Question

The Edison storage cell is represented as:
$Fe(s)|FeO(s)|KOH(aq.)|N{{i}_{2}}{{O}_{3}}(s)|Ni(s)$

The half-cell reactions are:
$N{{i}_{2}}{{O}_{3}}(s)+{{H}_{2}}O(l)+2{{e}^{-}}\to 2NiO(s)+2O{{H}^{-}};{{E}^{\circ }}=+0.40$
$FeO(s)+{{H}_{2}}O(l)+2{{e}^{-}}\to Fe(s)+2O{{H}^{-}};{{E}^{\circ }}=-0.87$
(a)- What is the cell reaction?
(b)- What is the emf of the cell? How does it depend on the concentration of $KOH$?
(c)- What is the maximum amount of energy that can be obtained from one mole of $N{{i}_{2}}{{O}_{3}}$?

Answer 1

Hint: The emf of the cell can be calculated by the formula ${{E}_{cell}}=E_{cell}^{\circ }-\dfrac{0.0591}{n}\log \dfrac{[\text{products}]}{[\text{reactants }\!\!]\!\!\text{ }}$, where n is the number of electrons involved in the cell reaction. The maximum amount of electrical energy can be calculated by the formula $G=nF{{E}^{\circ }}$ , where n is the number of electrons, F is the faraday’s constant whose value is 96500 C, and ${{E}^{\circ }}$ is the emf of the cell reaction.

Complete Solution :
(a)- The cell is represented as: $Fe(s)|FeO(s)|KOH(aq.)|N{{i}_{2}}{{O}_{3}}(s)|Ni(s)$
So, the Fe will be the anode or the oxidation of iron takes place and $N{{i}_{2}}{{O}_{3}}$ will be the cathode or reduction of $N{{i}_{2}}{{O}_{3}}$ takes place.
So, the actual cell reaction will be:
$Fe+2O{{H}^{-}}+FeO+{{H}_{2}}O+2{{e}^{-}}(Anode)$
$N{{i}_{2}}{{O}_{3}}+{{H}_{2}}O+2{{e}^{-}}\to 2NiO+2O{{H}^{-}}(Cathode)$
Hence, the cell reaction will be:
$Fe+N{{i}_{2}}{{O}_{3}}\to FeO+2NiO$

(b)- The emf of the cell can be calculated by the formula ${{E}_{cell}}=E_{cell}^{\circ }-\dfrac{0.0591}{n}\log \dfrac{[\text{products}]}{[\text{reactants }\!\!]\!\!\text{ }}$, where n is the number of electrons in the reaction. For the reaction:
$Fe+N{{i}_{2}}{{O}_{3}}\to FeO+2NiO$

The formula will be:
${{E}_{cell}}=E_{cell}^{\circ }-\dfrac{0.0591}{2}\log \dfrac{{{[NiO]}^{2}}[FeO]}{[Fe][N{{i}_{2}}{{O}_{3}}]}$
But all the compounds are in solid form, so, the value of $\dfrac{{{[NiO]}^{2}}[FeO]}{[Fe][N{{i}_{2}}{{O}_{3}}]}$ will be 1.
So,
${{E}_{cell}}=E_{cell}^{\circ }-\dfrac{0.0591}{2}\log \dfrac{{{[NiO]}^{2}}[FeO]}{[Fe][N{{i}_{2}}{{O}_{3}}]}=E_{cell}^{\circ }$
$E_{cell}^{\circ }=0.87+0.40=1.27\text{ V}$
Therefore, the emf of the cell will be 1.27 V. As you can see that in the formula the concentration of KOH is not used, so the emf of the cell is independent of the concentration of $KOH$.

(c)- The maximum amount of electrical energy can be calculated by the formula $G=nF{{E}^{\circ }}$, where n is the number of electrons, F is the faraday's constant whose value is 96500 C, and ${{E}^{\circ }}$ is the emf of the cell reaction.
$G=nF{{E}^{\circ }}$
$G=2\text{ x 96500 x 1}\text{.27}$
$G=245.11\text{ kJ}$
So, the maximum energy will be 245.11 kJ.

Note: The $E_{cell}^{\circ }$ of the reaction can be calculated by the formula $E_{cell}^{\circ }={{E}_{cathode}}-{{E}_{anode}}$, where ${{E}_{cathode}}$ is the emf of the reaction taking place at the cathode and ${{E}_{anode}}$ is the emf of the reaction taking place at the anode.