Calculate $E_{\text{cell}}^{\circ}$ for the reaction: $Mg_{(s)} + 2 Ag_{(aq)}^{+} \rightarrow Mg_{(aq)}^{2+} + 2 Ag_{(s)}$,given that $E_{Ag^{+}/Ag}^{\circ} = 0.8 \ V$ and $E_{Mg^{2+}/Mg}^{\circ} = -2.37 \ V$. (in $V$)
- A$-3.17$
- B$3.17$
- C$-1.57$
- D$1.57$
Explore More
Similar Questions
Assertion : Copper reacts with $HCl$ and liberates hydrogen.
Reason : Hydrogen is present above $Cu$ in the reactivity series.
Reason : Hydrogen is present above $Cu$ in the reactivity series.
DifficultAIIMS 2000
View SolutionIf $E^0_{Fe^{2+} / Fe} = -0.441 \ V$ and $E^0_{Fe^{3+} / Fe^{2+}} = 0.771 \ V$,the standard emf of the cell reaction $Fe_{(s)} + 2 Fe^{3+}_{(aq)} \longrightarrow 3 Fe^{2+}_{(aq)}$ is
$A$ standard hydrogen electrode has zero electrode potential because
MediumIIT 1997
View SolutionFor the following cell reaction,$Ag | Ag^{+} | AgCl | Cl^{-} | Cl_2, Pt$
$\Delta G_f^{\circ}(AgCl) = -109 \ kJ/mol$
$\Delta G_f^{\circ}(Cl^{-}) = -129 \ kJ/mol$
$\Delta G_f^{\circ}(Ag^{+}) = 78 \ kJ/mol$
$E^{\circ}$ of the cell is
$\Delta G_f^{\circ}(AgCl) = -109 \ kJ/mol$
$\Delta G_f^{\circ}(Cl^{-}) = -129 \ kJ/mol$
$\Delta G_f^{\circ}(Ag^{+}) = 78 \ kJ/mol$
$E^{\circ}$ of the cell is
DifficultAP EAMCET 2009
View SolutionFor four metals $A, B, C$ and $D$,the standard reduction potentials are $-3.05 \ V, -1.66 \ V, -0.40 \ V$ and $0.80 \ V$ respectively. Which metal exhibits the highest chemical reactivity?
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo