Electrochemistry -- Diagnostic Tests

Question

Manganate(VII) ions ($MnO_{4}^{-}$) oxidise iron(II) ions ($Fe^{2+}$) to iron(III) ions ($Fe^{3+}$) in acidic medium, being reduced to manganese(II) ions ($Mn^{2+}$).

(a) Write the half-equation for the reduction of $MnO_{4}^{-}$ to $Mn^{2+}$ in acidic medium. [2 marks]

(b) Write the half-equation for the oxidation of $Fe^{2+}$ to $Fe^{3+}$. [1 mark]

(c) Combine the half-equations to give the balanced overall ionic equation. [2 marks]

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Worked Solution

(a) Reduction (gain of electrons):

$$MnO_{4}^{-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2+} + 4H_{2}O$$

Balancing check: Mn: 1, O: 4, H: 8, charge: $(-1) + 8(+1) + 5(-1) = +2$ on left; $+2 + 0 = +2$ on right. Balanced.

(b) Oxidation (loss of electrons):

$$Fe^{2+} \rightarrow Fe^{3+} + e^{-}$$

(c) To balance electrons, multiply the oxidation half-equation by 5:

$$5Fe^{2+} \rightarrow 5Fe^{3+} + 5e^{-}$$

Add to the reduction half-equation and cancel electrons:

$$MnO_{4}^{-} + 8H^{+} + 5Fe^{2+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_{2}O$$

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Question

Consider the electrolysis of concentrated aqueous sodium chloride using inert electrodes.

(a) Identify the products at the anode and cathode, giving reasons for your choices. [4 marks]

(b) Write the overall equation for the electrolysis. [1 mark]

(c) How would the products change if dilute aqueous sodium chloride were electrolysed instead? Explain. [3 marks]

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Worked Solution

(a) Cathode (negative electrode): $Na^{+}$ and $H_{2}O$ are present. $H_{2}O$ is reduced in preference to $Na^{+}$ (since $Na$ is very reactive, below hydrogen in the reactivity series; $H^{+}$ from water is more reduced):

$$2H_{2}O(l) + 2e^{-} \rightarrow H_{2}(g) + 2OH^{-}(aq)$$

Product: hydrogen gas.

Anode (positive electrode): $Cl^{-}$ and $H_{2}O$ are present. In concentrated $NaCl$ solution, $Cl^{-}$ is discharged in preference to $H_{2}O$ (the overpotential of chlorine is lower than that of oxygen at the anode, and the high concentration of $Cl^{-}$ favours its discharge):

$$2Cl^{-}(aq) \rightarrow Cl_{2}(g) + 2e^{-}$$

Product: chlorine gas.

(b) Overall:

$$2NaCl(aq) + 2H_{2}O(l) \xrightarrow{\text{electrolysis}} Cl_{2}(g) + H_{2}(g) + 2NaOH(aq)$$

(c) In dilute aqueous $NaCl$:

• Cathode: Same as before — $H_{2}O$ is still reduced to $H_{2}$ (since $Na^{+}$ is never discharged in aqueous solution).
• Anode: The low concentration of $Cl^{-}$ means $H_{2}O$ is now preferentially discharged:

$$4OH^{-}(aq) \rightarrow O_{2}(g) + 2H_{2}O(l) + 4e^{-}$$

Or equivalently:

$$2H_{2}O(l) \rightarrow O_{2}(g) + 4H^{+}(aq) + 4e^{-}$$

Product at anode changes from chlorine to oxygen.

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Question

In the electrolysis of copper(II) sulphate solution using copper electrodes, a current of 0.500 A was passed for 30.0 minutes. The mass of the anode decreased by 0.296 g.

