Electrochemistry Essentials: Predicting Cell Reactions and Practical Outcomes

1) A practical workflow for predicting real electrochemical outcomes

This chapter focuses on turning a list of possible half-reactions into a concrete prediction: what reacts, what forms, which electrode does what, and whether the process is spontaneous under standard conditions. You will combine (i) candidate half-reactions, (ii) standard potentials, (iii) balanced overall equations, and (iv) qualitative “reality checks” (species present, phases, and competing reactions).

Step A — List plausible half-reactions from what is actually present

From the chemicals and phases given (metals, ions, gases, water), write 2–4 plausible reduction half-reactions that could occur at a cathode and 2–4 plausible oxidation half-reactions that could occur at an anode. In aqueous systems, always include water as a candidate when appropriate.

• Cathode candidates (reductions): metal-ion to metal, halogen to halide, dissolved oxygen to water/hydroxide, water to hydrogen.
• Anode candidates (oxidations): metal to metal-ion, halide to halogen, water to oxygen, hydroxide to oxygen.

Step B — Compute E°_cell for each pairing

Use reduction potentials from the table. For a chosen pairing:

• Keep the cathode half-reaction as a reduction with its listed E°.
• Choose an anode half-reaction by selecting a reduction from the table and reversing it (that makes it oxidation). Do not change the numerical E° when you reverse; instead use subtraction in the cell calculation.
• Compute: E°cell = E°(cathode, reduction) − E°(anode, reduction).

Interpretation under standard conditions: E°cell > 0 predicts a spontaneous galvanic reaction in that direction; E°cell < 0 predicts it will not proceed spontaneously (it would require electrolysis).

Step C — Write and balance the overall reaction

After choosing the most plausible cathode/anode pairing (often the one with the most positive E°cell among feasible options), combine the two half-reactions and balance electrons. Then add spectator ions only if you are asked for a molecular equation or net ionic equation.

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Tip: E° values are intensive; multiplying a half-reaction to balance electrons does not multiply E°.

Step D — Identify qualitative outcomes (what you would observe)

• Electrode changes: metal plating at the cathode; metal dissolving at the anode; gas evolution (H₂, O₂, Cl₂, Br₂) at electrodes.
• Solution changes: ion concentrations shift; pH can change near electrodes (OH⁻ near cathode if water is reduced; H⁺ near anode if water is oxidized).
• Limiting qualitative outcome: in a simple displacement, the “more easily oxidized” metal tends to dissolve while the “more easily reduced” ion plates or forms a reduced product—provided the species is present and not blocked by competing reactions.

2) Worked examples: displacement, halogens, and aqueous compatibility

Example 1 — Metal displacement: Zn(s) in CuSO₄(aq)

Given: Zn(s) placed in CuSO₄(aq) (so Cu²⁺(aq) is present). Predict whether Cu plates and Zn dissolves.

Candidate reductions (cathode): Cu2+ + 2e− → Cu(s) (E° = +0.34 V) and possibly 2H2O + 2e− → H2 + 2OH− (E° = −0.83 V, basic form) if water reduction competes.

Candidate oxidation (anode): Zn(s) can oxidize; use the reduction potential for the reverse half-reaction: Zn2+ + 2e− → Zn(s) (E° = −0.76 V).

Compute E°cell for Cu plating with Zn oxidation:

E°cell = E°(Cu2+/Cu) − E°(Zn2+/Zn) = (+0.34) − (−0.76) = +1.10 V

Overall reaction (balanced):

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

Practical outcome: copper metal deposits on zinc; zinc dissolves; blue Cu²⁺ color fades as Cu²⁺ is consumed. Water reduction is not favored here because Cu²⁺ reduction is much easier than water reduction under comparable conditions.

Example 2 — Metal displacement that fails: Cu(s) in ZnSO₄(aq)

Try the analogous pairing: Would Zn²⁺ reduce to Zn(s) while Cu oxidizes to Cu²⁺?

Compute:

E°cell = E°(Zn2+/Zn) − E°(Cu2+/Cu) = (−0.76) − (+0.34) = −1.10 V

Prediction: not spontaneous. Practically, copper will not displace zinc from Zn²⁺ solution.

