The standard reduction potentials at $298 \ K$ for the following half-cell reactions are given below:
$Zn^{2+}_{(aq)} + 2e^{-} \rightleftharpoons Zn_{(s)} : -0.762 \ V$
$Cr^{3+}_{(aq)} + 3e^{-} \rightleftharpoons Cr_{(s)} : -0.74 \ V$
$2H^{+}_{(aq)} + 2e^{-} \rightleftharpoons H_{2(g)} : 0.00 \ V$
$Fe^{3+}_{(aq)} + e^{-} \rightleftharpoons Fe^{2+}_{(aq)} : 0.77 \ V$
Which one of the following is the strongest reducing agent?
$Zn^{2+}_{(aq)} + 2e^{-} \rightleftharpoons Zn_{(s)} : -0.762 \ V$
$Cr^{3+}_{(aq)} + 3e^{-} \rightleftharpoons Cr_{(s)} : -0.74 \ V$
$2H^{+}_{(aq)} + 2e^{-} \rightleftharpoons H_{2(g)} : 0.00 \ V$
$Fe^{3+}_{(aq)} + e^{-} \rightleftharpoons Fe^{2+}_{(aq)} : 0.77 \ V$
Which one of the following is the strongest reducing agent?
- A$Zn_{(s)}$
- B$Cr_{(s)}$
- C$H_{2(g)}$
- D$Fe^{2+}_{(aq)}$
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The oxidation potentials of $Zn, Cu, Ag, H_2$ and $Ni$ are $0.76, -0.34, -0.80, 0$ and $0.25 \ V$ respectively. Which reaction provides the maximum voltage?
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View SolutionOn the basis of the following electrode potentials,which one is the strongest reducing agent?
$E^0_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \text{ V}$,$E^0_{MnO_4^-/Mn^{2+}} = 1.51 \text{ V}$,$E^0_{Br_2/Br^{-}} = 1.09 \text{ V}$,$E^0_{Zn^{2+}/Zn} = -0.76 \text{ V}$
$E^0_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \text{ V}$,$E^0_{MnO_4^-/Mn^{2+}} = 1.51 \text{ V}$,$E^0_{Br_2/Br^{-}} = 1.09 \text{ V}$,$E^0_{Zn^{2+}/Zn} = -0.76 \text{ V}$
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 ............ $V$.
MediumIIT 1988
View SolutionIf 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?
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View SolutionThe standard Gibbs energy change for the Daniel cell reaction is $Zn_{(s)} + Cu^{2+}_{(aq)} \longrightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$ where $E_{\text{cell}}^{\circ} = 1.1 \ V$.
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