∆G = -nFE
Where:
∆G is the Gibbs free energy change (-827 kJ/mol).
n is the number of moles of electrons transferred.
F is the Faraday constant (96500 C/mol).
E is the cell potential.
\(\frac 43 Al+O_2→\frac 23 Al_2O_2\)
It's clear that \(\frac 43\) moles of electrons are transferred for every 1 mole of O2 consumed. So, n = \(\frac 43\)
Now, plug these values into the equation:
-827 kJ = -nFE
-827 kJ = -(\(\frac 43\)) . (96500 C/mol) . E
E = \(\frac {(-827\ kJ) }{[-(\frac 43) . (96500 \ C/mol)] }\)E ≈ 2.14 V
So, the minimum e.m.f. required to carry out the electrolysis of Al2O3 is approximately 2.14 V.
Therefore, the correct option is (A): 2.14 V.


Electricity is passed through an acidic solution of Cu$^{2+}$ till all the Cu$^{2+}$ was exhausted, leading to the deposition of 300 mg of Cu metal. However, a current of 600 mA was continued to pass through the same solution for another 28 minutes by keeping the total volume of the solution fixed at 200 mL. The total volume of oxygen evolved at STP during the entire process is ___ mL. (Nearest integer)
Given:
$\mathrm{Cu^{2+} + 2e^- \rightarrow Cu(s)}$
$\mathrm{O_2 + 4H^+ + 4e^- \rightarrow 2H_2O}$
Faraday constant = 96500 C mol$^{-1}$
Molar volume at STP = 22.4 L
What is Microalbuminuria ?
The output (Y) of the given logic implementation is similar to the output of an/a …………. gate.
Galvanic cells, also known as voltaic cells, are electrochemical cells in which spontaneous oxidation-reduction reactions produce electrical energy. It converts chemical energy to electrical energy.
It consists of two half cells and in each half cell, a suitable electrode is immersed. The two half cells are connected through a salt bridge. The need for the salt bridge is to keep the oxidation and reduction processes running simultaneously. Without it, the electrons liberated at the anode would get attracted to the cathode thereby stopping the reaction on the whole.