Example 3 — Halogen displacement: Cl₂(aq) added to KI(aq)

Given: chlorine water mixed with potassium iodide solution. Predict whether iodine forms.

Cathode candidate (reduction): Cl2 + 2e− → 2Cl− (E° ≈ +1.36 V)

Anode candidate (oxidation): iodide can be oxidized; use reduction potential: I2 + 2e− → 2I− (E° ≈ +0.54 V)

Compute:

E°cell = E°(Cl2/Cl−) − E°(I2/I−) = 1.36 − 0.54 = +0.82 V

Overall reaction:

Cl2(aq) + 2I−(aq) → 2Cl−(aq) + I2(aq/s)

Practical outcome: iodide is oxidized to iodine (brown/purple depending on conditions); chloride forms. This is a classic halogen displacement: the stronger oxidizing halogen (higher E° for X₂/X⁻) oxidizes the halide of a weaker oxidizer.

Example 4 — Aqueous compatibility note: electrolysis of NaCl(aq) vs molten NaCl

In aqueous electrolysis, the “obvious” ionic half-reactions may be beaten by water reactions. Compare what is possible in each medium.

Case A: molten NaCl(l)

• Cathode: Na+ + e− → Na(l) (sodium metal forms)
• Anode: 2Cl− → Cl2(g) + 2e− (chlorine gas forms)

Case B: aqueous NaCl(aq)

Cathode candidates: Na+ + e− → Na(s) is extremely unfavorable in water; water reduction to H₂ is typically favored. So expect:

2H2O + 2e− → H2(g) + 2OH−(aq)

Anode candidates: chloride oxidation to Cl₂ competes with water oxidation to O₂. Which dominates depends on concentration, electrode material, and overpotential; in concentrated brine with suitable electrodes, Cl₂ is commonly produced. In dilute chloride, O₂ becomes more likely.

Practical outcome to predict qualitatively: in NaCl(aq), you generally do not plate sodium; you get H₂ at the cathode and either Cl₂ or O₂ at the anode, with local pH changes (more basic near cathode).

3) Interpreting diagrams and data: from E° tables and schematics to directions and signs

From an E° table: decide anode/cathode and electron direction

When you are given two half-cells (or two candidate couples), you can determine roles directly from their reduction potentials under standard conditions:

• The half-reaction with the more positive E° (as a reduction) is the more favorable reduction → it is the cathode in a spontaneous galvanic cell.
• The other half-cell runs in reverse (oxidation) → it is the anode.
• Electrons flow externally from anode → cathode.

Given a cell schematic: assign electrodes and predict the voltage sign

Suppose a schematic shows two beakers connected by a salt bridge and a voltmeter reading. If the cell is operating galvanically (no external power supply), the measured cell voltage should be positive when the voltmeter’s positive lead is connected to the cathode.

Mini-example: determine direction and E°cell from a partial schematic

Given: Half-cells: Fe(s)|Fe²⁺(aq) and Ag⁺(aq)|Ag(s). Use E° values: Ag+ + e− → Ag E° = +0.80 V; Fe2+ + 2e− → Fe E° = −0.44 V.

• Cathode: Ag⁺/Ag (more positive E°) → Ag plates.
• Anode: Fe/Fe²⁺ → Fe dissolves to Fe²⁺.
• E°cell: 0.80 − (−0.44) = +1.24 V.
• Electron direction: from Fe electrode to Ag electrode through the wire.

4) Troubleshooting unrealistic predictions: making E°-based results chemically plausible

Check 1 — Is the reacting species actually available in the stated phase?

• If the table suggests reducing a metal ion, confirm that ion is present in solution (not locked in an insoluble solid or absent entirely).
• If a gas is predicted (Cl₂, H₂, O₂), confirm there is a pathway: appropriate ions/water present and an electrode where gas can evolve.

Check 2 — Are you accidentally predicting an impossible aqueous product?

• Plating very reactive metals from water (e.g., Na, K, Ca) is typically unrealistic in aqueous solution because water reduction outcompetes metal-ion reduction.
• If your computed best cathode reaction is “Mⁿ⁺ → M(s)” for a very reactive metal in water, re-check water reduction as a competing cathode process.

