibility from $E^\circ$ values
Account for the fact that zinc displaces copper from copper sulphate solution, but copper cannot displace zinc from zinc sulphate solution. (Given: $E^\circ_{Zn^{2+}/Zn} = -0.76\,\text{V}$ , $E^\circ_{Cu^{2+}/Cu} = +0.34\,\text{V}$ )
Numerical $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$
Calculate the standard EMF of the cell $Mg\,|\,Mg^{2+}\,||\,Cu^{2+}\,|\,Cu$ , given $E^\circ_{Mg^{2+}/Mg} = -2.37\,\text{V}$ and $E^\circ_{Cu^{2+}/Cu} = +0.34\,\text{V}$ , and state whether the cell reaction is spontaneous.
Define / statestandard electrode potential and SHE
Define standard electrode potential. State the conditions under which the standard hydrogen electrode operates and explain why its potential is taken as $0.00\,\text{V}$ .
Predict the productsign of $E^\circ_{cell}$ from the electrochemical series
Using the electrochemical series, predict whether the reaction $Ni + 2Ag^+ \rightarrow Ni^{2+} + 2Ag$ is feasible. (Given: $E^\circ_{Ni^{2+}/Ni} = -0.25\,\text{V}$ , $E^\circ_{Ag^+/Ag} = +0.80\,\text{V}$ ) Justify your answer by calculating $E^\circ_{cell}$ .
Give reasonsreducing power and displacement of $H_2$ from acids
With reference to standard reduction potentials, explain why magnesium can displace hydrogen from dilute hydrochloric acid whereas copper cannot. (Given: $E^\circ_{Mg^{2+}/Mg} = -2.37\,\text{V}$ , $E^\circ_{Cu^{2+}/Cu} = +0.34\,\text{V}$ , $E^\circ_{H^+/H_2} = 0.00\,\text{V}$ )
Diagram / graphconstruction of the SHE
Draw a labelled diagram of the standard hydrogen electrode and write the half-cell reaction occurring at it.

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.