The relationship between charge (\( Q \)) and the number of electrons (\( n \)) is given by:
\(Q = ne\)
The total charge is also related to current and time:
\(I \times t = ne\)
To find the number of electrons, we can solve for \( n \):
\(n = \frac{I \times t}{e}\)
Substituting the given values (current = 1 A, time = 60 seconds, and the elementary charge \( e = 1.6 \times 10^{-19} \) C):
\(n = \frac{1 \times 60}{1.6 \times 10^{-19}}\)
Finally, calculating \( n \):
\(n = 3.75 \times 10^{20}\)
The number of electrons is \( 3.75 \times 10^{20} \).
If \( E^\circ_{Fe^{2+}/Fe} = -0.441 \, \text{V} \) and \( E^\circ_{Fe^{3+}/Fe^{2+}} = 0.771 \, \text{V} \),
the standard emf of the cell reaction \( Fe(s) + 2Fe^{3+}(aq) \rightarrow 3Fe^{2+}(aq) \) is:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] For the reaction, \( Fe^{3+} \) is reduced to \( Fe^{2+} \) (reduction at the cathode), and \( Fe \) is oxidized to \( Fe^{2+} \) (oxidation at the anode). So: \[ E^\circ_{\text{cell}} = E^\circ_{Fe^{3+}/Fe^{2+}} - E^\circ_{Fe^{2+}/Fe} \] \[ E^\circ_{\text{cell}} = 0.771 \, \text{V} - (-0.441 \, \text{V}) = 0.771 + 0.441 = 1.212 \, \text{V} \] Hence, the standard emf of the cell reaction is \( 1.212 \, \text{V} \).
Consider the following
Statement-I: Kolbe's electrolysis of sodium propionate gives n-hexane as product.
Statement-II: In Kolbe's process, CO$_2$ is liberated at anode and H$_2$ is liberated at cathode.
O\(_2\) gas will be evolved as a product of electrolysis of:
(A) an aqueous solution of AgNO3 using silver electrodes.
(B) an aqueous solution of AgNO3 using platinum electrodes.
(C) a dilute solution of H2SO4 using platinum electrodes.
(D) a high concentration solution of H2SO4 using platinum electrodes.
Choose the correct answer from the options given below :
A solution of aluminium chloride is electrolyzed for 30 minutes using a current of 2A. The amount of the aluminium deposited at the cathode is _________
A bob of heavy mass \(m\) is suspended by a light string of length \(l\). The bob is given a horizontal velocity \(v_0\) as shown in figure. If the string gets slack at some point P making an angle \( \theta \) from the horizontal, the ratio of the speed \(v\) of the bob at point P to its initial speed \(v_0\) is :
Electrolysis is the process by which an element is decomposed and undergoes some chemical change under the influence of any electric current. The first-ever electrolysis was executed out by Sir Humphrey Davey in the year 1808. Electrolysis can occur in both Galvanic cells and Electrolytic cells.
Read More: Products of Electrolysis