Question

\(E^{\circ}_{\mathrm{Cu^{2+}|Cu}} = +0.34\ \mathrm{V}\)
\(E^{\circ}_{\mathrm{Zn^{2+}|Zn}} = -0.76\ \mathrm{V}\)
Identify the incorrect statement from the options below for the above cell :

• A. If \(E_{ext} \lt 1.1\ \mathrm{V}\), Zn dissolves at anode and Cu deposits at cathode
• B. If \(E_{ext} = 1.1\ \mathrm{V}\), no flow of \(e^-\) or current occurs
• C. If \(E_{ext} \gt 1.1\ \mathrm{V}\), \(e^-\) flows from Cu to Zn
• D. If \(E_{ext} \gt 1.1\ \mathrm{V}\), Zn dissolves at Zn electrode and Cu deposits at Cu electrode

Hint:

This is a Daniell cell with E(cell) = 0.34 - (-0.76) = 1.10 V. An opposing external potential below 1.1 V leaves it galvanic, equal to 1.1 V stops the current, above 1.1 V reverses the reaction (electrolysis).

Answer 1

The Correct Option is D

\(E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.34 - (-0.76) = 1.10\ \mathrm{V}\). The external source in the figure opposes the cell.
(1) \(E_{ext} \lt 1.1\ \mathrm{V}\): the cell still works as a galvanic cell: Zn dissolves at the anode and Cu deposits at the cathode. Correct.
(2) \(E_{ext} = 1.1\ \mathrm{V}\): the opposing potential balances the cell emf, so no current flows. Correct.
(3) \(E_{ext} \gt 1.1\ \mathrm{V}\): the reaction is reversed (electrolytic cell) and electrons flow from Cu to Zn. Correct.
(4) \(E_{ext} \gt 1.1\ \mathrm{V}\): in the reversed reaction \(\mathrm{Zn^{2+}} + \mathrm{Cu} \to \mathrm{Zn} + \mathrm{Cu^{2+}}\), Zn is deposited on the Zn electrode and Cu dissolves, so this statement is wrong.
The incorrect statement is (4).
Answer: (4) If E_ext > 1.1 V, Zn dissolves at Zn electrode and Cu deposits at Cu electrode