JEE MAIN - Chemistry (2013 (Offline) - No. 15)
Given
$$E_{C{r^{2 + }}/Cr}^o$$ = -0.74 V; $$E_{MnO_4^ - /M{n^{2 + }}}^o$$ = 1.51 V
$$E_{C{r_2}O_7^{2 - }/C{r^{3 + }}}^o$$ = 1.33 V; $$E_{Cl/C{l^ - }}^o$$ = 1.36 V
Based on the data given above, strongest oxidising agent will be :
$$E_{C{r^{2 + }}/Cr}^o$$ = -0.74 V; $$E_{MnO_4^ - /M{n^{2 + }}}^o$$ = 1.51 V
$$E_{C{r_2}O_7^{2 - }/C{r^{3 + }}}^o$$ = 1.33 V; $$E_{Cl/C{l^ - }}^o$$ = 1.36 V
Based on the data given above, strongest oxidising agent will be :
Cr3+
Mn2+
$$MnO_4^ - $$
Cl-
Penjelasan
In electrochemistry, the strongest oxidizing agent will be the one with the highest standard electrode potential (E°), because a higher E° value means a greater tendency to gain electrons, i.e., get reduced. An oxidizing agent gains electrons and in doing so, oxidizes another species.
From the provided data, the species with the highest standard electrode potential (E°) is MnO₄⁻, with an E° of 1.51 V. This means that MnO₄⁻ has the greatest tendency to gain electrons and thus is the strongest oxidizing agent.
Therefore, the correct answer is :
Option C : $$MnO_4^ - $$ .