Q. The emf of the cell, $(E_{{Zn}^{2 +}\ /Zn}\ = \ - 0.76\ V)$ $Zn\ /\ {Zn}^{2 +}\ (1\ M)\ ||\ {Cu}^{2 +}\ (1\ M)\ |\ Cu$ $(E_{{Cu}^{2 +}\ /Cu}\ = \ + 0.34\ V)$ will be
A) +1.10 V
B) -1.10 V
C) +0.42 V
D) -0.42 V
Explanation
(a)
Given, that
$Zn\ /\ {Zn}^{2 +}\ ||\ {Cu}^{2 +}\ /\ Cu$
∴ Zn is anode and Cu is cathode.
Given,
${Zn}^{2 +}/\ Zn\ = \ - \ 0.76\ V$
${Cu}^{2 +}\ /\ Cu\ = \ + \ 0.34\ V$
$E_{cell} = \ E_{cathode}\ - \ E_{anode}$
= 0.34 – (- 0.76)
= 0.34 + 0.76
= 1.10 V
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.