(a) Calculate the charge that passed through the solution. [2 marks]

(b) Calculate the theoretical mass of copper that should have deposited at the cathode. [3 marks]

(c) Calculate the percentage difference between the theoretical and actual mass change, and suggest a reason for any discrepancy. [2 marks]

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Worked Solution

(a) $$Q = It = 0.500 \times 30.0 \times 60 = 900 \text{ C}

(b) At the cathode: $Cu^{2+} + 2e^{-} \rightarrow Cu$ Moles of electrons: $n(e^{-}) = \frac{Q}{F} = \frac{900}{96500} = 9.326 \times 10^{-3}$ mol Moles of $Cu$ deposited: $n(Cu) = \frac{9.326 \times 10^{-3}}{2} = 4.663 \times 10^{-3}$ mol

m(Cu) = 4.663 \times 10^{-3} \times 63.5 = 0.296 \text{ g}

(c) The theoretical mass (0.296 g) matches the actual mass change at the anode (0.296 g) exactly in this case. This makes sense because with copper electrodes, the anode dissolves ($Cu \rightarrow Cu^{2+} + 2e^{-}$) at the same rate as copper deposits at the cathode. The concentration of $Cu^{2+}$ in solution remains constant, and the mass transfer is equal. If there were a discrepancy, possible reasons would include: - Current not being constant - Side reactions (e.g., if the current density is too high, $H_{2}$ may be evolved at the cathode) - Impurities in the copper electrodes - Oxidation of the anode not involving just copper dissolution --- <!-- Breadcrumb Schema for SEO --> <script type="application/ld+json"> { "@context": "https://schema.org", "@type": "BreadcrumbList", "itemListElement": [{"name": "Home", "url": "https://wyattau.com"}, {"name": "dse", "url": "https://dse.wyattau.com"}, {"name": "Chemistry", "url": "https://dse.wyattau.com/chemistry"}, {"name": "Diagnostics", "url": "https://dse.wyattau.com/chemistry/diagnostics"}, {"name": "Diag Electrochemistry", "url": "https://dse.wyattau.com/chemistry/diagnostics/diag-electrochemistry"}] } </script> ## Intuition **Electricity from chemistry:** Electrochemistry is like a chemical battery, redox reactions transfer electrons from one substance to another, and we can harness that electron flow as electricity. **Why it matters:** From car batteries to rechargeable phones, electrochemistry powers modern life. Understanding redox helps you predict which reactions occur and how to control them. **The key insight:** The electrochemical series ranks substances by their tendency to gain or lose electrons, higher in the series means more likely to be oxidized. ## Integration Test 1: Electrochemical Series Predictions **Question** Given the following standard electrode potentials: <div class="table-wrap"> | Half-equation | $E^{\circ}$ (V) | | -------------------------------------------- | --------------- | | $Zn^{2+} + 2e^{-} \rightleftharpoons Zn$ | $-0.76$ | | $Fe^{2+} + 2e^{-} \rightleftharpoons Fe$ | $-0.44$ | | $2H^{+} + 2e^{-} \rightleftharpoons H_{2}$ | $0.00$ | | $Cu^{2+} + 2e^{-} \rightleftharpoons Cu$ | $+0.34$ | | $Ag^{+} + e^{-} \rightleftharpoons Ag$ | $+0.80$ | | $Br_{2} + 2e^{-} \rightleftharpoons 2Br^{-}$ | $+1.07$ | </div> (a) Predict whether zinc will react with aqueous copper(II) sulphate. Write the equation and calculate the cell EMF. [3 marks] (b) Predict whether silver will react with dilute hydrochloric acid. Explain. [2 marks] (c) A student sets up a cell using $Zn|Zn^{2+}$ and $Ag|Ag^{+}$ half-cells. Calculate the standard cell EMF and state the direction of electron flow in the external circuit. [3 marks] --- <!-- Breadcrumb Schema for SEO --> <script type="application/ld+json"> { "@context": "https://schema.org", "@type": "BreadcrumbList", "itemListElement": [{"name": "Home", "url": "https://wyattau.com"}, {"name": "dse", "url": "https://dse.wyattau.com"}, {"name": "Chemistry", "url": "https://dse.wyattau.com/chemistry"}, {"name": "Diagnostics", "url": "https://dse.wyattau.com/chemistry/diagnostics"}, {"name": "Diag Electrochemistry", "url": "https://dse.wyattau.com/chemistry/diagnostics/diag-electrochemistry"}] } </script> **Worked Solution** (a) Yes, zinc will react with aqueous $CuSO_{4}$. $Zn$ is a stronger reducing agent than $Cu$ (more negative $E^{\circ}$). Oxidation: $Zn \rightarrow Zn^{2+} + 2e^{-}$ ($E^{\circ} = +0.76$ V as oxidation) Reduction: $Cu^{2+} + 2e^{-} \rightarrow Cu$ ($E^{\circ} = +0.34$ V)

