Question

If ${{E}^{0}}$ ($Z{{n}^{+2}}$, Zn) = - 0.7363V and ${{E}^{0}}$ ($F{{e}^{+2}}$,Fe) = 0.44V , then the emf of the cell ,
\[Zn|Z{{n}^{+2}}(a=0.001)||F{{e}^{+2}}(a=0.05)|Fe\] is:
(a) equal to 0.323V
(b) less than 0.323V
(C) Greater than 0.323V
(D) equal to 1.103V

Answer 1

Hint: To solve this question we should be aware of EMF and the formula to calculate the E.M.F of the given cell. Firstly, writing the given part may help us to not get confused thus, lead us to the correct answer.

Complete step by step solution:
The maximum difference or net voltage between two electrodes is known as the electromotive force of a cell. In case of half-cell reaction.
E is representation of E.M.F
The formula to calculate the E.M.F is:
\[E={{E}^{0}}+\dfrac{2.303RT}{nF}\log \dfrac{[product]}{[Reactant]}\]
Where, ${{E}^{0}}$= standard electrode potential
R = Gas constant (8.314 J${{K}^{=1}}$ $mo{{l}^{-1}}$)
T = Temperature in kelvin (298K)
n = number of electrons
F = Faraday's constant (96487 C $mo{{l}^{-1}}$)
The potential difference between the electrode and the electrolyte is known as the potential of the electrode.
When all the substance or electrons involved in the reaction in a standard state where the concentration is 1M , pressure is 1atm for gas and solids as well as pure liquids formed at 298K, the measure of potential in these reaction is standard potential electrode (${{E}^{0}}$).
Given, ${{E}^{0}}$ ($Z{{n}^{+2}}$, Zn) = - 0.7363V
${{E}^{0}}$ ($F{{e}^{+2}}$,Fe) = 0.44V
[$Z{{n}^{+2}}$] = 0.001M
[$F{{e}^{+2}}$] = 0.05M
Cell reactions are as follows:
R.H.S electrode: $F{{e}^{+2}}+2{{e}^{-}}\to Fe$$$ $$ (reduction)
L.H.S electrode: $Zn\to Z{{n}^{+2}}+2{{e}^{-}}$$$ $$ (Oxidation)
Now, let's calculate the emf of given cell, Reactant
\[E={E}_{right{electrode}} - {E}_{left{\text{electrode}}}\]
Firstly, let's calculate \[{{E}_{right\text{ }electrode}}\]
\[{{E}_{right\text{ }electrode}}={{E}^{0}}+\dfrac{2.303\times 8.314\times 298}{2\times 96487}\ln [F{{e}^{+2}}]\]
\[{{E}_{right\text{ }electrode}}={{E}^{0}}+\dfrac{0.0591}{2}\ln [F{{e}^{+2}}]\]
\[{{E}_{right\text{ }electrode}}=0.44+\dfrac{0.0591}{2}\ln [0.05]\]
\[{{E}_{right\text{ }electrode}}=0.44+0.02955\times \ln [0.05]\]
\[{{E}_{right\text{ }electrode}}\] = 0.3514V
Let's calculate \[{{E}_{\text{left }electrode}}\]
\[{{E}_{\text{left }electrode}}={{E}^{0}}+\dfrac{2.303\times 8.314\times 298}{2\times 96487}\ln [Z{{n}^{+2}}]\]
 \[{{E}_{\text{left }electrode}}=(-0.7363)+0.02955\ln [0.001]\]
\[{{E}_{\text{left }electrode}}\] = -0.8723
\[E={{E}_{right\text{ }electrode}}-{{E}_{left\text{ }electrode}}\]
= 0.3514 - (-0.8723)
= 1.2237V
Thus, the E.M.F of the cell is 1.2237V.

Option C is the correct answer.

Note: Be careful with the sign while calculating. Option D is equal to 1.103V, so only we can choose that option when our answer is exactly equal to 1.103V but the answer we arrived at is 1.2237V. Hence, option C is the correct answer. The E.M.F of the cell is positive so the above equation is spontaneous.