JEE MAIN - Chemistry (2017 (Offline) - No. 20)
Given
$$E_{C{l_2}/C{l^ - }}^o$$ = 1.36 V, $$E_{C{r^{3 + }}/Cr}^o$$ = - 0.74 V
$$E_{C{r_2}{O_7}^{2 - }/C{r^{3 + }}}^o$$ = 1.33 V, $$E_{Mn{O_4}^ - /Mn ^{2+}}^o$$ = 1.51 V
Among the following, the strongest reducing agent is :
$$E_{C{l_2}/C{l^ - }}^o$$ = 1.36 V, $$E_{C{r^{3 + }}/Cr}^o$$ = - 0.74 V
$$E_{C{r_2}{O_7}^{2 - }/C{r^{3 + }}}^o$$ = 1.33 V, $$E_{Mn{O_4}^ - /Mn ^{2+}}^o$$ = 1.51 V
Among the following, the strongest reducing agent is :
Mn2+
Cr3+
Cl–
Cr
Explanation
$$E_{C{l_2}/C{l^ - }}^o$$ = 1.36 V, $$E_{C{r^{3 + }}/Cr}^o$$ = - 0.74 V
$$E_{C{r_2}{O_7}^{2 - }/C{r^{3 + }}}^o$$ = 1.33 V, $$E_{Mn{O_4}^ - /Mn ^{2+}}^o$$ = 1.51 V
More negative the E° value of the species, more stronger is the reducing agent. Since Cr3+ is having least reducing potential, so Cr would be strongest reducing agent.
$$E_{C{r_2}{O_7}^{2 - }/C{r^{3 + }}}^o$$ = 1.33 V, $$E_{Mn{O_4}^ - /Mn ^{2+}}^o$$ = 1.51 V
More negative the E° value of the species, more stronger is the reducing agent. Since Cr3+ is having least reducing potential, so Cr would be strongest reducing agent.
Comments (0)
