For a Galvanic cell consisting of a zinc electrode and a standard hydrogen electrode,$E^{\circ}(Zn^{+2}_{(aq)} \mid Zn_{(s)}) = -0.76 \ V$. Identify the reaction that takes place at the positive electrode during the working of the cell.
- A$Zn_{(s)} \longrightarrow Zn^{+2}_{(aq)} + 2e^{-}$
- B$Zn^{+2}_{(aq)} + 2e^{-} \longrightarrow Zn_{(s)}$
- C$H_{2(g)} \longrightarrow 2H^{+}_{(aq)} + 2e^{-}$
- D$2H^{+}_{(aq)} + 2e^{-} \longrightarrow H_{2(g)}$
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Electrode potential data are given below :
$Fe^{3+}_{(aq)} + e^- \to Fe^{2+}_{(aq)}; \, E^o = +0.77 \, V$
$Al^{3+}_{(aq)} + 3e^- \to Al_{(s)}; \, E^o = -1.66 \, V$
$Br_{2(aq)} + 2e^- \to 2Br^{-}_{(aq)}; \, E^o = +1.08 \, V$
Based on the data given above,the reducing power of $Fe^{2+}$,$Al$ and $Br^{-}$ will increase in the order:
$Fe^{3+}_{(aq)} + e^- \to Fe^{2+}_{(aq)}; \, E^o = +0.77 \, V$
$Al^{3+}_{(aq)} + 3e^- \to Al_{(s)}; \, E^o = -1.66 \, V$
$Br_{2(aq)} + 2e^- \to 2Br^{-}_{(aq)}; \, E^o = +1.08 \, V$
Based on the data given above,the reducing power of $Fe^{2+}$,$Al$ and $Br^{-}$ will increase in the order:
Easy
View SolutionCalculate 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 $V$)
Arrange the following metals in the order of their decreasing standard electrode potential: $Mg, K, Ba, Ca$.
Medium
View SolutionThe $E^{0}_{Red}$ values for $P, Q, R$ and $S$ are $-2.90 \, V, +0.34 \, V, +1.20 \, V$ and $-0.76 \, V$ respectively. The decreasing order of their reactivity is:
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View SolutionArrange the following metals in the order in which they displace each other from the solution of their salts: $Al, Cu, Fe, Mg$ and $Zn$.
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