Standard electrode potential for the cell with cell reaction $Zn_{(s)} + Cu^{2+}_{(aq)} \rightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$ is $1.1 \ V$. Calculate the standard Gibbs energy change for the cell reaction. (Given $F = 96487 \ C \ mol^{-1}$)
- A$-200.27 \ kJ \ mol^{-1}$
- B$-212.27 \ kJ \ mol^{-1}$
- C$-212.27 \ J \ mol^{-1}$
- D$-200.27 \ J \ mol^{-1}$
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Which of the following reactions exhibits the minimum standard reduction potential?
If the standard reduction potentials for four divalent elements $X, Y, Z,$ and $W$ are $-1.46 \ V, -0.36 \ V, 0.15 \ V,$ and $-1.24 \ V$ respectively,then:
Medium
View SolutionFor 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?
DifficultJEE MAIN 2019
View SolutionGiven $E^{\circ} Mn^{7+} / Mn^{2+} = 1.51 \ V$ and $E^{\circ} Mn^{4+} / Mn^{2+} = 1.23 \ V$. Calculate $E^{\circ} Mn^{7+} / Mn^{4+}$. (in $V$)
For $Mg^{2+} + 2e^{-} \rightarrow Mg_{(s)}$,$E^{0} = -2.37 \, V$ and for $Cu^{2+} + 2e^{-} \rightarrow Cu_{(s)}$,$E^{0} = +0.33 \, V$. The $emf$ of the cell $(E^{0}_{Cell})$ is .......... $V$.
Medium
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