Which of the Phosphorus oxoacid can create silver mirror from $AgNO _3$ solution?
- $H _4 P _2 O _7$
- $\left( HPO _3\right)_n$
- $H _4 P _2 O _6$
- $H _4 P _2 O _5$
The Correct Option is D
Solution and Explanation
Phosphorus oxoacids with a \( \text{P-H} \) bond act as reducing agents and can reduce \( \text{AgNO}_3 \) to silver (\( \text{Ag} \)), producing a silver mirror.
Among the given options, \( \text{H}_4\text{P}_2\text{O}_5 \) contains a \( \text{P-H} \) bond, which makes it capable of reducing \( \text{AgNO}_3 \).
The reaction mechanism involves the oxidation of the oxoacid, with the \( \text{P-H} \) bond breaking and silver (\( \text{Ag} \)) being reduced.
The silver forms a mirror on the surface of the reaction vessel.
Top Questions on Balancing a redox reaction
- Which of the following is an example of a redox reaction?
- VITEEE - 2025
- Chemistry
- Balancing a redox reaction
- How many moles of electrons are required to reduce 0.5 moles of Fe\(^{3+}\) to Fe\(^{2+}\)?
- AP EAPCET - 2025
- Chemistry
- Balancing a redox reaction
- What is the oxidation state of chlorine in \( \text{Cl}_2 \text{O}_7 \)?
- MHT CET - 2025
- Chemistry
- Balancing a redox reaction
- In the reaction: \[ \text{Zn} + \text{CuSO}_4 \rightarrow \text{ZnSO}_4 + \text{Cu} \] Which substance is oxidized?
- MHT CET - 2025
- Chemistry
- Balancing a redox reaction
- The standard free energy change (\( \Delta G^\circ \)) for the following reaction (in kJ) at 25°C is \[ 3Ca(s) + 2 Au^{+}(aq, 1M) \rightleftharpoons 3Ca^{2+}(aq, 1M) + 2Au(s) \] (given: \( E^\circ_{{Au}^{3+/2+}} = +1.50 \, V \), \( E^\circ_{{Ca}^{2+/Ca}} = -2.87 \, V \), \( F = 96500 \, {C mol}^{-1} \))
- AP EAMCET - 2024
- Chemistry
- Balancing a redox reaction
Questions Asked in JEE Main exam
The term independent of $ x $ in the expansion of $$ \left( \frac{x + 1}{x^{3/2} + 1 - \sqrt{x}} \cdot \frac{x + 1}{x - \sqrt{x}} \right)^{10} $$ for $ x>1 $ is:
- JEE Main - 2025
- Binomial theorem
- The value of current \( I \) in the electrical circuit as given below, when the potential at \( A \) is equal to the potential at \( B \), will be ________________________ A.
- JEE Main - 2025
- Electrical Circuits
Two cells of emf 1V and 2V and internal resistance 2 \( \Omega \) and 1 \( \Omega \), respectively, are connected in series with an external resistance of 6 \( \Omega \). The total current in the circuit is \( I_1 \). Now the same two cells in parallel configuration are connected to the same external resistance. In this case, the total current drawn is \( I_2 \). The value of \( \left( \frac{I_1}{I_2} \right) \) is \( \frac{x}{3} \). The value of x is 1cm.
- JEE Main - 2025
- Current electricity
If $ \theta \in [-2\pi,\ 2\pi] $, then the number of solutions of $$ 2\sqrt{2} \cos^2\theta + (2 - \sqrt{6}) \cos\theta - \sqrt{3} = 0 $$ is:
- JEE Main - 2025
- Trigonometric Equations
- If \( \int \left( x \sin^{-1} x + \sin^{-1} x (1 - x^2)^{3/2} + \frac{x}{1 - x^2} \right) dx = g(x) + C \), where C is the constant of integration, then \( g\left(\frac{1}{2}\right) \) equals:
- JEE Main - 2025
- Integration by Parts
Concepts Used:
Balancing a redox reaction
The method that is based on the difference in oxidation number of oxidizing agent and the reducing agent is known as oxidation number. The half-reaction method entirely depends on the division of the redox reactions into oxidation half and reduction half. It entirely depends on the individual which method to choose and use.
Oxidation Number Method:
Like various other reactions, it is very important to write the correct compositions and formulas. A very important thing to keep in mind while writing oxidation-reduction reactions is to write the compositions and formulas of the substances and the products present in the chemical reaction in a very correct manner.
Half-Reaction Method:
In this procedure, we decouple the equation into two halves. After that, we balance both the parts of the reaction separately. Finally, we add them together to get a balanced equation.