Q.How would you determine the standard electrode potential of the system $Mg^{2+}|Mg$?

Concept: Cell Representation & Nernst Equation

To find the standard electrode potential of $Mg^{2+}|Mg$, you set up a cell with a reference electrode (e.g., SHE) and measure the cell potential under standard conditions.

Steps

1. Construct the cell: $Mg(s) | Mg^{2+}(aq, 1M) || H^+(aq, 1M) | H_2(g, 1 bar) | Pt(s)$.
2. Measure the cell potential ($E^\circ_{cell}$) at 298 K with a voltmeter. Since SHE is assigned $0\ \text{V}$, the measured value equals $E^\circ_{Mg^{2+}|Mg}$ (with sign determined by the direction of spontaneous reaction).
3. For $Mg^{2+}|Mg$, the reduction reaction is $Mg^{2+} + 2e^- \rightarrow Mg$. The measured $E^\circ_{cell}$ is negative (Mg is more reactive than H₂), so $E^\circ_{Mg^{2+}|Mg} = -2.36\ \text{V}$.

The standard electrode potential of a system like $Mg^{2+}|Mg$ is defined as the potential of the half-cell $Mg^{2+}(aq, 1\ \text{M}) | Mg(s)$ measured against the Standard Hydrogen Electrode (SHE) at 298 K. But here’s the catch: magnesium metal is so reactive that it displaces hydrogen from water even in neutral solutions. If you dip a magnesium rod into a 1 M $Mg^{2+}$ solution, you’ll see bubbles of hydrogen gas — the Mg is actually reacting with water, not sitting quietly at equilibrium. That makes a direct measurement impossible.

So how do we get the value? We use a clever indirect method: construct a cell where the Mg half-cell is combined with a reference electrode whose potential is known, measure the cell potential under carefully controlled conditions, and then back-calculate the Mg potential.

Let’s walk through the actual experimental procedure step by step.

1. Choose a reference electrode that doesn’t interfere. The Standard Hydrogen Electrode (SHE) is the universal reference, but it’s impractical for routine lab work. Instead, we often use a calomel electrode (saturated calomel electrode, SCE) or a silver-silver chloride electrode — these have stable, well-known potentials. For this explanation, let’s assume we use the SHE as the reference, since that’s the definition.

2. Set up the cell carefully. We cannot simply put Mg metal in a 1 M $Mg^{2+}$ solution and connect it to the SHE, because the Mg will react with water. Instead, we use a non-aqueous solvent or a very carefully deoxygenated aqueous solution, and we work quickly. In practice, the measurement is done in a solution of $MgSO_4$ or $MgCl_2$ at exactly 1 M concentration, with the Mg electrode freshly polished and the solution thoroughly purged of dissolved oxygen. The cell is:

3. Measure the cell potential. A voltmeter (or potentiometer) connected between the two electrodes gives the electromotive force (emf) of the cell. Because the Mg electrode is more negative than the SHE, the measured cell potential will be positive if we connect the SHE as the cathode and Mg as the anode. Let’s say we measure $E_{\text{cell}} = 2.36\ \text{V}$ at 298 K.

4. Apply the cell potential equation. For a cell written as:

The overall cell reaction is:

The cell potential is the difference between the cathode and anode potentials:

Here, $E_{\text{cathode}} = E^\circ_{H^+/H_2} = 0\ \text{V}$ (by definition), and $E_{\text{anode}} = E^\circ_{Mg^{2+}/Mg}$ (the value we want). So:

Therefore:

5. Is it really that simple? In practice, the measurement isn’t done at exactly 1 M concentrations because of the reactivity issue. Instead, we measure the cell potential at a known, lower concentration of $Mg^{2+}$ (say $10^{-3}\ \text{M}$) and then use the Nernst equation to extrapolate to standard conditions. The Nernst equation for the Mg half-cell is:

(at 298 K, using base-10 log). If we measure the cell potential at $[Mg^{2+}] = 10^{-3}\ \text{M}$, we get a different $E_{\text{cell}}$, and we solve for $E^\circ_{Mg^{2+}/Mg}$.

6. Confirm with multiple concentrations. To be rigorous, we measure $E_{\text{cell}}$ at several different $Mg^{2+}$ concentrations, plot $E_{\text{cell}}$ vs. $\log[Mg^{2+}]$, and extrapolate to $\log[Mg^{2+}] = 0$ (i.e., 1 M). The intercept gives $E^\circ_{Mg^{2+}/Mg}$ directly. This also checks that the system obeys the Nernst equation (the slope should be $0.0296\ \text{V}$), confirming that the electrode is behaving reversibly.

7. The accepted value. Through such careful experiments, the standard electrode potential of the $Mg^{2+}|Mg$ system is determined to be $-2.36\ \text{V}$ (vs. SHE). This large negative value reflects magnesium’s strong tendency to lose electrons — it’s a powerful reducing agent.

Method: Using the Nernst Equation with a Known Reference Electrode

This is a standard method for determining the standard electrode potential of a half-cell that cannot be measured directly (like Mg²⁺|Mg, which reacts with water).

