What is the observation we get when a gold ring is placed in the solution of $AgNO_3$? $[E_{Ag^{+}|Ag}^o = 0.80 \ V$ and $E_{Au^{3+}|Au}^o = 1.40 \ V]$
- A$A$. $Ag$ is deposited on the gold ring
- B$B$. Gold ring dissolves
- C$C$. $Au$ is deposited on $AgNO_3$ solution
- D$D$. No reaction occurs
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Similar Questions
From the given values,which one is a strong oxidizing agent?
$[Fe(CN)_6]^{4-} \to [Fe(CN)_6]^{3-} + e^-, E^o = -0.35 \ V$
$Fe^{2+} \to Fe^{3+} + e^-, E^o = -0.77 \ V$
$[Fe(CN)_6]^{4-} \to [Fe(CN)_6]^{3-} + e^-, E^o = -0.35 \ V$
$Fe^{2+} \to Fe^{3+} + e^-, E^o = -0.77 \ V$
Easy
View SolutionOn the basis of the following $E^o$ values,the strongest oxidizing agent is:
$[Fe(CN)_6]^{4-} \to [Fe(CN)_6]^{3-} + e^-; E^o = -0.35 \ V$
$Fe^{2+} \to Fe^{3+} + e^-; E^o = -0.77 \ V$
$[Fe(CN)_6]^{4-} \to [Fe(CN)_6]^{3-} + e^-; E^o = -0.35 \ V$
$Fe^{2+} \to Fe^{3+} + e^-; E^o = -0.77 \ V$
MediumAIPMT 2008
View SolutionThe $EMF$ of a cell in terms of the reduction potential of its left and right electrodes is:
MediumAIEEE 2002
View SolutionWhich one of the following metals cannot evolve $H_2$ from acids,$H_2O$,or from its compounds?
Easy
View SolutionThe $EMF$ of a cell whose half-cell reactions are given below is .......... $V$.
$Mg^{2+} + 2e^- \to Mg_{(s)}; E^o = -2.37 \ V$
$Cu^{2+} + 2e^- \to Cu_{(s)}; E^o = +0.34 \ V$ (in $V$)
$Mg^{2+} + 2e^- \to Mg_{(s)}; E^o = -2.37 \ V$
$Cu^{2+} + 2e^- \to Cu_{(s)}; E^o = +0.34 \ V$ (in $V$)
Medium
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