For the cell $Zn_{(s)} | Zn^{2+}_{(aq)} || M^{x+}_{(aq)} | M_{(s)}$,different half cells and their standard electrode potentials are given below:
$M^{x+}_{(aq)} / M_{(s)}$ $Au^{3+}_{(aq)} / Au_{(s)}$ $Ag^{+}_{(aq)} / Ag_{(s)}$ $Fe^{3+}_{(aq)} / Fe^{2+}_{(aq)}$ $Fe^{2+}_{(aq)} / Fe_{(s)}$ $E^o M^{x+} / M (V)$ $1.40$ $0.80$ $0.77$ $-0.44$
If $E^o Zn^{2+}/Zn = -0.76 \ V$,which cathode will give a maximum value of $E^o_{cell}$ per electron transferred?
| $M^{x+}_{(aq)} / M_{(s)}$ | $Au^{3+}_{(aq)} / Au_{(s)}$ | $Ag^{+}_{(aq)} / Ag_{(s)}$ | $Fe^{3+}_{(aq)} / Fe^{2+}_{(aq)}$ | $Fe^{2+}_{(aq)} / Fe_{(s)}$ |
| $E^o M^{x+} / M (V)$ | $1.40$ | $0.80$ | $0.77$ | $-0.44$ |
If $E^o Zn^{2+}/Zn = -0.76 \ V$,which cathode will give a maximum value of $E^o_{cell}$ per electron transferred?
- A$Ag^{+}/Ag$
- B$Fe^{3+}/Fe^{2+}$
- C$Au^{3+}/Au$
- D$Fe^{2+}/Fe$
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Similar Questions
Consider the following ${E^0}$ values:
${E^0}_{Fe^{3+}/Fe^{2+}} = + 0.77 \ V$
${E^0}_{Sn^{2+}/Sn} = - 0.14 \ V$
Under standard conditions,the potential for the reaction $Sn_{(s)} + 2Fe^{3+}_{(aq)} \to 2Fe^{2+}_{(aq)} + Sn^{2+}_{(aq)}$ is ............ $V$.
${E^0}_{Fe^{3+}/Fe^{2+}} = + 0.77 \ V$
${E^0}_{Sn^{2+}/Sn} = - 0.14 \ V$
Under standard conditions,the potential for the reaction $Sn_{(s)} + 2Fe^{3+}_{(aq)} \to 2Fe^{2+}_{(aq)} + Sn^{2+}_{(aq)}$ is ............ $V$.
MediumAIEEE 2004
View SolutionGiven below are half-cell reactions:
$MnO_{4}^{-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2+} + 4H_{2}O$,
$E^{o}_{MnO_{4}^{-} / Mn^{2+}} = +1.510 \, V$
$\frac{1}{2} O_{2} + 2H^{+} + 2e^{-} \rightarrow H_{2}O$,
$E^{o}_{O_{2} / H_{2}O} = +1.223 \, V$
Will the permanganate ion,$MnO_{4}^{-}$,liberate $O_{2}$ from water in the presence of an acid?
$MnO_{4}^{-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2+} + 4H_{2}O$,
$E^{o}_{MnO_{4}^{-} / Mn^{2+}} = +1.510 \, V$
$\frac{1}{2} O_{2} + 2H^{+} + 2e^{-} \rightarrow H_{2}O$,
$E^{o}_{O_{2} / H_{2}O} = +1.223 \, V$
Will the permanganate ion,$MnO_{4}^{-}$,liberate $O_{2}$ from water in the presence of an acid?
MediumNEET 2022
View SolutionBased on the given data,which is the strongest oxidizing agent? $E_{Li^+|Li}^o = -3.03 \ V$,$E_{Ba^{2+}|Ba}^o = -2.73 \ V$,$E_{Na^+|Na}^o = -2.71 \ V$,and $E_{Mg^{2+}|Mg}^o = -2.37 \ V$.
Easy
View SolutionGiven the standard electrode potentials:
$K^{+} / K = -2.93 \, V$,$Ag^{+} / Ag = 0.80 \, V$
$Hg^{2+} / Hg = 0.79 \, V$
$Mg^{2+} / Mg = -2.37 \, V$,$Cr^{3+} / Cr = -0.74 \, V$
Arrange these metals in their increasing order of reducing power.
$K^{+} / K = -2.93 \, V$,$Ag^{+} / Ag = 0.80 \, V$
$Hg^{2+} / Hg = 0.79 \, V$
$Mg^{2+} / Mg = -2.37 \, V$,$Cr^{3+} / Cr = -0.74 \, V$
Arrange these metals in their increasing order of reducing power.
Medium
View SolutionIn a cell,the following reactions take place:
$Fe^{2+} \rightarrow Fe^{3+} + e^{-}$ $\quad$ $E^{\circ}_{Fe^{3+} / Fe^{2+}} = 0.77 \, V$
$2I^{-} \rightarrow I_{2} + 2e^{-}$ $\quad$ $E^{\circ}_{I_{2} / I^{-}} = 0.54 \, V$
The standard electrode potential for the spontaneous reaction in the cell is $x \times 10^{-2} \, V$ at $298 \, K$. The value of $x$ is .... (Nearest Integer)
$Fe^{2+} \rightarrow Fe^{3+} + e^{-}$ $\quad$ $E^{\circ}_{Fe^{3+} / Fe^{2+}} = 0.77 \, V$
$2I^{-} \rightarrow I_{2} + 2e^{-}$ $\quad$ $E^{\circ}_{I_{2} / I^{-}} = 0.54 \, V$
The standard electrode potential for the spontaneous reaction in the cell is $x \times 10^{-2} \, V$ at $298 \, K$. The value of $x$ is .... (Nearest Integer)
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