Question

The standard reduction potential $E^{o}$ for the half reactions are as:
$Z n \rightarrow Z n^{2+}+2 e^{-}, E^{o}=0.76\, V$
$C u \rightarrow C u^{2+}+2 e^{-}, E^{o}=0.34\, V$
The emf for the cell reaction: $Z n+C u^{2+} \rightarrow Z n^{2}+C u$

• A. $ 0.42\,V $
• B. $ -0.42\,\,V $
• C. $ -1.1\,V $
• D. $ 1.1\,V $

Answer 1

Answer: (A)

(i)Decide cathode and anode
(ii) $E_{c e l l}^{o}=E^{o}{ }_{C}-E^{o}{ }_{A}$ of electrode potential are substituted in terms of reduction potential.
Given, $E_{Z n / Z n^{2+}}^{o}=0.76 \,V$
$E_{C u / C u^{2+}}^{o}=0.34 \,V$
$\therefore Zn$ is anode ( $\because$ it has higher oxidation potential)
$\therefore E_{Z n^{2+} / Z n}^{o}=-0.76\, V$
and $E_{C u^{2+} / C u}^{o}=-0.34\, V$
$E_{c e l l}^{o}=E^{o}{ }_{C}-E^{o}{ }_{A}$
$=-0.34\, V-(-0.76\, V)$
$=0.34\, V+0.76\, V$
$=0.42 \,V$