The standard electrode potential $(E^{\circ})$ values of $Al^{3+}/Al, Ag^{+}/Ag, K^{+}/K$ and $Cr^{3+}/Cr$ are $-1.66 \ V, 0.80 \ V, -2.93 \ V$ and $-0.74 \ V,$ respectively. The correct decreasing order of reducing power of the metal is:
- A$Ag > Cr > Al > K$
- B$K > Al > Cr > Ag$
- C$K > Al > Ag > Cr$
- D$Al > K > Ag > Cr$
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Similar Questions
Match the following:
List-$I$ List-$II$ $(A)$ Potential of hydrogen electrode at $pH = 10$ $(I)$ $0.76 \ V$ $(B)$ $Cu^{2+} | Cu$ $(II)$ $0.059$ $(C)$ $Zn | Zn^{2+}$ $(III)$ $-0.591 \ V$ $(D)$ $\frac{2.303 RT}{F}$ $(IV)$ $0.337 \ V$ $(V)$ $-0.76 \ V$
$A$ $B$ $C$ $D$
$(a)$ $(III)$ $(I)$ $(II)$ $(V)$
$(b)$ $(II)$ $(V)$ $(I)$ $(IV)$
$(c)$ $(III)$ $(IV)$ $(I)$ $(II)$
$(d)$ $(V)$ $(I)$ $(IV)$ $(II)$
| List-$I$ | List-$II$ |
|---|---|
| $(A)$ Potential of hydrogen electrode at $pH = 10$ | $(I)$ $0.76 \ V$ |
| $(B)$ $Cu^{2+} | Cu$ | $(II)$ $0.059$ |
| $(C)$ $Zn | Zn^{2+}$ | $(III)$ $-0.591 \ V$ |
| $(D)$ $\frac{2.303 RT}{F}$ | $(IV)$ $0.337 \ V$ |
| $(V)$ $-0.76 \ V$ |
$A$ $B$ $C$ $D$
$(a)$ $(III)$ $(I)$ $(II)$ $(V)$
$(b)$ $(II)$ $(V)$ $(I)$ $(IV)$
$(c)$ $(III)$ $(IV)$ $(I)$ $(II)$
$(d)$ $(V)$ $(I)$ $(IV)$ $(II)$
Consider the half-cell reduction reactions:
$Mn^{2+} + 2e^{-} \rightarrow Mn$,$E^{\circ} = -1.18 \ V$
$Mn^{3+} + e^{-} \rightarrow Mn^{2+}$,$E^{\circ} = +1.51 \ V$
The $E^{\circ}$ for the reaction $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$,and the possibility of the forward reaction are respectively:
$Mn^{2+} + 2e^{-} \rightarrow Mn$,$E^{\circ} = -1.18 \ V$
$Mn^{3+} + e^{-} \rightarrow Mn^{2+}$,$E^{\circ} = +1.51 \ V$
The $E^{\circ}$ for the reaction $3Mn^{2+} \rightarrow Mn + 2Mn^{3+}$,and the possibility of the forward reaction are respectively:
The standard electrode potential $(E^{\circ}_{M^{3+}/M^{2+}})$ for $V$,$Cr$,$Mn$,and $Co$ are $-0.26 \ V$,$-0.41 \ V$,$+1.57 \ V$,and $+1.97 \ V$,respectively. The metal ions which can liberate $H_2$ from a dilute acid are
DifficultJEE MAIN 2023
View SolutionWhat is the standard reduction potential of $Cu^{2+} \mid Cu_{(s)}$ if the $E^{\circ}$ of the following cell is $0.46 \ V$ (in $V$)?
$Cu_{(s)} \mid Cu_{(aq)}^{2+} \parallel Ag_{(aq)}^{+} \mid Ag_{(s)}$ ; $E_{Ag^{+} / Ag}^{\circ} = 0.80 \ V$
$Cu_{(s)} \mid Cu_{(aq)}^{2+} \parallel Ag_{(aq)}^{+} \mid Ag_{(s)}$ ; $E_{Ag^{+} / Ag}^{\circ} = 0.80 \ V$
Calculate the $emf$ of the cell $Cu_{(s)} | Cu^{2+}_{(aq)} || Ag^+_{(aq)} | Ag_{(s)}$. Given: $E^0_{Cu^{2+}/Cu} = +0.34 \ V$,$E^0_{Ag^+/Ag} = +0.80 \ V$.
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