Question

Consider the cell:
$Zn|Z{n}^{2+}\left(aq\right)\left(1.0M\right)||C{u}^{2+}\left(aq\right)\left(1.0M\right)||Cu$
The standard reduction potential are $0.350$ V for
$C{u}^{2+}\left(aq\right)+2e^{-}\to Cu$ and $-0.763$ V for
$Z{n}^{2+}\left(aq\right)+2e^{-}\to Zn$

The cell reaction is:

• A. $Zn+C{u}^{3+}\to Z{n}^{+}+C{u}^{2+}$
  
• B. $Zn+C{u}^{2+}\to Z{n}^{2+}+Cu$
• C. $Z{n}^{2+}+C{u}^{3+}\to Z{n}^{2+}+C{u}^{2+}$
• D. $Zn+C{u}^{+}\to Z{n}^{2+}+Cu$

Answer 1

Answer: (B)

$Zn+C{u}^{2+}\to Z{n}^{2+}+Cu$

In the cell
$Zn|Z{n}^{2+}\left(aq\right)\left(1.0M\right)||C{u}^{2+}\left(aq\right)\left(1.0M\right)||Cu$
the oxidation half reaction is

(at anode) $Zn$ $\to Z{n}^{2+}+2e^{-}$ ....(i)

the reduction half reaction is
(at cathode) $C{u}^{2+}+2e^{-}\to Cu$ ....(ii)

On adding equations (i) and (ii), we get

$Zn+C{u}^{2+}\to Z{n}^{2+}+Cu$

This is the cell reaction.