Arrange the following metals in the order in which they displace each other from the solution of their salts: $Al$,$Cu$,$Fe$,$Mg$,and $Zn$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) metal with a stronger reducing power displaces a metal with a weaker reducing power from its salt solution.
The order of increasing reducing power (based on standard electrode potentials) for the given metals is $Cu < Fe < Zn < Al < Mg$.
Therefore,a metal higher in this series can displace any metal lower than it from its salt solution.
Thus,the order in which the metals can displace each other is $Mg > Al > Zn > Fe > Cu$.

Explore More

Similar Questions

If the $E^{0}$ values for $Mg^{+2} | Mg$,$Zn^{+2} | Zn$,and $Fe^{+2} | Fe$ are $-2.37 \ V$,$-0.76 \ V$,and $-0.44 \ V$ respectively,which statement is correct?

Calculate the $emf$ of the cell $Cu_{(s)} | Cu^{2+}_{(aq)} || Ag^+_{(aq)} | Ag_{(s)}$. Given: $E^0_{Cu^{2+}/Cu} = +0.34 \ V$,$E^0_{Ag^+/Ag} = +0.80 \ V$.

If the cell potential of the cell at $298 \ K$ is $2.36 \ V$,write the cell reaction and calculate the standard electrode potential of the $Mg^{2+} \mid Mg$ half-cell.
$Mg_{(s)} \mid Mg^{2+}_{(1 \ M)} \parallel H^{+}_{(1 \ M)} \mid H_{2(g)} (1 \ bar) \mid Pt_{(s)}$

The $emf$ of a galvanic cell,with electrode potentials of silver $= +0.80 \ V$ and that of copper $= +0.34 \ V$,is ........... $V$.

If the standard potential of the electrochemical cell for the reaction $2Ag_{(aq)}^+ + Cd_{(s)} \to Cd_{(aq)}^{2+} + 2Ag_{(s)}$ is $1.20 \ V$,calculate the standard Gibbs free energy change in $kJ$.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo