I_2 + 2e^{-} to 2I^{-}; E^{o} = 0.54 VCl_2 + | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) substance with a higher reduction potential acts as a stronger oxidizing agent and can oxidize a substance with a lower reduction potential.Compar

Vedclass · Accepted Answer

Iodide ion is oxidised by chlorine

Vedclass · Answer

(B) substance with a higher reduction potential acts as a stronger oxidizing agent and can oxidize a substance with a lower reduction potential.Comparing the reduction potentials:$E^{o}(Cl_2/Cl^-) = 1.36 \ V > E^{o}(I_2/I^-) = 0.54 \ V$. Therefore,$Cl_2$ can oxidize $I^-$ to $I_2$.$E^{o}(Mn^{3+}/Mn^{2+}) = 1.50 \ V > E^{o}(Fe^{3+}/Fe^{2+}) = 0.77 \ V$. Therefore,$Mn^{3+}$ can oxidize $Fe^{2+}$.Option $B$ states that the iodide ion $(I^-)$ is oxidised by chlorine $(Cl_2)$,which is correct because $E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode} = 1.36 \ V - 0.54 \ V = 0.82 \ V > 0$.

$I_2 + 2e^{-} \to 2I^{-}$; $E^{o} = 0.54 \ V$
$Cl_2 + 2e^{-} \to 2Cl^{-}$; $E^{o} = 1.36 \ V$
$Mn^{3+} + e^{-} \to Mn^{2+}$; $E^{o} = 1.50 \ V$
$Fe^{3+} + e^{-} \to Fe^{2+}$; $E^{o} = 0.77 \ V$
Which of the following is a correct statement?

Similar Questions

Discuss the method to determine the cell potential of any cell when a standard hydrogen electrode is considered as the cathode with a suitable example.

$E^{\circ}$ of $Cu$ is $+0.34 \ V$ while that of $Zn$ is $-0.76 \ V$. Explain.

For the reaction $Cr_2O_7^{2-} + I^- \to I_2 + Cr^{3+}$,if $E_{cell}^o = 0.79 \ V$ and $E_{Cr_2O_7^{2-}}^o = 1.33 \ V$,then $E_{I_2}^o = $ ............. $V$.

Aluminium displaces hydrogen from dilute $HCl$ whereas silver does not. The e.m.f. of a cell prepared by combining $Al/Al^{3+}$ and $Ag/Ag^{+}$ is $2.46 \ V$. The reduction potential of silver electrode is $+0.80 \ V$. The reduction potential of aluminium electrode is $........... \ V$.

Given,for $Sn^{4+} / Sn^{2+}$,standard reduction potential is $0.15 \ V$ and for $Au^{3+} / Au$,standard reduction potential is $1.5 \ V$. For the reaction,$3 Sn^{2+} + 2 Au^{3+} \longrightarrow 3 Sn^{4+} + 2 Au$,the value of $E_{\text{cell}}^{\circ}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$I_2 + 2e^{-} \to 2I^{-}$; $E^{o} = 0.54 \ V$$Cl_2 + 2e^{-} \to 2Cl^{-}$; $E^{o} = 1.36 \ V$$Mn^{3+} + e^{-} \to Mn^{2+}$; $E^{o} = 1.50 \ V$$Fe^{3+} + e^{-} \to Fe^{2+}$; $E^{o} = 0.77 \ V$Which of the following is a correct statement?