–<\/sup><\/p>\nQuestion 2.
\nAn oxidising agent is a substance which can
\na) Gain electrons
\nb) Lose an electronegative radical
\nc) Undergo decrease in the oxidation number of one of its atoms
\nd) Undergo any one of the above changes
\nAnswer:
\nd) Undergo any one of the above changes<\/p>\n
Question 3.
\nThe arrangement of metals in the order of decreasing tendency to lose electrons is called _________ .
\nAnswer:
\nActivity series<\/p>\n
Question 4.
\nWhen KMnO4<\/sub>, reacts with acidified FeSO4<\/sub>
\na) Only FeSO4<\/sub> is oxidised
\nb) Only KMnO4<\/sub> is oxidised
\nc) FeSO4<\/sub> is oxidised and KMnO4<\/sub> is reduced
\nd) KMnO4<\/sub> is oxidised and FeSO4<\/sub> is reduced
\nAnswer:
\nc) FeSO4<\/sub> is oxidised and KMnO4<\/sub> is reduced<\/p>\nQuestion 5.
\nIn the disproportionation reaction, which of the following statements is not true?
\na) The same species is simultaneously oxidised as well as reduced
\nb) The reacting species must contain an element having at least three oxidation states
\nc) The element in the reacting species is present in the lowest oxidation state
\nd) The element in the reacting species is present in the intermediate oxidation state
\nAnswer:
\nc) The element in the reacting species is present in the lowest oxidation state<\/p>\n
Question 6.
\nFind the oxidation state of oxygen in OF2<\/sub>.
\nAnswer:
\nThe oxidation number of fluorine in its compounds is always taken as -1.
\nIn OF2<\/sub>
\nX+ (-1 \u00d7 2) = 0
\nX = +2<\/p>\nQuestion 7.
\nThe oxidation numbers of chlorine atoms in bleaching powder is _________ .
\nAnswer:
\n-1<\/p>\n
Question 8
\nSO2<\/sub> can act as
\na) Oxidising agent only
\nb) Reducing agent only
\nc) Both oxidising and reducing agents
\nd) Acid and a reducing agent only
\nAnswer:
\nc) Both oxidising and reducing agents<\/p>\nQuestion 9.
\nIn the reaction
\n2KMnO4<\/sub> +16HCl \u2192 5Cl2<\/sub> + MnCl2<\/sub> + 2KCl + 8H2<\/sub>O the reduction product is _________ .
\nAnswer:
\nMnCl2<\/sub><\/p>\nQuestion 10.
\nThe strongest reducing agent is .
\na) K
\nb) Ba
\nc) Li
\nd) Na
\nAnswer:
\nc) Li<\/p>\n
Question 11.
\nOxidation state of oxygen in H2<\/sub>O2<\/sub> is _________ .
\nAnswer:
\n-1<\/p>\nPlus One Chemistry Redox Reactions Two Mark Questions and Answers<\/h3>\n
Question 1.
\nBalance the following equation using oxidation number method:
\nMnO2<\/sub> + Cl–<\/sup> \u2192 Mn2+<\/sup> + Cl2<\/sub>
\nAnswer:
\nAssigning oxidation numbers:
\n
\nEquating the increase and decrease in oxidation number:
\nMnO2<\/sub> + 2Cl–<\/sup> \u2192 Mn2+<\/sup> + Cl2<\/sub>
\nBalancing hydrogen and oxygen atoms:
\nMnO2<\/sub> + 2Cl–<\/sup> + 4H+<\/sup> \u2192 Mn2+<\/sup> + Cl2<\/sub> + 2H2<\/sub>O<\/p>\nQuestion 2.
