Calculate the standard free energy change for the reaction $\frac{1}{2}Cu_{(s)} + \frac{1}{2}Cl_{2(g)} \rightleftharpoons \frac{1}{2}Cu^{2+} + Cl^-$ taking place at $25\ ^oC$ in a cell whose standard e.m.f. is $1.02 \ V$ (in $J$).
- A$-98430$
- B$98430$
- C$96500$
- D$-49215$
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If $Cu^{2+} + 2e^{-} \rightarrow Cu, E^{0} = 0.337 \ V$ and $Cu^{2+} + e^{-} \rightarrow Cu^{+}, E^{0} = 0.153 \ V$,then for the reaction $Cu^{+} + e^{-} \rightarrow Cu$,$E^{0}_{cell} =$ .............. $V$.
Medium
View SolutionThe standard Gibbs energy change for the Daniel cell reaction is $Zn_{(s)} + Cu^{2+}_{(aq)} \longrightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$ where $E_{\text{cell}}^{\circ} = 1.1 \ V$.
On the basis of the following electrode potentials,which one is the strongest reducing agent?
$E^0_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \text{ V}$,$E^0_{MnO_4^-/Mn^{2+}} = 1.51 \text{ V}$,$E^0_{Br_2/Br^{-}} = 1.09 \text{ V}$,$E^0_{Zn^{2+}/Zn} = -0.76 \text{ V}$
$E^0_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \text{ V}$,$E^0_{MnO_4^-/Mn^{2+}} = 1.51 \text{ V}$,$E^0_{Br_2/Br^{-}} = 1.09 \text{ V}$,$E^0_{Zn^{2+}/Zn} = -0.76 \text{ V}$
Consider the cell given below:
$Ag_{(s)} | Ag^{\oplus} || Cu^{2+} | Cu_{(s)}$
Given:
$Ag^{\oplus} + e^{-} \to Ag; E^{o} = x$
$Cu^{2+} + 2e^{-} \to Cu; E^{o} = y$
The value of $E^{o}_{cell}$ is:
$Ag_{(s)} | Ag^{\oplus} || Cu^{2+} | Cu_{(s)}$
Given:
$Ag^{\oplus} + e^{-} \to Ag; E^{o} = x$
$Cu^{2+} + 2e^{-} \to Cu; E^{o} = y$
The value of $E^{o}_{cell}$ is:
Medium
View SolutionWhat is the value of the standard electrode potential of a standard hydrogen electrode $(SHE)$ (in $V$)?
Easy
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