Question
Redox reactions can be used to produce electricity.
(a) State the oxidation state of sulfur in copper(II) sulfate.
▶️Answer/Explanation
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Ans:
(a) The oxidation state of sulfur in copper(II) sulfate (CuSO4) is \(+6\). This can be determined by considering the oxidation states of the other elements in the compound:
The oxidation state of copper (\(Cu\)) is \(+2\) (given by the Roman numeral in the compound name, copper(II)).
The oxidation state of oxygen (\(O\)) in sulfate ion (\(SO_4^{2-}\)) is \(-2\) (given by the charge of the sulfate ion).
By assigning variable \(x\) to the oxidation state of sulfur, we can write the equation:
\[ (+2) + x + 4(-2) = 0 \]
Solving for \(x\) yields:
\[ x = +6 \]
Question
(b) A voltaic cell was constructed using a copper(II) sulfate/copper half-cell and a zinc sulfate/zinc half-cell.
(i) Outline why electrons flow from zinc to copper when these half cells are connected with a wire. Use section 25 of the data booklet.
▶️Answer/Explanation
Ans:
Electrons flow from zinc to copper when these half-cells are connected with a wire because zinc has a higher tendency to lose electrons (higher reduction potential) compared to copper. According to the electrochemical series, zinc has a more negative standard reduction potential (\(E^\circ\)) than copper. Therefore, zinc undergoes oxidation (loses electrons) at the anode, while copper undergoes reduction (gains electrons) at the cathode, resulting in the flow of electrons through the external wire from zinc to copper.
Question
(ii) Formulate equations for the reactions taking place at each electrode.
Anode (negative electrode): . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Cathode (positive electrode): . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
▶️Answer/Explanation
Ans:
Anode (negative electrode): \( Zn(s) \rightarrow Zn^{2+}(aq) + 2e^- \)
Cathode (positive electrode): \( Cu^{2+}(aq) + 2e^- \rightarrow Cu(s) \)
Question
(c) (i) Calculate the standard cell potential for the voltaic cell in part (b). Use section 24 of the data booklet.
▶️Answer/Explanation
Ans:
To calculate the standard cell potential (\(E^\circ_{cell}\)), we use the equation:
\[ E^\circ_{cell} = E^\circ_{cathode} – E^\circ_{anode} \]
Given standard reduction potentials (\(E^\circ_{cathode}\) for copper and \(E^\circ_{anode}\) for zinc from the electrochemical series, we substitute the values:
\[ E^\circ_{cell} = (+0.34 \, \text{V}) – (-0.76 \, \text{V}) = 1.10 \, \text{V} \]
Question
(ii) Calculate the standard Gibbs free energy change, \(ΔG^{\theta}\), in kJ \(mol^{-1}\) , for this reaction. Use section 1 of the data booklet. (If you did not answer part (c)(i) use 1.05V, but this is not the correct value.)
▶️Answer/Explanation
Ans:
To calculate the standard Gibbs free energy change (\(ΔG^\circ\)) for the reaction, we use the equation:
\[ ΔG^\circ = -nFE^\circ_{cell} \]
where \(n\) is the number of moles of electrons transferred (from the balanced equation) and \(F\) is the Faraday constant (96485 C/mol).
Given the standard cell potential (\(E^\circ_{cell}\)), we substitute the values:
\[ ΔG^\circ = -2 \times 96485 \, \text{C/mol} \times (1.10 \, \text{V}) \]
\[ ΔG^\circ = -212.267 \, \text{kJ/mol} \]
Therefore, the correct standard cell potential for the voltaic cell in part (c)(i) is \(1.10 \, \text{V}\), and the correct standard Gibbs free energy change for the reaction in part (c)(ii) is \(-212.267 \, \text{kJ/mol}\).