\nBalance the following equation using the half-reaction method:
\nCu + NO3<\/sub>–<\/sup> \u2192 Cu2+<\/sup> + NO2<\/sub>
\nAnswer:
\nSeparating into half-reactions:
\nOxidation half: Cu \u2192 Cu2+<\/sup>
\nReduction half: NO3<\/sub>–<\/sup> \u2192 NO2<\/sub>
\nBalancing oxygen and hydrogen atoms:
\nNO3<\/sub>–<\/sup> + 2H+<\/sup> \u2192 NO2<\/sub> + H2<\/sub>O
\nBalancing charge by adding electrons and making the number of electrons equal in the two half-reactions:
\nCu \u2192 Cu2+<\/sup> + 2e–<\/sup>
\n2NO3<\/sub>–<\/sup> + 4H+<\/sup> + 2e–<\/sup> \u2192 2NO2<\/sub> + 2H2<\/sub>O
\nAdding the two half-reactions to achieve the overall reaction:
\nCu + 2NO3<\/sub> + 4H+<\/sup> \u2192 Cu2+<\/sup> + 2NO2<\/sub> + 2H2<\/sub>O<\/p>\nQuestion 3.
\nComplete the following ionic equations:<\/p>\n
\n- Al3+<\/sup> + 3e–<\/sup> \u2192 …………….<\/li>\n
- MnO4<\/sub>2-<\/sup> \u2192 + e–<\/sup><\/li>\n
- K \u2192 K+<\/sup> + ……………<\/li>\n
- Fe2+<\/sup> \u2192 Fe3+<\/sup> +<\/li>\n<\/ol>\n
Answer:<\/p>\n
\n- Al3+<\/sup> + 3e \u2192 Al<\/li>\n
- MnO4<\/sub>2-<\/sup> \u2192 MnO4<\/sub>–<\/sup>+ e–<\/sup><\/li>\n
- K \u2192 K+<\/sup> + e–<\/sup><\/li>\n
- Fe2+<\/sup> \u2192 Fe3+<\/sup> + e–<\/sup><\/li>\n<\/ol>\n
Question 4.
\nFind the oxidation number of P in the following compounds:<\/p>\n
\n- Na2<\/sub>PO4<\/sub><\/li>\n
- H3<\/sub>P2<\/sub>O7<\/sub><\/li>\n
- PH3<\/sub><\/li>\n
- H3<\/sub>PO4<\/sub><\/li>\n<\/ol>\n
Answer:<\/p>\n
\n- Na2<\/sub>PO4<\/sub>, Oxidation state of P = +6<\/li>\n
- H4<\/sub>P2<\/sub>O7<\/sub>, Oxidation state of P = +5<\/li>\n
- PH3<\/sub>, Oxidation state of P = -3<\/li>\n
- H3<\/sub>PO4<\/sub>, Oxidation state of P = +5<\/li>\n<\/ol>\n
Question 5.
\nChoose the correct oxidation number of sulphur in the compounds in column Afrom column B.<\/p>\n
\n\n\nColumn A<\/td>\n | Column B<\/td>\n<\/tr>\n |
\nNa2<\/sub>SO4<\/sub><\/td>\n-2<\/td>\n<\/tr>\n | \nH2<\/sub>sO3<\/sub><\/td>\n+7<\/td>\n<\/tr>\n | \nH2<\/sub>S<\/td>\n+6<\/td>\n<\/tr>\n | \nH2<\/sub>S2<\/sub>O7<\/sub><\/td>\n+4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Answer:<\/p>\n \n\n\nColumn A<\/td>\n | Column B<\/td>\n<\/tr>\n | \nNa2<\/sub>SO4<\/sub><\/td>\n+6<\/td>\n<\/tr>\n | \nH2<\/sub>SO3<\/sub><\/td>\n+4<\/td>\n<\/tr>\n | \nH2<\/sub>S<\/td>\n-2<\/td>\n<\/tr>\n | \nH2<\/sub>S2<\/sub>O8<\/sub><\/td>\n+7<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Question 6. \nExplain oxidation number and valency. \nAnswer: \nValency of an atom is its combining capacity and is denoted by a number without sign. The valency of an element is always a whole number.<\/p>\n Oxidation number is a net charge which an atom has or appears to have when the other atoms from the molecule are removed as ions assuming that the shared pair of electrons is with more electronegative atom.<\/p>\n Question 7. \nSome rules related to oxidation number are given below. Correct the mistakes.<\/p>\n \n- Oxidation number of alkali metals and alkaline earth metals is +2.<\/li>\n
- Oxidation number of hydrogen is always +1.<\/li>\n
- Algebraicsum of oxidation number of all the atoms in an ion is not equal to the charge on the ion.<\/li>\n<\/ul>\n
Answer:<\/p>\n | | | | | | | | | |