Consider the following:
$Zn^{2+} + 2e^- \longrightarrow Zn_{(s)} ; E^o = -0.76 \, V$
$Ca^{2+} + 2e^- \longrightarrow Ca_{(s)} ; E^o = -2.87 \, V$
$Mg^{2+} + 2e^- \longrightarrow Mg_{(s)} ; E^o = -2.36 \, V$
$Ni^{2+} + 2e^- \longrightarrow Ni_{(s)} ; E^o = -0.25 \, V$
The reducing power of the metals increases in the order:
$Zn^{2+} + 2e^- \longrightarrow Zn_{(s)} ; E^o = -0.76 \, V$
$Ca^{2+} + 2e^- \longrightarrow Ca_{(s)} ; E^o = -2.87 \, V$
$Mg^{2+} + 2e^- \longrightarrow Mg_{(s)} ; E^o = -2.36 \, V$
$Ni^{2+} + 2e^- \longrightarrow Ni_{(s)} ; E^o = -0.25 \, V$
The reducing power of the metals increases in the order:
- A$Ni < Zn < Mg < Ca$
- B$Ni < Zn < Ca < Mg$
- C$Zn < Mg < Ni < Ca$
- D$Ca < Mg < Zn < Ni$
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Similar Questions
The electrode potentials are given as follows:
$Fe_{(aq)}^{3+} + e^- \to Fe_{(aq)}^{2+}$; $E^o = 0.771 \, V$
$I_{2(s)} + 2e^- \to 2I_{(aq)}^-$; $E^o = 0.536 \, V$
For the cell reaction $2Fe_{(aq)}^{3+} + 2I_{(aq)}^- \to 2Fe_{(aq)}^{2+} + I_{2(s)}$,the value of $E^o_{cell}$ is:
$Fe_{(aq)}^{3+} + e^- \to Fe_{(aq)}^{2+}$; $E^o = 0.771 \, V$
$I_{2(s)} + 2e^- \to 2I_{(aq)}^-$; $E^o = 0.536 \, V$
For the cell reaction $2Fe_{(aq)}^{3+} + 2I_{(aq)}^- \to 2Fe_{(aq)}^{2+} + I_{2(s)}$,the value of $E^o_{cell}$ is:
Medium
View SolutionIf 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 Solution$(i)$ Copper metal dissolves in $1 \ M$ silver nitrate solution and crystals of silver metal get deposited.
$(ii)$ Silver metal does not react with $1 \ M$ zinc nitrate solution.
$(iii)$ Zinc metal dissolves in $1 \ M$ copper sulphate solution and copper metal gets deposited.
Hence,the order of decreasing strength of the three metals as reducing agents will be:
$(ii)$ Silver metal does not react with $1 \ M$ zinc nitrate solution.
$(iii)$ Zinc metal dissolves in $1 \ M$ copper sulphate solution and copper metal gets deposited.
Hence,the order of decreasing strength of the three metals as reducing agents will be:
Easy
View SolutionReview the $SRP$ (at $25\,\text{°C}$) data in acidic medium:
$Ti^{4+} + e^- \to Ti^{3+}, \, E^o = -x \text{ V}$
$Fe^{3+} + e^- \to Fe^{2+}, \, E^o = -y \text{ V}$
where $x < y$,point out the wrong statement.
$Ti^{4+} + e^- \to Ti^{3+}, \, E^o = -x \text{ V}$
$Fe^{3+} + e^- \to Fe^{2+}, \, E^o = -y \text{ V}$
where $x < y$,point out the wrong statement.
Medium
View Solution$A$ reaction,$Ni_{(s)} + Cu^{2+}_{(aq)} \rightarrow Ni^{2+}_{(aq)} + Cu_{(s)}$ occurs in a cell. Calculate $E^0_{cell}$ if $E^0_{Cu} = 0.337 \ V$ and $E^0_{Ni} = -0.257 \ V$. (in $V$)
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