The standard reduction potentials at $25^o C$ are given below. Which is the strongest reducing agent?
$Zn^{2+}_{(aq)} + 2e^{-} \rightleftharpoons Zn_{(s)}$,$E^o_{RP} = -0.762 \, V$
$Cr^{3+}_{(aq)} + 3e^{-} \rightleftharpoons Cr_{(s)}$,$E^o_{RP} = -0.740 \, V$
$2H^{+}_{(aq)} + 2e^{-} \rightleftharpoons H_{2(g)}$,$E^o_{RP} = 0.00 \, V$
$Fe^{3+}_{(aq)} + e^{-} \rightleftharpoons Fe^{2+}_{(aq)}$,$E^o_{RP} = 0.77 \, V$
$Zn^{2+}_{(aq)} + 2e^{-} \rightleftharpoons Zn_{(s)}$,$E^o_{RP} = -0.762 \, V$
$Cr^{3+}_{(aq)} + 3e^{-} \rightleftharpoons Cr_{(s)}$,$E^o_{RP} = -0.740 \, V$
$2H^{+}_{(aq)} + 2e^{-} \rightleftharpoons H_{2(g)}$,$E^o_{RP} = 0.00 \, V$
$Fe^{3+}_{(aq)} + e^{-} \rightleftharpoons Fe^{2+}_{(aq)}$,$E^o_{RP} = 0.77 \, V$
- A$Zn$
- B$Cr$
- C$H_{2(g)}$
- D$Fe^{2+}_{(aq)}$
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Similar Questions
Standard electrode potentials are given:
$A^{+}/A = -2.93 \ V$ $B^{+}/B = 0.80 \ V$
$C^{2+}/C = -2.37 \ V$ $D^{3+}/D = -0.74 \ V$
Increasing order of reducing power of these metals:
$A^{+}/A = -2.93 \ V$ $B^{+}/B = 0.80 \ V$
$C^{2+}/C = -2.37 \ V$ $D^{3+}/D = -0.74 \ V$
Increasing order of reducing power of these metals:
Medium
View SolutionThe standard reduction potentials of $Zn^{2+}|Zn$,$Cu^{2+}|Cu$ and $Ag^{+}|Ag$ are respectively $-0.76 \ V$,$0.34 \ V$ and $0.80 \ V$. The following cells were constructed:
$(1)$ $Zn|Zn^{2+}||Cu^{2+}|Cu$
$(2)$ $Zn|Zn^{2+}||Ag^{+}|Ag$
$(3)$ $Cu|Cu^{2+}||Ag^{+}|Ag$
What is the correct order of $E_{\text{cell}}^{\circ}$ of these cells?
$(1)$ $Zn|Zn^{2+}||Cu^{2+}|Cu$
$(2)$ $Zn|Zn^{2+}||Ag^{+}|Ag$
$(3)$ $Cu|Cu^{2+}||Ag^{+}|Ag$
What is the correct order of $E_{\text{cell}}^{\circ}$ of these cells?
The standard electrode potentials are: $K^{+}/K = -2.93 \ V$,$Ag^{+}/Ag = 0.80 \ V$,$Hg^{2+}/Hg = 0.79 \ V$,$Mg^{2+}/Mg = -2.37 \ V$,$Cr^{3+}/Cr = -0.74 \ V$. Arrange these metals in the increasing order of their reducing power.
Medium
View SolutionCalculate the cell potential in volts for the cell reaction given below:
$Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$
Given:
$Zn \rightarrow Zn^{2+} + 2e^{-}, E^{0} = 0.76 \ V$
$Fe \rightarrow Fe^{2+} + 2e^{-}, E^{0} = 0.41 \ V$ (in $V$)
$Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$
Given:
$Zn \rightarrow Zn^{2+} + 2e^{-}, E^{0} = 0.76 \ V$
$Fe \rightarrow Fe^{2+} + 2e^{-}, E^{0} = 0.41 \ V$ (in $V$)
Medium
View SolutionGiven: $E^{0}_{Mn^{+7} \mid Mn^{+2}} = 1.5 \ V$ and $E^{0}_{Mn^{+4} \mid Mn^{+2}} = 1.2 \ V$,then $E^{0}_{Mn^{+7} \mid Mn^{+4}}$ is (in $V$)
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