E^{\circ}_{\text{cell}} = 0.34 - (-0.76) = +1.10 \text{ V}

Since $E^{\circ}_{\text{cell}} \gt 0$ The reaction is feasible:

Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)

(b) Silver will **not** react with dilute HCl. $Ag$ has $E^{\circ} = +0.80$ V, which is more positive than $H^{+}/H_{2}$ ($E^{\circ} = 0.00$ V). Silver is **below** hydrogen in the electrochemical series, meaning $Ag$ is a weaker reducing agent than $H_{2}$. Silver cannot reduce $H^{+}$ to $H_{2}$.

E^{\circ}_{\text{cell}} = 0.00 - 0.80 = -0.80 \text{ V} \lt 0

The reaction is not feasible. (c) Oxidation at anode: $Zn \rightarrow Zn^{2+} + 2e^{-}$ Reduction at cathode: $Ag^{+} + e^{-} \rightarrow Ag$

E^{\circ}{\text{cell}} = E^{\circ}{\text{cathode}} - E^{\circ}_{\text{anode}} = 0.80 - (-0.76) = +1.56 \text{ V}

Electrons flow from the **zinc electrode (anode)** through the external circuit to the **silver electrode (cathode)**. --- <!-- Breadcrumb Schema for SEO --> <script type="application/ld+json"> { "@context": "https://schema.org", "@type": "BreadcrumbList", "itemListElement": [{"name": "Home", "url": "https://wyattau.com"}, {"name": "dse", "url": "https://dse.wyattau.com"}, {"name": "Chemistry", "url": "https://dse.wyattau.com/chemistry"}, {"name": "Diagnostics", "url": "https://dse.wyattau.com/chemistry/diagnostics"}, {"name": "Diag Electrochemistry", "url": "https://dse.wyattau.com/chemistry/diagnostics/diag-electrochemistry"}] } </script> ## Integration Test 2: Electrolysis + Mass Calculation **Question** A student electrolysed molten aluminium oxide ($Al_{2}O_{3}$) using graphite electrodes with a current of 5.00 A for 2.00 hours. (a) Write the half-equations occurring at the anode and cathode. [2 marks] (b) Calculate the mass of aluminium produced at the cathode. [3 marks] (c) The actual mass obtained was 2.80 g. Calculate the current efficiency. [2 marks] --- <!-- Breadcrumb Schema for SEO --> <script type="application/ld+json"> { "@context": "https://schema.org", "@type": "BreadcrumbList", "itemListElement": [{"name": "Home", "url": "https://wyattau.com"}, {"name": "dse", "url": "https://dse.wyattau.com"}, {"name": "Chemistry", "url": "https://dse.wyattau.com/chemistry"}, {"name": "Diagnostics", "url": "https://dse.wyattau.com/chemistry/diagnostics"}, {"name": "Diag Electrochemistry", "url": "https://dse.wyattau.com/chemistry/diagnostics/diag-electrochemistry"}] } </script> **Worked Solution** (a) Cathode (reduction): $Al^{3+} + 3e^{-} \rightarrow Al$ Anode (oxidation): $2O^{2-} \rightarrow O_{2} + 4e^{-}$ (b) $$Q = It = 5.00 \times 2.00 \times 3600 = 36000 \text{ C}

