Question

The standard reduction potential data at ${25}^{o}C$ is given below:
$E^o(F{e}^{3+},F{e}^{2+})=+0.77V$;
$E^o(F{e}^{2+},Fe)=-0.44V$;
$E^o(C{u}^{2+},Cu)=+0.34V$;
$E^o(C{u}^{+},Cu)=+0.52V$;
$E^o\left(O_2\right(g)+4H^{+}+4e^{-}\to 2H_{2}O)=+1.23V$;
$E^o\left(O_2\right(g)+2H_{2}O+4e^{-}\to 4O{H}^{-})=+0.4V$;
$E^o(C{r}^{3+},Cr)=-0.74V$;
$E^o(C{r}^{2+},Cr)=-0.91V$;
Which of the following statements is/are correct on the basis of above data:

• A. $O_2$ is a better oxidising agent in acidic medium
• B. $C{r}^{2+}$ shows disproportionation to $Cr$ and $C{r}^{3+}$ in water
• C. $O_2$ oxidises $F{e}^{2+}$ to $F{e}^{3+}$ in acidic medium
• D. $C{u}^{+}$ oxidises $Fe$ to $F{e}^{2+}$ and $H_{2}O$ to $O_2$ in acidic medium

Answer 1

Answer: (A), (C)

The correct options are
A $O_2$ is a better oxidising agent in acidic medium
C $O_2$ oxidises $F{e}^{2+}$ to $F{e}^{3+}$ in acidic medium

a) As $E_{O_2/H_{2}O,H^{+}}^o>E_{O_2/O{H}^{-}}^o$
So, $O_2$ is a better oxidising agent in acidic medium.

b) $C{r}^{2+}\to 2C{r}^{3+}+Cr$


$x\times 1+2(-0.91)=-0.74\times 3$
$x=-2.22+1.82=-0.4V$
$E_{cell}^o=E_{C{r}^{2+}/Cr}^o-E_{C{r}^{3+}/C{r}^{2+}}^o$
$E_{cell}^o=-0.91-(-0.4)$
$E_{cell}^o=-0.51V$
As this is negative, so reaction is not spontaneous.

c) As standard reduction potential of $O_2$ in acidic medium is more than that of $F{e}^{3+}$ so, $O_2$ oxidises $F{e}^{2+}$ to $F{e}^{3+}$.

d) As standard reduction potential of $C{u}^{+}/Cu$ is more than that of $F{e}^{2+}/Fe$ so, $C{u}^{+}$ oxidises $Fe$.
But standard reduction potential of $C{u}^{+}/Cu$ is less than that of $O_2/H_{2}O$ in an acidic medium, so $C{u}^{+}$ does not oxidise $H_{2}O$ to $O_2$ in an acidic medium.