Question

${H^ + }$ ions are reduced at platinum electrode prior to $Z{n^{2 + }}$
A.$Z{n^{2 + }}$
B.$C{u^{2 + }}$
C.$A{g^ + }$
D.${I_2}$

Answer 1

The ions which get reduced first in presence of another ions have higher reduction potential . Higher reduction potential means which have a higher tendency to be reduced in presence of other ions and it is experimentally determined .

Complete step by step answer:
At first we have to understand what reduction is.
Reduction is defined as gain at electrons or decrease in oxidation number .
$2{H^ + } + 2{e^ - } \to {H_{2(g)}}$
In this example hydronium ions get reduced because it accepts electrons and there is a decrease in oxidation number from +1 to 0.
After that we have to decide what factor is important when one ion gets reduced in presence of another ion .
The ions which are first reduced have higher reducing tendencies and have higher reduction potential .
$E^\circ {H^ + }/{H_2} = 0$ $v$
$E^\circ Z{n^{2 + }}/Zn = - 0.76v$
$E^\circ C{u^{2 + }}/Cu = 0.34v$
$E^\circ A{g^ + }/Ag = 0.2v$
$E^\circ {I_2}/{I^ - } = 0.54v$
From these experimentally determined value only reduction potential of $Z{n^{2 + }}/Zn$ is less than $E^\circ {H^ + }/{H_2}$
So higher the reduction potential value higher is the reducing tendency of that ions in presence of other ions .
So ${H^ + }$ ion is reduced at platinum electrode prior to $Z{n^{2 + }}$.

Note:
Electrochemical series also sometimes referred to as activity series is a list that describes the arrangement of elements in order of their increasing electrode potential values. The series has been established by measuring the potential of various electrodes versus standard hydrogen electrodes (SHE).So There is no need to learn the value of reductional potential but only we have to learn the electrochemical series where u get which ions is reduced first .