Question

Identify the incorrect statement from the options below for the above cell: Given : \( E _{ Cu ^{2}+ Cu }^{0}=+0.34 V E _{ Zn ^{2} \mid Zn }^{0}=-0.76 V \) a) If \( E _{\text {ext }}>1.1 V , Zn \) dissolves at \( Zn \) electrode and \( Cu \) deposits at \( Cu \) electrode b) If \( E _{ ext }>1.1 V \), e flows from \( Cu \) to \( Zn \) c) If \( E_{e x t}=1.1 V \), no flow of \( e ^{-} \)or current occurs d) If \( E _{ ext }<1.1 V , Zn \) dissolves at anode and \( Cu \) deposits at cathode

Answer 1

Correct option is (a)

E^oCu²⁺/Cu = +0.34 V

E^oZn⁺²/Zn = -0.76 V

Zn + Cu⁺² → Zn⁺² + Cu

E^o = 0.34 - (-0.76)

E^o = 1.1 V

Using nernst equation

E_cell = 1.1 V  - \(\frac{0.0591}2log\frac{Zn^{+2}}{Cu^{+2}}\)

E_cell = 1.1 V  - \(\frac{0.0591}2logK_{eq}=0\) 

1.1 = \(\frac{0.0591}2log\frac{[Zn^{+2}]}{[Cu^{+2}]}\) 

(a) If E_ext > 1.1V, means reaction gas backward direction . Cu dissolve at copper electrode and Zn deposits at Zn electrode.

(b) If E_ext > 1.1V, Cu oxidize and Zn reduced it means, electron flows rom Cu to Zn.

(c) If E_ext > 1.1V, it means there is no oxidation or reduction take place therefore no flow of current occur.

(d) If E_ext > 1.1V, it means oxidization of Zn and reduction of Cu take place. Zn dissolves at anode and Cu deposits at cathode.