Check 3 — Competing reactions at the anode in aqueous electrolysis

At the anode, halide oxidation can compete with water oxidation. A purely E°-based choice can be misleading because kinetics and overpotential matter. A practical rule-of-thumb for many classroom predictions:

• Concentrated Cl⁻/Br⁻ solutions: halogen evolution is often observed at the anode.
• Dilute halide solutions or sulfate/nitrate solutions: oxygen evolution from water is often observed.

When a problem explicitly says “inert electrodes in aqueous solution” and does not specify concentration, state both possibilities and what would push the system toward one (e.g., high [Cl⁻] favors Cl₂).

Check 4 — Are you mixing up E°cell arithmetic or sign conventions?

• Do not add E° values after reversing a half-reaction; use E°cell = E°cath − E°anode with both taken as reduction potentials from the table.
• Do not multiply E° when scaling half-reactions for electron balance.
• If you get a negative E°cell for a reaction you “expect,” verify you didn’t swap anode/cathode or accidentally used oxidation potentials.

Check 5 — Are you ignoring complexation/precipitation that removes ions?

Some ions may be tied up by complex formation or precipitation, reducing their effective availability. In many introductory predictions you will not compute equilibrium shifts, but you should at least ask: does the problem mention complexing agents (e.g., NH₃) or insoluble salts (e.g., AgCl)? If so, the “free” ion concentration may be low, changing what is feasible.

5) Practice set: mixed-format prompts (tables, diagrams, partial notation)

Data table (use for questions 1–6)

1) Metal displacement (net ionic)

Predict whether a reaction occurs and write the net ionic equation (if any):

• (a) Fe(s) placed in Cu²⁺(aq)
• (b) Pb(s) placed in Zn²⁺(aq)
• (c) Zn(s) placed in Pb²⁺(aq)

2) Halogen displacement (products and observations)

For each mixture, predict whether a displacement occurs and identify the halogen formed (if any):

• (a) Cl₂ + 2Br⁻
• (b) Br₂ + 2Cl⁻
• (c) Br₂ + 2I⁻

3) Compute E°cell and label anode/cathode

For each pairing below, compute E°cell and label anode/cathode under standard conditions:

• (a) Zn(s)|Zn²⁺ || Cu²⁺|Cu(s)
• (b) Fe(s)|Fe²⁺ || Pb²⁺|Pb(s)
• (c) Ag⁺|Ag(s) || Cu²⁺|Cu(s) (decide which side must be anode for a spontaneous cell)

4) Diagram interpretation (text schematic)

A student draws a galvanic cell and labels the left electrode as “Cu(s) in Cu²⁺” and the right electrode as “Zn(s) in Zn²⁺.” The student also draws an arrow for electron flow from left to right.

• (a) Is the electron-flow arrow correct? If not, correct it.
• (b) Which electrode is the anode? Which is the cathode?
• (c) If the voltmeter’s positive lead is on the left electrode, should the reading be positive or negative?

5) Partial cell notation completion

Complete the missing parts and indicate the expected sign of E°cell:

Fe(s) | Fe2+(aq) || ________(aq) | ________(s)

Choose from: Ag⁺, Ag, Zn²⁺, Zn, Pb²⁺, Pb. Then compute E°cell.

6) Aqueous electrolysis reality check

You are electrolyzing an aqueous solution containing Na⁺, Cl⁻, and water using inert electrodes.

• (a) Name a likely cathode reaction and product.
• (b) List two plausible anode reactions (one involving chloride, one involving water) and state one condition that would favor each.
• (c) Predict how pH changes near the cathode.

7) Troubleshooting prompt: “My prediction says sodium plates out”

A classmate calculates a very negative E° for Na+ + e− → Na but still writes sodium metal as the cathode product in aqueous NaCl electrolysis because “Na⁺ is the cation present.” Identify at least two checks from the troubleshooting section that show why this is unrealistic, and propose the corrected cathode reaction.

8) Mixed prompt with competing species

A solution contains Cu²⁺(aq), H⁺(aq), and Cl⁻(aq). A metal strip of Zn(s) is added.

• (a) List two possible reductions that could occur on the metal surface.
• (b) Using E° values, predict which reduction is more favorable under standard conditions.
• (c) State one practical observation that would distinguish the two outcomes (metal plating vs gas evolution).