Question

The standard reduction potential data at ${25}^{\circ }C$ is given below.

$E^{\circ }\left(F{e}^{3+}/F{e}^{2+}\right)=0.77V;E^{\circ }\left(F{e}^{2+}/Fe\right)=-0.44V;E^{\circ }\left(C{u}^{2+}/Cu\right)=0.34V;E^{\circ }\left(C{u}^{+}/Cu\right)=+0.52V;E^{\circ }\left(O_2\right(g)+4H^{+}+4e^{-})\to 2H_{2}O=+1.23V;E^{\circ }\left(O_2\right(g)+2H_{2}O+4e^{-})\to 4O{H}^{-}=+0.40V{E}^{\circ }\left(C{r}^{3+}/Cr\right)=-0.74V;E^{\circ }\left(C{r}^{2+}/Cr\right)=+0.91V$

Match $E^{\circ }$ of the rebox pair in Column I with the values given in Column II and select the correct answer using the code given below the lists.

$\begin{array}{cc}ColumnI& ColumnII\\ P.E^{\circ }\left(F{e}^{3+}/Fe\right)& 1.-0.18V\\ Q.E^{\circ }(4H_{2}O\rightleftharpoons 4H^{+}+4O{H}^{-})& 2.-0.4V\\ R.E^{\circ }(C{u}^{2+}+Cu\to 2C{u}^{+})& 3.-0.04V\\ S.E^{\circ }\left(C{r}^{3+}/C{r}^{2+}\right)& 4.-0.83V\end{array}$

• A. P - (4) Q - (1) R - (2) S - (3)
• B. P - (2) Q - (3) R - (4) S - (1)
• C. P - (1) Q - (2) R - (3) S - (4)
• D. P - (3) Q - (4) R - (1) S - (2)

Answer 1

The correct option is D

P - (3) Q - (4) R - (1) S - (2)

PLAN When different number of electrons are involved in a redox reaction.
$\Delta G_{net}^{\circ }=\Delta G_1^{\circ }+\Delta G_2^{\circ }-n_{3}F{E}_3^{\circ }=-n_{1}F{E}_1^{\circ }-n_{2}F{E}_2^{\circ }\therefore E_3^{\circ }=\frac{n_1{E}^{\circ }+n_2{E}_2^{\circ }}{n_3}$
$\left(P\right)E_3^{\circ }\left(F{e}^{3+}/Fe\right)$
Net reaction $F{e}^{3+}\to Fe$
is obtained from
$F{e}^{3+}+e^{-}\to F{e}^{2+}\underset{–}{F{e}^{2+}+2e^{-}\to Fe}\because \underset{–}{F{e}^{3+}+3e^{-}\to Fe}\underset{–}{n{E}^{\circ }}{n}_1=1E_1^o=0.77V{n}_2=2E_2^o=0.44V{n}_3=3E_3^o?$
$E_3^{\circ }=\frac{n_1{E}_1^{\circ }+n_2{E}_2^{\circ }}{n_3}=\frac{0.77+2(-0.44)}{3}=\frac{-0.11}{3}=-0.04V$
Thus, P - (3)
Net reaction
$4H_{2}O\rightleftharpoons 4H^{+}+4H^{+}+4O{H}^{-}$
is obtained from
$2H_{2}O\to O_2+4H^{+}+4e^{-}\underset{–}{2H_{2}O+O_2+4e^{-}\to O{H}^{-}}\underset{–}{4H_{2}O\to 4H^{+}+4e^{-}}$
$\underset{–}{n{E}^{\circ }}{n}_1=4-1.23V{n}_2=4+0.40V{n}_3=4?E_3^{\circ }=\frac{n_1{E}^{\circ }+n_2{E}_2^{\circ }}{n_3}=E_1^{\circ }+E_2^{\circ }=-1.23+0.40=-0.83V$
Thus, Q - (4)
$\left(R\right)C{u}^{2+}+Cu\to 2C{u}^{+}OxidationReduction$
$E^{\circ }ofC{u}^{2+}\to C{u}^{+}$ is also required
$C{u}^{2+}+2e^{-}\to Cu\underset{–}{Cu\to C{u}^{+}+e^{-}}\underset{–}{C{u}^{2+}+e^{-}\to C{u}^{+}}$
$\underset{–}{n{E}^{\circ }}20.34V1-0.52V{E}_3^{\circ }?E_3^{\circ }=\frac{n_1{E}_1^{\circ }+n_2{E}_2^{\circ }}{n_3}=\frac{2\times 0.34+1\times (-0.52)}{1}=0.16V$
Also,
$Cu\to C{u}^{+}+e^{-}\underset{–}{C{u}^{2+}+e^{-}\to C{u}^{+}}\underset{–}{n{E}^{\circ }}n=1,0.52V{n}_2=1-0.10VC{u}^{2+}+Cu\to 2C{u}^{+}{E}^{\circ }=-0.52+0.16=-0.36V$
Thus, (R) - (1)
$\left(S\right)C{r}^{3+}\to C{r}^{2+}$ is obtained from
$C{r}^{(}3+)+3e^{-}\to Cr\underset{–}{Cr\to C{r}^{(}2+)+2e^{-}}\underset{–}{C{r}^{3+}+e^{-}\to C{r}^{2+}}n{E}^{\circ }3-0.74V2+0.91V1?E_3^{\circ }=\frac{-0.74\times 3+2\times 0.91}{1}=-0.4V$
Thus, S = (2)
P - (3), (Q) - (4), R - (1), S - (2)