Question

A solution is one molar in each of ${\text{NaCl, CdC}}{{\text{l}}_2},{\text{ ZnC}}{{\text{l}}_2}{\text{ and PbC}}{{\text{l}}_2}$. To this, tin metal is added. Which of the following is true? Given:
${\text{E}}_{{\text{P}}{{\text{b}}^{{\text{2 + }}}}{\text{/Pb}}}^{\text{o}} = - 0.126{\text{ V}}$, ${\text{E}}_{{\text{S}}{{\text{n}}^{{\text{2 + }}}}{\text{/Sn}}}^{\text{o}} = - 0.136{\text{ V}}$, ${\text{E}}_{{\text{C}}{{\text{d}}^{{\text{2 + }}}}{\text{/Cd}}}^{\text{o}} = - 0.40{\text{ V}}$ ,
${\text{E}}_{{\text{Z}}{{\text{n}}^{{\text{2 + }}}}{\text{/Zn}}}^{\text{o}} = - 0.763{\text{ V}}$ , ${\text{E}}_{{\text{N}}{{\text{a}}^{\text{ + }}}{\text{/Na}}}^{\text{o}} = - 2.71{\text{ V}}$
A.${\text{Sn}}$ can reduce ${\text{N}}{{\text{a}}^ + }$ to ${\text{Na}}$
B.${\text{Sn}}$ can reduce ${\text{Z}}{{\text{n}}^{2 + }}$ to ${\text{Zn}}$
C.${\text{Sn}}$ can reduce ${\text{C}}{{\text{d}}^{2 + }}$ to ${\text{Cd}}$
D.${\text{Sn}}$ can reduce ${\text{P}}{{\text{b}}^{2 + }}$ to ${\text{Pb}}$

Answer 1

we must have the knowledge of reduction potential of elements and its application. We have been given the reduction potential; the element having highest reduction potential will get reduced. In negative, the bigger number is actually small.

Complete step by step solution:
We have been given the standard reduction potential of each various metals. The higher the reduction potential, the more tendencies the element has to reduce and the higher is oxidation potential then the element will tend to oxidize.
Oxidation refers to the loss of electron and formation of a cation while reduction refers to gain of electron and formation of an anion. In the given question we can see that all metals gain electrons and convert into their neutral states.
We have to check that tin can reduce which metal:
So the element which has to get reduced must have higher reduction potential. Among all the given options, $P{b^{2 + }}$ has the highest reduction potential, because in negative the smallest one is actually the highest value. So tin can reduce ${\text{P}}{{\text{b}}^{2 + }}$to ${\text{Pb}}$ by gaining electron and itself get oxidized by losing electron.

Hence the correct option is option D.

Note: The standard electrode potential is represented as standard reduction potential. The value of standard reduction potential or standard electrode potential does not depend upon concentration of solution. It only depends upon the nature of the species. All the values of standard electrode potential are calculated by taking standard hydrogen electrodes as a reference.