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Balancing redox reactions Quiz 2

chemistry

 Which of the following elements can neither exhibit a positive nor negative oxidation state? 

 Generally, inert gases or noble gases have a filled octet, because of which they are very stable and do not react, therefore, there is no change in their oxidation state.

 Xenon has high ionization enthalpy, high electron gain enthalpy, and has a stable electronic configuration. Therefore, they have neither positive nor negative oxidation. 

 Select the correct application of redox reaction.

  Redox reactions occur almost everywhere, where oxidation and reduction take place simultaneously.

 By using suitable reducing agents, metals can be extracted by the reduction of metal oxide. Whereas in batteries, the redox reaction takes place. In green plants, carbon dioxide and water convert into carbohydrates in the presence of sunlight. 

 Below are the steps for balancing the equation by oxidation number method. What will be the correct order of balancing an equation by the oxidation number method? $ \\ $(i) Balance all atoms other than H and O. $ \\ $  (ii) Finally balance H and O by adding water molecules. $ \\ $(iii) Multiply an integer to the oxidising or reducing agent to balance the increase or decrease in oxidation state. $ \\ $(iv) Calculate the increase or decrease in the oxidation state of an atom. $ \\ $(v) Determine the oxidation of each atom. $ \\ $(vi) Write the skeletal equation of the reaction.

 First, the skeletal equation is to be written, and then by hit and trial, or by using the electron balance method, the reaction needs to be balanced.

 For balancing an ionic equation by oxidation number method, we have to write the skeletal equation.  After that, we have to calculate the oxidation state of each atom and analyse the decrease and increase in the oxidation number. Now, balance the oxidation state by adding an integer. After balancing the oxidation state, we have to balance all other atoms and then hydrogen and oxygen atoms. 

 What will be the balanced chemical reaction by Ion - electron or Half-reaction method of the given equation? $\text{A}{{\text{s}}_{2}}{{\text{S}}_{3}}\text{ + NO}_{3}^{-}\text{ + }{{\text{H}}^{+}}\text{ }\to \text{ AsO}_{4}^{-3}\text{ + S + NO + }{{\text{H}}_{2}}\text{O}$

 The skeletal equation is given in the question, analyse the places where electrons are added or taken away and follow the steps used in the half electron method to find the balanced equation.

  In the skeletal equation, the oxidation state of Arsenic increases from +3 to +5, S from -2 to 0 while that of N has decreased from +5 to +2. Balance the skeletal equation: $\text{ 3 A}{{\text{s}}_{2}}{{\text{S}}_{3}}\text{ + 10 NO}_{3}^{-}\to \text{ 6 AsO}_{4}^{-3}\text{ + 9 S + 10 NO }{} \text{ and} \\ \text{Now, balance the oxygen atoms and charges  by adding water and H$^+$ ion}\\ \text{3A}{{\text{s}}_{2}}{{\text{S}}_{3}}\,\text{+ 10NO}_{3}^{-}\text{ +}\text{ 4}{{\text{H}}_{2}}\text{O}\to \text{ 6AsO}_{4}^{-3}\text{ + 9S + 10NO + 8}{{\text{H}}^{+}}$

 In the given reaction, ${{\text{K}}_{2}}\text{C}{{\text{r}}_{2}}{{\text{O}}_{7}}\text{(aq) + 4}{{\text{H}}_{2}}{{\text{O}}_{2}}\text{(aq) + }{{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}\text{(aq) }\to \text{ }{{\text{K}}_{2}}\text{S}{{\text{O}}_{4}}\text{(aq) + 2Cr}{{\text{O}}_{5}}\text{(ether) + 5}{{\text{H}}_{2}}\text{O(l)} \\ $ Select the correct statement about the reaction from the following options.

 Find out the oxidation state of potassium and chromium from the two compounds. Each potassium atom has a charge of $+1$ and each oxygen atom has a charge of $(-)2$.

In the above reaction, the oxidation state of chromium in potassium dichromate & chromium oxide peroxide is 6, that is no change in the oxidation state, it means there is neither oxidation nor reducing reaction. Whereas the oxidation state of oxygen from hydrogen peroxide to hydrogen decreases, which means it undergoes reduction reaction. $\\ {{\text{K}}_{2}}\text{C}{{\text{r}}_{2}}{{\text{O}}_{7}} \\  2\text{ + 2x -14 = 0} \\  \text{2x = 12} \\ \text{x = +6} \\ \text{Cr}{{\text{O}}_{5}} \\ \text{x -10 = 0} \\ \text{x = 10} $