Calculate the standard potential of a cell having the following electrode reactions:
$Cd_{(aq)}^{2+} + 2e^{-} \rightarrow Cd_{(s)}$ $E^{\circ} = -0.403 \ V$
$Zn_{(aq)}^{2+} + 2e^{-} \rightarrow Zn_{(s)}$ $E^{\circ} = -0.763 \ V$ (in $V$)
$Cd_{(aq)}^{2+} + 2e^{-} \rightarrow Cd_{(s)}$ $E^{\circ} = -0.403 \ V$
$Zn_{(aq)}^{2+} + 2e^{-} \rightarrow Zn_{(s)}$ $E^{\circ} = -0.763 \ V$ (in $V$)
- A$0.201$
- B$0.360$
- C$0.481$
- D$1.166$
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Similar Questions
The standard electrode potential for the Daniell cell is $1.1 \ V$. What will be the value of standard Gibbs energy for the reaction?
$Zn_{(s)} + Cu_{(aq)}^{2+} \rightarrow Zn_{(aq)}^{2+} + Cu_{(s)}$
$(1 \ F = 96487 \ C \ mol^{-1})$
$Zn_{(s)} + Cu_{(aq)}^{2+} \rightarrow Zn_{(aq)}^{2+} + Cu_{(s)}$
$(1 \ F = 96487 \ C \ mol^{-1})$
$I_2 + 2e^{-} \to 2I^{-}$; $E^{o} = 0.54 \ V$
$Cl_2 + 2e^{-} \to 2Cl^{-}$; $E^{o} = 1.36 \ V$
$Mn^{3+} + e^{-} \to Mn^{2+}$; $E^{o} = 1.50 \ V$
$Fe^{3+} + e^{-} \to Fe^{2+}$; $E^{o} = 0.77 \ V$
Which of the following is a correct statement?
$Cl_2 + 2e^{-} \to 2Cl^{-}$; $E^{o} = 1.36 \ V$
$Mn^{3+} + e^{-} \to Mn^{2+}$; $E^{o} = 1.50 \ V$
$Fe^{3+} + e^{-} \to Fe^{2+}$; $E^{o} = 0.77 \ V$
Which of the following is a correct statement?
Medium
View SolutionGiven:
$E^o_{Fe^{3+} /Fe} = -0.036 \ V, E^o_{Fe^{2+} /Fe} = -0.439 \ V$
The value of standard electrode potential for the change,
$Fe^{3+}_{(aq)} + e^- \rightarrow Fe^{2+}_{(aq)}$ will be ........ $V$.
$E^o_{Fe^{3+} /Fe} = -0.036 \ V, E^o_{Fe^{2+} /Fe} = -0.439 \ V$
The value of standard electrode potential for the change,
$Fe^{3+}_{(aq)} + e^- \rightarrow Fe^{2+}_{(aq)}$ will be ........ $V$.
Given,for $Sn^{4+} / Sn^{2+}$,standard reduction potential is $0.15 \ V$ and for $Au^{3+} / Au$,standard reduction potential is $1.5 \ V$. For the reaction,$3 Sn^{2+} + 2 Au^{3+} \longrightarrow 3 Sn^{4+} + 2 Au$,the value of $E_{\text{cell}}^{\circ}$ is:
For the following cell reaction,$Ag | Ag^{+} | AgCl | Cl^{-} | Cl_2, Pt$
$\Delta G_f^{\circ}(AgCl) = -109 \ kJ/mol$
$\Delta G_f^{\circ}(Cl^{-}) = -129 \ kJ/mol$
$\Delta G_f^{\circ}(Ag^{+}) = 78 \ kJ/mol$
$E^{\circ}$ of the cell is
$\Delta G_f^{\circ}(AgCl) = -109 \ kJ/mol$
$\Delta G_f^{\circ}(Cl^{-}) = -129 \ kJ/mol$
$\Delta G_f^{\circ}(Ag^{+}) = 78 \ kJ/mol$
$E^{\circ}$ of the cell is
DifficultAP EAMCET 2009
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