Chapter U5, Problem 18STP

Interpretation Introduction

Interpretation:

The correct option for the electrochemical cell needs to be determined for the following two half-reactions:

$\begin{array}{l}{\text{Li}}^{\text{+}}+{\text{ e}}^{\text{-}}\to \text{ Li - 3}\text{.05 V}\\ {\text{Zn}}^{\text{2+}}+{\text{ 2e}}^{\text{-}}\to \text{ Zn - 0}\text{.76 V}\end{array}$

Concept introduction:

For a chemical species, reduction potential is defined as the tendency to gain an electron and get reduced.

The potential of each redox couple under the standard state (concentration = 1 M, T = 298 K and P = 1 atm) for all regents is said to be the standard reduction potential, ${\text{E}}^{\text{o}}$.

Answer to Problem 18STP

The correct option is (c) as $\text{Zn}$ is reduced, $\text{Li}$ is oxidized and ${\text{E}}_{\text{cell}}\text{= }2.46\text{ V}$.

Explanation of Solution

The two half-reactions and their reduction potential values are:

$\begin{array}{l}{\text{Li}}^{\text{+}}+{\text{ e}}^{\text{-}}\to \text{ Li - 3}\text{.05 V}\\ {\text{Zn}}^{\text{2+}}+{\text{ 2e}}^{\text{-}}\to \text{ Zn - 0}\text{.76 V}\end{array}$

The higher value of reduction potential indicates that it has a greater tendency to gain electron and reduce then the other species which has lower reduction potential. Since, the reduction potential of zinc is greater than lithium so, the zinc will undergo reduction and lithium will undergo oxidationfor the given electrochemical cell.

The reduction reaction will take place at cathode and oxidation at anode. The half-cell reactions are:

$\begin{array}{l}\text{Anode: Li }\to {\text{ Li}}^{\text{+}}+{\text{ e}}^{\text{-}}\text{ + 3}\text{.05 V}\\ {\text{Cathode: Zn}}^{\text{2+}}+{\text{ 2e}}^{\text{-}}\to \text{ Zn - 0}\text{.76 V}\end{array}$

The cell potential is determined by difference between the reduction potential of two electrodes (cathode and anode). The formula is:

${\text{E}}_{\text{cell}}{\text{= E}}_{\text{cathode(reduction potential)}}{\text{- E}}_{\text{anode(reduction potential)}}$

The value of ${\text{ E}}_{\text{cathode(reduction potential)}}=\text{- 0}\text{.76 V}$ and ${\text{ E}}_{\text{anode(reduction potential)}}=\text{- 3}\text{.05 V}$. Substituting the values,

$\begin{array}{l}{\text{E}}_{\text{cell}}\text{= -0}\text{.76 V - (- 3}\text{.05 V)}\\ {\text{E}}_{\text{cell}}\text{= -0}\text{.76 V + 3}\text{.05 V}\\ {\text{E}}_{\text{cell}}\text{= }2.46\text{ V}\end{array}$

Hence, the correct option is (c) as $\text{Zn}$ is reduced, $\text{Li}$ is oxidized and ${\text{E}}_{\text{cell}}\text{= }2.46\text{ V}$.