Q. $E_{Cu}^{{^\circ}}\ = 0.34\ V\ ,\ E_{Zn}^{{^\circ}}\ = \ 0.76\ V.$ ADaniel cell contains 0.1 M $Zn{SO}_{4}$ solution and 0.01 M $Cu{SO}_{4}$ solution at its electrodes. EMF of the cell is
A) 1.10 V
B) 1.04 V
C) 1.16 V
D) 1.07 V
Explanation
(a)
${Cu}^{2 +} + \ {2e}^{-}\ \rightarrow \ Cu,$ $E^{{^\circ}}$ = 0.34
${Zn}^{2 +} + \ {2e}^{-}\ \rightarrow \ Zn$, $E^{{^\circ}}$ = 0.76
In the cell,
$Cu\ |\ {Cu}^{2 +}\ ||\ {Zn}^{2 +}\ |\ Zn$
anode cathode
In the cell ,
$E_{cell}^{{^\circ}} = \ E_{cathode}^{{^\circ}}\ - \ \ E_{anode}^{{^\circ}}$
= 0.76 – (-0.34)
= 1.10 V
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.