Question

$Ca,Na,Mg,Zn,Fe,H,Cu,Hg,Ag,Au$
Given elements are in order of decreasing reactivity according to electrochemical series. Half-reaction possible for the shown electrochemical cell at the anode is :

• A. ${Cu}^{2+}+e^{-}\to Cu+$
• B. $Zn\left(s\right)\to {Zn}^{2+}+2e^{-}$
• C. ${Zn}^{2+}+2e^{-}\to Zn\left(s\right)$
• D. $Cu\left(s\right)\to {Cu}^{2+}+2e^{-}$
• E. ${Cu}^{2+}+2e^{-}\to Cu\left(s\right)$

Answer 1

Answer: (B)

$Zn\left(s\right)\to {Zn}^{2+}+2e^{-}$

Half-reaction possible for the shown electrochemical cell at the anode is the oxidation of zinc to Zn(II) ions. $Zn\left(s\right)\to {Zn}^{2+}+2e^{-}$
According to electrochemical series, $Zn$ has high reactivity. Hence, it undergoes oxidation.
At cathode reduction of $Cu\left(II\right)$ ions to $Cu$ will takes place.
${Cu}^{2+}+{2e}^{-}\to Cu$