{"id":40015,"date":"2024-02-17T08:22:46","date_gmt":"2024-02-17T02:52:46","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=40015"},"modified":"2024-02-17T16:20:32","modified_gmt":"2024-02-17T10:50:32","slug":"plus-one-chemistry-chapter-wise-questions-answers-chapter-8","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/plus-one-chemistry-chapter-wise-questions-answers-chapter-8\/","title":{"rendered":"Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions"},"content":{"rendered":"

Plus One Chemistry Chapter Wise Questions and Answers\u00a0Chapter 8 Redox Reactions <\/strong>is part of Kerala Plus One Chemistry Chapter Wise Questions and Answers<\/a>. Here we have given Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions.<\/p>\n

Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions<\/h2>\n

Plus One Chemistry Redox Reactions One Mark Questions and Answers<\/h3>\n

Question 1.
\nIn which of the following, oxidation number of chlorine is +5?
\na) Cl–<\/sup>
\nb) ClO–<\/sup>
\nc) ClO2<\/sub>–<\/sup>
\nd) ClO3<\/sub>–<\/sup>
\nAnswer:
\nd) ClO3<\/sub>–<\/sup><\/p>\n

Question 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>\n

Question 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>\n

Question 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>\n

Question 9.
\nIn the reaction
\n2KMnO4<\/sub> +16HCl \u2192 5Cl2<\/sub> + MnCl2<\/sub> + 2KCl + 8H2<\/sub>O the reduction product is _________ .
\nAnswer:
\nMnCl2<\/sub><\/p>\n

Question 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>\n

Plus 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\"Plus
\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>\n

Question 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>\n

Question 3.
\nComplete the following ionic equations:<\/p>\n

    \n
  1. Al3+<\/sup> + 3e–<\/sup> \u2192 …………….<\/li>\n
  2. MnO4<\/sub>2-<\/sup> \u2192 + e–<\/sup><\/li>\n
  3. K \u2192 K+<\/sup> + ……………<\/li>\n
  4. Fe2+<\/sup> \u2192 Fe3+<\/sup> +<\/li>\n<\/ol>\n

    Answer:<\/p>\n

      \n
    1. Al3+<\/sup> + 3e \u2192 Al<\/li>\n
    2. MnO4<\/sub>2-<\/sup> \u2192 MnO4<\/sub>–<\/sup>+ e–<\/sup><\/li>\n
    3. K \u2192 K+<\/sup> + e–<\/sup><\/li>\n
    4. 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
      1. Na2<\/sub>PO4<\/sub><\/li>\n
      2. H3<\/sub>P2<\/sub>O7<\/sub><\/li>\n
      3. PH3<\/sub><\/li>\n
      4. H3<\/sub>PO4<\/sub><\/li>\n<\/ol>\n

        Answer:<\/p>\n

          \n
        1. Na2<\/sub>PO4<\/sub>, Oxidation state of P = +6<\/li>\n
        2. H4<\/sub>P2<\/sub>O7<\/sub>, Oxidation state of P = +5<\/li>\n
        3. PH3<\/sub>, Oxidation state of P = -3<\/li>\n
        4. 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\n\n\n\n\n
          Column A<\/td>\nColumn B<\/td>\n<\/tr>\n
          Na2<\/sub>SO4<\/sub><\/td>\n-2<\/td>\n<\/tr>\n
          H2<\/sub>sO3<\/sub><\/td>\n+7<\/td>\n<\/tr>\n
          H2<\/sub>S<\/td>\n+6<\/td>\n<\/tr>\n
          H2<\/sub>S2<\/sub>O7<\/sub><\/td>\n+4<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

          Answer:<\/p>\n\n\n\n\n\n\n\n
          Column A<\/td>\nColumn B<\/td>\n<\/tr>\n
          Na2<\/sub>SO4<\/sub><\/td>\n+6<\/td>\n<\/tr>\n
          H2<\/sub>SO3<\/sub><\/td>\n+4<\/td>\n<\/tr>\n
          H2<\/sub>S<\/td>\n-2<\/td>\n<\/tr>\n
          H2<\/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

              \n
            • Oxidation number of alkali metals is +1.<\/li>\n
            • Oxidation number of alkaline earth metals is +2.<\/li>\n
            • Oxidation number of H is +1 except in metallic hydrides.<\/li>\n<\/ul>\n