Moles of electrons: $n(e^{-}) = \frac{36000}{96500} = 0.3731$ mol

Moles of $Al$: $n(Al) = \frac{0.3731}{3} = 0.1244$ mol

$$m(Al) = 0.1244 \times 27.0 = 3.36 \text{ g}$$

(c)

$$\text{Current efficiency} = \frac{\text{actual mass}}{\text{theoretical mass}} \times 100\% = \frac{2.80}{3.36} \times 100\% = 83.3\%$$

The efficiency is less than 100% due to:

• Some current being used for side reactions (e.g., oxidation of the graphite anode to $CO_{2}$)
• Heat losses in the electrolytic cell
• Some aluminium re-oxidising before it is collected

Forgetting to balance electrons before combining half-equations: When combining oxidation and reduction half-equations, you must multiply them so the number of electrons lost equals electrons gained. Don’t just add the half-equations directly.

Confusing anode and cathode in electrolysis vs galvanic cells: In electrolysis, the anode is positive (oxidation) and cathode is negative (reduction). In galvanic cells, it’s reversed, anode is negative (oxidation) and cathode is positive (reduction).

Assuming Na⁺ is discharged in aqueous NaCl electrolysis: Na⁺ is never discharged in aqueous solution because water is reduced instead (Na is very reactive). At the anode, Cl⁻ is discharged in concentrated solution, but OH⁻ from water is discharged in dilute solution.

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Question

Consider the following cell:

$$Sn(s) | Sn^{2+}(aq, 1.0 \text{ mol/dm}^{3}) || Cu^{2+}(aq, 1.0 \text{ mol/dm}^{3}) | Cu(s)$$

Standard electrode potentials: $Sn^{2+}/Sn = -0.14$ V; $Cu^{2+}/Cu = +0.34$ V.

(a) Identify the anode and cathode. Calculate the standard cell EMF. [2 marks]

(b) Write the overall cell reaction. [1 mark]

(c) If the concentration of $Sn^{2+}$ is increased to 5.0 mol/dm$^{3}$ while $[Cu^{2+}]$ remains at 1.0 mol/dm$^{3}$Use the Nernst equation to predict the effect on the cell EMF. [4 marks]

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Worked Solution

(a) Anode: $Sn$ electrode (oxidation occurs; $Sn$ has the more negative $E^{\circ}$).

Cathode: $Cu$ electrode (reduction occurs).

$$E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}} = 0.34 - (-0.14) = +0.48 \text{ V}$$

(b) $Sn(s) + Cu^{2+}(aq) \rightarrow Sn^{2+}(aq) + Cu(s)$

(c) Using the Nernst equation for the cell reaction $Sn + Cu^{2+} \rightarrow Sn^{2+} + Cu$ (2 electrons transferred):

$$E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{RT}{nF}\ln Q$$

At 298 K:

$$E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0592}{n}\log Q$$ $$Q = \frac{[Sn^{2+}]}{[Cu^{2+}]} = \frac{5.0}{1.0} = 5.0$$ $$E_{\text{cell}} = 0.48 - \frac{0.0592}{2}\log(5.0)$$ $$= 0.48 - 0.0296 \times 0.699$$ $$= 0.48 - 0.0207 = 0.459 \text{ V}$$

The cell EMF decreases from 0.48 V to 0.459 V. Increasing $[Sn^{2+}]$ (product concentration) shifts the equilibrium towards the reactants, reducing the driving force for the forward reaction.

Note: In the DSE context, the Nernst equation is often presented as $E = E^{\circ} + \frac{0.0592}{n}\log\frac{[\text{oxidised}]}{[\text{reduced}]}$ applied to each half-cell separately. The conclusion is the same: increasing $[Sn^{2+}]$ makes the tin half-cell potential more positive, reducing the overall cell potential.

• Atomic Structure: Atomic structure is foundational
• Equilibrium: Equilibrium connects topics
• Organic Chemistry: Organic chemistry is a major area