Why can't we measure it directly?

If you dip Mg metal into Mg²⁺ solution, the system is not at equilibrium under standard conditions — Mg reacts with water. So we use an indirect electrochemical cell method.

Steps

1. Set up a galvanic cell Combine the Mg²⁺|Mg half-cell with a standard hydrogen electrode (SHE) as the reference. Cell notation:

2. Measure the cell potential ($E_{\text{cell}}$)

   Using a voltmeter (high impedance), measure the emf of this cell under standard conditions (298 K, 1 M, 1 atm).

   For Mg²⁺|Mg, the measured value is approximately +2.36 V (Mg acts as the anode).

3. Write the half-reactions

   • Anode (oxidation):

• Cathode (reduction):

4. Apply the cell potential equation

Here, $E_{\text{cathode}}^\circ = 0.00 \, \text{V}$ (SHE).

So:

5. Solve for the unknown

Key Result

Standard electrode potential of Mg²⁺|Mg is $\mathbf{-2.36 \, V}$ (at 298 K).

Why this works

The Nernst equation isn't directly used here for calculation — it's the principle that allows us to relate measured cell potential to standard potentials. The method relies on:

• The SHE having a defined $E^\circ = 0.00 \, \text{V}$
• The cell potential being the difference between two half-cell potentials
• Standard conditions ensuring $Q = 1$, so $E_{\text{cell}} = E_{\text{cell}}^\circ$

Common Mistakes: Determining the Standard Electrode Potential of Mg²⁺|Mg

✗ Mistake 1: Thinking You Can Measure It by Simply Dipping Mg in 1 M Mg²⁺ Solution

The error: Students assume the standard electrode potential can be measured the same way as a less-reactive metal — just immerse the metal in its 1 M salt solution and read a voltmeter against the SHE.

Why it's wrong: Magnesium is extremely reactive. In an aqueous Mg²⁺ solution, metallic Mg spontaneously reacts with water (or dissolved oxygen), evolving hydrogen gas. This side reaction disturbs the electrode equilibrium, so a stable, meaningful potential cannot be read directly under true standard conditions.

How to avoid: Always ask, "Is this electrode reactive enough to react with the solvent itself?" For very electropositive metals (Mg, Na, K, Ca), direct aqueous measurement is unreliable — an indirect method is required.

✗ Mistake 2: Forgetting That a Reference Electrode (SHE) Is Required

The error: Students try to quote "the potential of Mg" as if it is an absolute number measured in isolation.

Why it's wrong: Electrode potentials are always defined RELATIVE to a reference — the Standard Hydrogen Electrode (SHE), assigned $E^\circ = 0$ V by convention. You cannot measure a single electrode's potential without pairing it with a reference half-cell to complete a circuit.

How to avoid: Always set up a full cell: $Mg(s) \mid Mg^{2+}(aq) \parallel H^+(aq) \mid H_2(g) \mid Pt(s)$, then use $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}$ with the SHE's potential fixed at 0 V.

✗ Mistake 3: Mixing Up the Sign of the Measured emf

The error: Students read a positive cell emf and then also write $E^\circ_{Mg^{2+}/Mg}$ as positive.

Why it's wrong: Since Mg is the anode (it is oxidised) in the cell against SHE, a positive cell emf actually means $E^\circ_{Mg^{2+}/Mg}$ (a reduction potential) is negative: $E^\circ_{cell} = E^\circ_{cathode}(SHE, 0\ V) - E^\circ_{anode}(Mg)$, so $E^\circ_{Mg} = -E^\circ_{cell}$.

How to avoid: Always write out which electrode is the cathode and which is the anode before assigning a sign to the unknown reduction potential.

✗ Mistake 4: Ignoring the Need to Extrapolate to True Standard Conditions

The error: Students assume any single measurement, at any concentration, directly gives the standard potential.

Why it's wrong: Because a literal 1 M Mg²⁺/Mg half-cell is hard to maintain cleanly (Mistake 1), the practical measurement is often done at a lower, more controlled concentration, then the Nernst equation is used to extrapolate the result to 1 M (i.e. $\log[Mg^{2+}] = 0$).

How to avoid: Remember the two-part method: (1) measure $E_{cell}$ at a convenient, known concentration; (2) use the Nernst equation $E = E^\circ + \frac{0.059}{2}\log[Mg^{2+}]$ to solve for $E^\circ$, or better, measure at several concentrations and extrapolate the $E$ vs. $\log[Mg^{2+}]$ plot to zero.

✓ Correct Answer (for reference)

Construct the cell $Mg(s) \mid Mg^{2+}(aq,\ 1\ M) \parallel H^+(aq,\ 1\ M) \mid H_2(g,\ 1\ atm) \mid Pt(s)$, measure its emf (correcting for non-ideal concentrations via the Nernst equation if needed), and use $E^\circ_{cell} = E^\circ_{SHE} - E^\circ_{Mg^{2+}/Mg} = 0 - E^\circ_{Mg^{2+}/Mg}$ to obtain $E^\circ_{Mg^{2+}/Mg} = -2.36\ V$.

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