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Q. For $Zn^{2+}$ |Zn, $E^{\circ}$ = - 0.76 V then EMF of the cell
$Zn/Zn^{2+}$ (1M)|2H+ (1M)| $H_2 (1 atm)$ will be

AIIMSAIIMS 2012Electrochemistry

Solution:

$E^{\circ}_{cell}=E^{\circ}_{right}-E^{\circ }_{left}$
$E^{\circ }_{cell}=0-\left(0-0.76\right)=0.76 V$