The standard electrode potential (E^{circ}_{M | vedclass

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

Question

(C) metal ion $M^{2+}$ can liberate $H_2$ from a dilute acid if it can reduce $H^{+}$ to $H_2$ while itself getting oxidized to $M^{3+}$.This requires

Vedclass · Accepted Answer

$V^{2+}$ and $Cr^{2+}$

Vedclass · Answer

(C) metal ion $M^{2+}$ can liberate $H_2$ from a dilute acid if it can reduce $H^{+}$ to $H_2$ while itself getting oxidized to $M^{3+}$.This requires the reduction potential of the $M^{3+}/M^{2+}$ couple to be lower than the reduction potential of $H^{+}/H_2$ $(0.00 \ V)$.Alternatively,the oxidation potential of $M^{2+}/M^{3+}$ must be positive.Given $E^{\circ}(V^{3+}/V^{2+}) = -0.26 \ V$ and $E^{\circ}(Cr^{3+}/Cr^{2+}) = -0.41 \ V$,both are negative,meaning $V^{2+}$ and $Cr^{2+}$ are strong reducing agents that can reduce $H^{+}$ to $H_2$.$Mn^{2+}$ and $Co^{2+}$ have positive reduction potentials,meaning they are stable and cannot reduce $H^{+}$ to $H_2$.

The standard electrode potential $(E^{\circ}_{M^{3+}/M^{2+}})$ for $V$,$Cr$,$Mn$,and $Co$ are $-0.26 \ V$,$-0.41 \ V$,$+1.57 \ V$,and $+1.97 \ V$,respectively. The metal ions which can liberate $H_2$ from a dilute acid are

Similar Questions

If $Cu^{2+} + 2e^{-} \rightarrow Cu, E^{0} = 0.337 \ V$ and $Cu^{2+} + e^{-} \rightarrow Cu^{+}, E^{0} = 0.153 \ V$,then for the reaction $Cu^{+} + e^{-} \rightarrow Cu$,$E^{0}_{cell} =$ .............. $V$.

The $emf$ of a galvanic cell constituted with the electrodes $Zn^{2+} | Zn$ $(-0.76 \ V)$ and $Fe^{2+} | Fe$ $(-0.41 \ V)$ is

The standard reduction potentials for $Zn^{2+}/Zn$,$Ni^{2+}/Ni$ and $Fe^{2+}/Fe$ are $-0.76 \ V$,$-0.23 \ V$ and $-0.44 \ V$ respectively.
The reaction $X + Y^{2+} \rightarrow X^{2+} + Y$ will be spontaneous when:

For the reaction taking place in the cell:
$Pt_{(s)} | H_{2(g)} | H^{+}_{(aq)} || Ag^{+}_{(aq)} | Ag_{(s)}$
$E_{Cell}^0 = +0.5332 \ V$.
The value of $\Delta_f G^0$ in $kJ \ mol^{-1}$ (nearest integer) is:

Calculate the standard cell potential for a cell having the following reaction: $2 Al_{(s)} + 3 Ni^{2+} \rightarrow 2 Al^{3+} + 3 Ni_{(s)}$ given that $E_{Ni^{2+}/Ni}^{\circ} = -0.25 \ V$ and $E_{Al^{3+}/Al}^{\circ} = -1.66 \ V$. (in $V$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module