Bubbles of gas z are evolved from a solution | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The reduction potential indicates the tendency of a species to get reduced. $A$ higher reduction potential means a stronger oxidizing agent.Given

Vedclass · Accepted Answer

$y$ oxidizes $z$ but not $x$.

Vedclass · Answer

(A) The reduction potential indicates the tendency of a species to get reduced. $A$ higher reduction potential means a stronger oxidizing agent.Given the order of reduction potential: $x > y > z$.Since $z$ has the lowest reduction potential,it acts as the strongest reducing agent and is easily oxidized.Species with higher reduction potential can oxidize species with lower reduction potential.Therefore,$y$ (having higher reduction potential than $z$) can oxidize $z$,but $y$ cannot oxidize $x$ (as $x$ has a higher reduction potential than $y$).

Bubbles of gas $z$ are evolved from a solution containing $x$ and $y$. If the order of reduction potential is $x > y > z$,then:

Similar Questions

The value of $x$ from the given data is

The standard oxidation potentials for the half-reactions are given as $Zn \to Zn^{2+} + 2e^{-}; E^o = +0.76 \ V$ and $Fe \to Fe^{2+} + 2e^{-}; E^o = +0.41 \ V$. The $EMF$ for the cell reaction $Fe^{2+} + Zn \to Zn^{2+} + Fe$ is ............ $V$.

$E^{\circ}_{Fe^{2+}/Fe} = -0.441 \ V$ and $E^{\circ}_{Fe^{3+}/Fe^{2+}} = 0.771 \ V$,the standard $EMF$ of the reaction $Fe + 2Fe^{3+} \rightarrow 3Fe^{2+}$ will be $.......... \ V$.

Calculate $E^o_{cell}$ for the following cell in $V$:
$Zn_{(s)} | Zn^{2+}_{(aq.)} || Ag^{+}_{(aq.)} | Ag_{(s)}$
Given: $E^o_{Zn^{2+}/Zn} = -0.76 \ V$ ; $E^o_{Ag^{+}/Ag} = 0.80 \ V$

Given that the standard potentials $(E^{\circ})$ of $Cu^{2+}/Cu$ and $Cu^{+}/Cu$ are $0.34 \ V$ and $0.522 \ V$ respectively,the $E^{\circ}$ of $Cu^{2+}/Cu^{+}$ is (in $V$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module