Solveeit Logo
Login

Question

Question: \(E^{o}\) for \(Cl_{2}/Cl^{-} = + 1.36,I_{2}/I^{-} = + 0.53,Ag^{+}/Ag = + 0.79\), \(Na^{+}/Na = - ...

EoE^{o} for

Cl2/Cl−=+1.36,I2/I−=+0.53,Ag+/Ag=+0.79Cl_{2}/Cl^{-} = + 1.36,I_{2}/I^{-} = + 0.53,Ag^{+}/Ag = + 0.79,

Na+/Na=−2.71Na^{+}/Na = - 2.71 and Li+/Li=−3.04V.Li^{+}/Li = - 3.04V. What is the order of decreasing reducing strength for these species?

A

Li > Cl−Cl^{-}> Ag > I−I^{-}> Na

B

Li > Na > I−I^{-}> Ag > Cl−Cl^{-}

C

Cl−Cl^{-}> Ag I−I^{-}>Na > Li

D

Na > Li > Ag > Cl−Cl^{-} > I−I^{-}

Answer

Li > Na > I−I^{-}> Ag > Cl−Cl^{-}

Explanation

Solution

: The order is Li > Na > I−I^{-}> Ag > Cl−Cl^{-}