Chemistry · Ch 4 — The d- and f-Block Elements

Trends in the M²⁺/M Standard Electrode Potentials

4.3.5

Trends in the M²⁺/M Standard Electrode Potentials

The M²⁺/M reduction process

Forming a divalent metal ion M²⁺(aq) from the solid metal can be broken down into a chain of thermochemical steps: atomisation of the solid into gaseous atoms, successive removal of two electrons (the first and second ionisation), and hydration of the resulting gaseous ion. The measured standard electrode potential $E^{\ominus}$ for the M²⁺/M couple is essentially the net energy balance of this whole sequence — the enthalpy of atomisation, the sum of the first two ionisation enthalpies, and the strongly exothermic hydration enthalpy of M²⁺ acting together. Because an $E^{\ominus}$ rebuilt purely from these thermochemical quantities matches the experimentally observed values closely across the row from Ti to Zn, the electrode potential of a 3d metal can be understood almost entirely as a consequence of these three energy terms.

The general trend across the series

Moving from Ti to Zn, the $E^{\ominus}$ (M²⁺/M) values become steadily less negative. This shift runs parallel to the general rise in the combined first and second ionisation enthalpies across the row: as a metal atom becomes progressively harder to strip of two electrons, less energy is recovered on the reduction step, and the electrode potential is pulled toward less negative — and, in one case, positive — values.

Copper — the exception to the trend

Copper stands apart from every other first-row metal in having a positive $E^{\ominus}$ (M²⁺/M). This single fact explains why copper cannot displace hydrogen from dilute acids the way most other metals of the series can — only oxidising acids, such as nitric acid or hot concentrated sulphuric acid, attack copper metal, and even then it is the acid itself that undergoes reduction, not H⁺. The underlying reason is energetic: the enthalpy required to convert solid copper into gaseous Cu²⁺ is unusually high, and copper's hydration enthalpy is not large enough to compensate for it — leaving the overall process for Cu(s) → Cu²⁺(aq) endothermic on balance.

Figure 4.4Observed and calculated values for the standard electrode potentials (M²⁺ → M⁰) of the elements Ti to Zn

Fig. 4.4 — Observed and calculated values for the standard electrode potentials (M²⁺ → M⁰) of the elements Ti to Zn

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your NCERT textbook's own diagram.

The figure plots standard electrode potentials ( $E^\ominus$ for $\ce{M^{2+} + 2e- -> M}$ ) against the elements from titanium (Ti) to zinc (Zn) in the first transition series. The x-axis lists the elements in order of increasing atomic number: Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn. The y-axis shows $E^\ominus$ in volts, ranging from approximately −2.0 V to +0.5 V.

Two curves are drawn on the same axes:

Observed values (marked with squares): the actual measured $E^\ominus$ for each metal.
Calculated values (marked with diamonds): the $E^\ominus$ predicted from a simple thermochemical cycle using only the sum of the first two ionisation enthalpies ( $\Delta_i H_1^\ominus + \Delta_i H_2^\ominus$ ), the enthalpy of atomisation ( $\Delta_a H^\ominus$ ), and the hydration enthalpy of $\ce{M^{2+}}$ ( $\Delta_{\text{hyd}} H^\ominus$ ).

Both curves rise together irregularly from Ti toward a positive peak at Cu, then drop sharply at Zn. They track each other closely, but there are notable bumps — deviations where the observed value is more negative than the calculated trend — at Mn and Zn. The text explains that the general upward trend (from negative to less negative, then positive) is due to the increase in the sum of the first and second ionisation enthalpies across the series. The extra stability of the half-filled $d^5$ subshell in $\ce{Mn^{2+}}$ and the completely filled $d^{10}$ subshell in $\ce{Zn^{2+}}$ makes these ions harder to reduce, shifting their $E^\ominus$ to more negative values than the simple calculation predicts. For Ni, the more negative observed value is linked to its highly negative hydration enthalpy ( $\Delta_{\text{hyd}} H^\ominus$ ).

The key formula the textbook develops with this figure is the thermochemical cycle that relates $E^\ominus$ to the enthalpy changes:

E^\ominus(\ce{M^{2+}/M}) = -\frac{\Delta_a H^\ominus(\ce{M}) + \Delta_i H_1^\ominus(\ce{M}) + \Delta_i H_2^\ominus(\ce{M}) + \Delta_{\text{hyd}} H^\ominus(\ce{M^{2+}})}{2F}

where:

$\Delta_a H^\ominus(\ce{M})$ = enthalpy of atomisation of the metal (kJ mol⁻¹)
$\Delta_i H_1^\ominus(\ce{M})$ = first ionisation enthalpy (kJ mol⁻¹)
$\Delta_i H_2^\ominus(\ce{M})$ = second ionisation enthalpy (kJ mol⁻¹)
$\Delta_{\text{hyd}} H^\ominus(\ce{M^{2+}})$ = hydration enthalpy of the $\ce{M^{2+}}$ ion (kJ mol⁻¹)
$F$ = Faraday constant (96 485 C mol⁻¹) …

Departures from the smooth trend: Mn, Ni and Zn

Although the trend across the series is broadly smooth, three metals show $E^{\ominus}$ values noticeably more negative than the general trend would predict.

Manganese's deviation traces to the special stability of the half-filled $3d^5$ configuration of Mn²⁺, and zinc's deviation traces to the fully filled $3d^{10}$ configuration of Zn²⁺ — in both cases, the ion sits in an especially settled electronic arrangement that the smooth trend line does not otherwise account for. …

Table 4.4Thermochemical data (kJ mol⁻¹) for the first row Transition Elements and the Standard Electrode Potentials for the Reduction of Mᴵᴵ to M

Element (M)	$\Delta_aH^\ominus$ (M)	$\Delta_iH_1^\ominus$	$\Delta_iH_2^\ominus$	$\Delta_{hyd}H^\ominus(M^{2+})$	$E^\ominus$ /V
Ti	469	656	1309	−1866	−1.63
V	515	650	1414	−1895	−1.18
Cr	398	653	1592	−1925	−0.90
Mn	279	717	1509	−1862	−1.18
Fe	418	762	1561	−1998	−0.44
Co	427	758	1644	−2079	−0.28
Ni	431	736	1752	−2121	−0.25
Cu	339	745	1958	−2121	0.34
Zn	130	906	1734	−2059	−0.76

Start your 14-day free trial to unlock the full solution →

← Previous · 4.3.44.3 — Which of the 3d series of the transition metals exhibits the largest n…Next · 4.3.5 →Example 4.4 — Why is Cr²⁺ reducing and Mn³⁺ oxidising when both have d⁴ configuratio…