Q. The standard reduction potential $E^{{^\circ}}$ for the half reactions are as $Zn\ \rightarrow \ \ {Zn}^{2 +}\ + \ {2e}^{-}\ ,\ \ \ \ \ \ E^{{^\circ}}\ = \ 0.76\ V$ $Cu\ \rightarrow \ \ {Cu}^{2 +}\ + \ {2e}^{-}\ ,\ \ \ \ \ \ E^{{^\circ}}\ = \ 0.34\ V$ The emf for the cell reaction, $Zn\ + \ {Cu}^{2 +}\ \rightarrow \ {Zn}^{2 +} + \ Cu$
A) 0.42 V
B) -0.42 V
C) -1.1 V
D) 1.1 V
Explanation
(a)
Given, $E_{Zn\ /{Zn}^{2 +}}^{{^\circ}}\ = \ 0.76\ V$
$E_{Cu\ /{Cu}^{2 +}}^{{^\circ}}\ = \ 0.34\ V$
∴ Zn is anode (∵ It has higher oxidation potential)
∴ $E_{{Zn}^{2 +}\ /Zn}^{{^\circ}}\ = \ \ - \ 0.76\ V$
and $E_{{Cu}^{2 +}\ /Cu}^{{^\circ}}\ = \ \ - \ 0.34\ V$
$E_{cell}^{{^\circ}} = \ E_{cathode}^{{^\circ}}\ - \ \ E_{anode}^{{^\circ}}$
= - 0.34 V – ( - 0.76 V)
= 0.34 V + 0.76 V
= 0.42 V
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.