Oxidation of succinate ion produces ethylene and carbon dioxide gases. On passing $0.2$ Faraday electricity through an aqueous solution of potassium succinate, the total volume of gases (at both cathode and anode) at STP ($1\, atm$ and $273\, K$) is :
- $2.24\, L$
- $4.48\, L$
- $6.72\, L$
- $8.96\, L$
The Correct Option is D
Solution and Explanation
Top Questions on Electrolysis
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 _________
- JEE Main - 2025
- Chemistry
- Electrolysis
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 :- JEE Main - 2025
- Chemistry
- Electrolysis
Electrolysis of 600 mL aqueous solution of NaCl for 5 min changes the pH of the solution to 12. The current in Amperes used for the given electrolysis is ….. (Nearest integer).
- JEE Main - 2025
- Chemistry
- Electrolysis
- In electrolysis of dilute H\(_2\)SO\(_4\), what is liberated at anode?
- BCECE - 2025
- Chemistry
- Electrolysis
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 \( 1.212 \, \text{V} \).
- AP EAPCET - 2025
- Chemistry
- Electrolysis
Questions Asked in JEE Main exam
Let \( a \in \mathbb{R} \) and \( A \) be a matrix of order \( 3 \times 3 \) such that \( \det(A) = -4 \) and \[ A + I = \begin{bmatrix} 1 & a & 1 \\ 2 & 1 & 0 \\ a & 1 & 2 \end{bmatrix} \] where \( I \) is the identity matrix of order \( 3 \times 3 \).
If \( \det\left( (a + 1) \cdot \text{adj}\left( (a - 1) A \right) \right) \) is \( 2^m 3^n \), \( m, n \in \{ 0, 1, 2, \dots, 20 \} \), then \( m + n \) is equal to:
- JEE Main - 2025
- Determinants
Rate law for a reaction between $A$ and $B$ is given by $\mathrm{R}=\mathrm{k}[\mathrm{A}]^{\mathrm{n}}[\mathrm{B}]^{\mathrm{m}}$. If concentration of A is doubled and concentration of B is halved from their initial value, the ratio of new rate of reaction to the initial rate of reaction $\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)$ is
- JEE Main - 2025
- Chemical Kinetics
- Preparation of potassium permanganate from MnO\(_2\) involves a two-step process in which the 1st step is a reaction with KOH and KNO\(_3\) to produce:
- JEE Main - 2025
- Preparation
- Let $ A = \left\{ \theta \in [0, 2\pi] : \Re\left( \frac{2 \cos \theta + i \sin \theta}{\cos \theta - 3i \sin \theta} \right) = 0 \right\} $. Then $ \sum_{\theta \in A} \theta^2 $ is equal to:
- JEE Main - 2025
- Trigonometric Equations
- An aqueous solution of 0.1 M HA shows depression in freezing point of 0.2°C. If $ K_f $ (H$_2$O) = 1.86 K kg mol$^{-1}$ and assuming molarity = molality, find the dissociation constant of HA.
- JEE Main - 2025
- Solutions
Concepts Used:
Electrolysis
Electrolysis:
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.
Applications of Electrolysis:
- Electrolytic cells are used for the production of Hydrogen and Oxygen gas from water.
- They are used for the large-scale production of pure metals like sodium, copper, magnesium, etc.
- They are used to extract Aluminium from Bauxite.
- Electrolytic cells are used for electroplating, a process by which a thin protective layer of one metal is formed over another metal, mostly for protection.
- Electrolysis is used for the electrorefining of non-ferrous metals.
Read More: Products of Electrolysis