Question

Write the Nernst equation and emf of the following cells at 298K:
(A) \[Mg(s)|M{{g}^{2+}}(0.001M)||C{{u}^{+2}}(0.0001M)|Cu(s)\]
(B) $Fe(s)|F{{e}^{2+}}(0.001M)||{{H}^{+}}(1M)|{{H}_{2}}(g)(bar)|Pt(s)$
(C) $Sn(s)|S{{n}^{2+}}(0.50M)||{{H}^{+}}(0.020M)|{{H}_{2}}(g)(1bar)|Pt(s)$
(D) $Pt(s)|B{{r}^{-}}(0.010M)|B{{r}_{2}}(l)||{{H}^{+}}(0.030M)|{{H}_{2}}(g)(1bar)|Pt(s)$

Answer 1

Hint: EMF can be calculated using Nernst Equation. Nernst Equation is as follows:
$\text{E=}{{\text{E}}^{\text{O}}}\text{-}\dfrac{\text{0}\text{.0591}}{\text{n}}\text{lo}{{\text{g}}_{\text{10}}}\dfrac{\text{ }\!\![\!\!\text{ Product }\!\!]\!\!\text{ }}{\text{ }\!\![\!\!\text{ Reactant }\!\!]\!\!\text{ }}$
Number of electrons involved are denoted by n. Concentration of reactant and products their power raised to stoichiometric coefficient in balanced redox reaction must be written.

Complete step by step solution:
In the Nernst Equation, Number of electrons involved is denoted by n. The concentration of reactant and products their power raised to the stoichiometric coefficient in a balanced redox reaction must be written.
(A) For \[Mg(s)|M{{g}^{2+}}(0.001M)||C{{u}^{+2}}(0.0001M)|Cu(s)\],
Anode reaction is written on the left side and cathode reaction is written on the right side,
So magnesium undergoes oxidation and loses two electrons. Copper undergoes reduction by gaining two electrons to form copper, so n for the reaction is 2.
So Nernst equation can be written as:
$E={{E}^{o}}_{cell}-\dfrac{0.0591}{n}\log \dfrac{[M{{g}^{2+}}]}{[C{{u}^{2+}}]}$
Standard cell potential is calculated as follows:
${{E}^{o}}_{Cell}={{E}^{o}}_{cathode}-{{E}^{O}}_{anode}$
$=0.34-(-2.36)$
$=2.7V$
Emf of cell can be calculated as follows:
\[{{E}_{cell}}=2.7-\dfrac{0.0591}{2}\log \dfrac{0.001}{0.0001}\]
$=2.7-0.02955$
$=2.67V$
(B) For, \[Fe(s)|F{{e}^{2+}}(0.001M)||{{H}^{+}}(1M)|{{H}_{2}}(g)(bar)|Pt(s)\]
Nernst Equation can be written as:
${{E}_{cell}}={{E}^{O}}-\dfrac{0.0591}{n}\log \dfrac{[F{{e}^{2+}}]}{{{[{{H}^{+}}]}^{2}}}$
\[{{E}^{O}}\]for cell is 0.44V, so emf can be calculated as follows:
\[E=0.44V-\dfrac{0.0591}{2}\log \dfrac{0.001}{{{1}^{2}}}\]
$=0.52865V$
(C) For, \[Sn(s)|S{{n}^{2+}}(0.50M)||{{H}^{+}}(0.020M)|{{H}_{2}}(g)(1bar)|Pt(s)\]
Nernst Equation can be written as follows:

$E={{E}^{O}}-\dfrac{0.0591}{n}\log \dfrac{[S{{n}^{2+}}]}{{{[{{H}^{+}}]}^{2}}}$
\[{{E}^{O}}\] for the standard cell is 0.14V, so emf can be calculated as follows:
$E={{E}^{O}}-\dfrac{0.0591}{2}\log \dfrac{0.050}{{{(0.020)}^{2}}}$
$=0.14-0.062$
$= 0.078V$
(D) For, $Pt(s)|B{{r}^{-}}(0.010M)|B{{r}_{2}}(l)||{{H}^{+}}(0.030M)|{{H}_{2}}(g)(1bar)|Pt(s)$
So Nernst Equation can be written as follows:
\[E={{E}^{O}}-\dfrac{0.0591}{2}\log \dfrac{1}{{{[B{{r}^{-}}]}^{2}}{{[{{H}^{+}}]}^{2}}}\]
\[{{E}^{O}}\]for cells is -1.09V.
So, emf of cell can be calculated as follows:
${{E}_{cell}}={{E}^{o}}_{cell}-\dfrac{0.0591}{2}\log \dfrac{1}{{{(0.010)}^{2}}{{(0.030)}^{2}}}$
$= -1.09V- 0.02955 (0.0453+7)$
$= -1.298V$

Note: Nernst Equation can be used to calculate the emf of a cell. Standard cell potential can be calculated by taking standard cell potential of cathode and anode respectively. In the Nernst equation, n indicates the number of electrons involved in the reaction. The concentration of reactant and products their power raised to the stoichiometric coefficient in a balanced redox reaction must be written.