JEE Main 2020 · 4 September · morning · Single correct · +4 / −1
E_(Cu²⁺|Cu)⁰ = +0.34 V
E_(Zn²⁺|Zn)⁰ = -0.76 V
Identify the incorrect statement from the option below for the above cell :
- (a) If Eₑₓₜ < 1.1 V, Zn dissolves at anode and Cu deposits at cathode
- (b) If Eₑₓₜ = 1.1 V, no flow of e – or current occurs
- (c) If Eₑₓₜ > 1.1 V, e – flows from Cu to Zn
- (d) If Eₑₓₜ > 1.1 V, Zn dissolves at Zn electrode and Cu deposits at Cu electrode
Answer and solution
Answer (d) If Eₑₓₜ > 1.1 V, Zn dissolves at Zn electrode and Cu deposits at Cu electrode
Solution
Chemistry ma’am · 4 steps
Compare Eₑₓₜ with 1.1 V: once it is bigger, the outside source runs the whole cell backwards.
- Find the cell emf. Copper has the higher reduction potential, so it is the cathode and zinc is the anode. The cell pushes with E°_(cell) = 0.34 - (-0.76) = 1.10 V.
- Two opposing pushes. The external source is connected to oppose the cell. Whichever voltage is larger decides which way the electrons move and which reactions run.
Eₑₓₜ Electron flow (outer circuit) At Zn electrode At Cu electrode < 1.1 V Zn to Cu Zn dissolves Cu deposits = 1.1 V none no reaction no reaction > 1.1 V Cu to Zn Zn deposits Cu dissolves - Check option (d). Above 1.1 V the cell becomes electrolytic and both reactions reverse: Zn²⁺ + 2e⁻ → Zn and Cu → Cu²⁺ + 2e⁻. Option (d) describes the galvanic reactions, which no longer happen.
- So. Options (a), (b) and (c) are all correct statements, and (d) is the incorrect one.
The idea to keep. An opposing external voltage only stops the cell at exactly E_(cell). Beyond that, the electron flow and both electrode reactions reverse, and the galvanic cell becomes an electrolytic cell.
Answer from the NTA final answer key. Solution by PrepWiser.