              Question 8.
              \nMatch the following:<\/p>\n\n\n\n\n\n\n\n\n
              Oxidation number of Cl in Cl2<\/sub>O7<\/sub><\/td>\nCu<\/td>\n<\/tr>\n
              Oxidant<\/td>\nZn<\/td>\n<\/tr>\n
              Stannous Chloride, SnCl2<\/sub><\/td>\n+7<\/td>\n<\/tr>\n
              Oxidation number of C in diamond<\/td>\nGet reduced easily<\/td>\n<\/tr>\n
              The metal which can\u2019t displace H from dil.HCl<\/td>\nZero<\/td>\n<\/tr>\n
              <\/td>\nReducing agent for mercuric chloride<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

              Answer:<\/p>\n\n\n\n\n\n\n\n
              Oxidation number of Cl in Cl2<\/sub>O7<\/sub><\/td>\n+7<\/td>\n<\/tr>\n
              Oxidant<\/td>\nGet reduced I easily<\/td>\n<\/tr>\n
              Stannous Chloride, SnCl2<\/sub><\/td>\nReducing agent for mercuric chloride<\/td>\n<\/tr>\n
              Oxidation number of C in diamond<\/td>\nZero<\/td>\n<\/tr>\n
              The metal which can\u2019t displace H from dil.HCl<\/td>\nCu<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

              Question 9.
              \n1. Calculate the oxidation number of oxygen in OF2<\/sub> and KO2<\/sub>.
              \n2. When Zn rod is dipped in blue CuSO4<\/sub> solution ‘ the blue colour of CuSO4<\/sub> fades due to displacement reaction. Write the reaction and identify the following:
              \ni) The substance oxidised and the substance reduced.
              \nii) The oxidant and the reductant.
              \nAnswer:
              \n1. OF2<\/sub>: x + (-1 \u00d7 2) = 0
              \nx – 2 = 0
              \nx = +2
              \nKO2<\/sub>: (+1 \u00d7 1) + 2x = 0
              \n1 + 2x = 0
              \n2x = -1
              \nx = –\\(\\frac{1}{2}\\)<\/p>\n

              2. Zn(s) + CuSO4<\/sub> (aq) \u2192 ZnSO4<\/sub>(aq) + Cu(s)
              \ni) Substance oxidised – Zn
              \nSubstance reduced – Cu
              \nii) Oxidant-Cu
              \nReductant – Zn<\/p>\n

              Question 10.
              \na) Calculate the oxidation number of C in CH4<\/sub> and in CH3<\/sub>Cl.
              \nb) The sum of oxidation numbers of all atoms in a molecule is …………
              \nAnswer:
              \na) CH4<\/sub>:
              \nx + (1 \u00d7 4) = 0
              \nx + 4 = 0
              \nx = -4
              \nOxidation number of C in CH4<\/sub> is -4.<\/p>\n

              CH3<\/sub>Cl:
              \nx + (3 \u00d7 1) + -1 = 0
              \nx + 3 – 1= 0
              \nx + 2 = 0
              \nx = -2
              \nOxidation number of C in CH3<\/sub>Cl is -2.<\/p>\n

              b) Zero<\/p>\n

              Question 11.
              \n1. Write the oxidation state of each element and identify the oxidising agent and reducing agent in the following reaction:
              \nH2<\/sub>S(g) + Cl2<\/sub>(g) \u2192 2HCl(g) + S(s)
              \n2. Fill in the blanks and classify the following reactions into oxidation and reduction:
              \ni) Mn7+<\/sup> + 5e–<\/sup> \u2192 ……………
              \nii) Sn4+<\/sup> + …………… \u2192 Sn2+<\/sup>
              \niii) Na \u2192 Na+<\/sup> + ……………
              \niv) Fe3+<\/sup> +…………… \u2192 Fe2+<\/sup>
              \nAnswer:
              \n1.\"Plus
              \nReducing agent – H2<\/sub>S
              \nOxidising agent -Cl2<\/sub><\/p>\n

              2. i) Mn7+<\/sup> + 5e–<\/sup> \u2192 Mn2+<\/sup>
              \nii) Sn4+<\/sup> + 2e–<\/sup> \u2192 Sn2+<\/sup>
              \niii) Na \u2192 Na+ + e–<\/sup>
              \niv) Fe3+<\/sup> + e–<\/sup> \u2192 Fe2+<\/sup>
              \nOxidation: Reaction (iii)
              \nReduction: Reactions (i), (ii) and (iv)<\/p>\n

              Question 12.
              \nDihydrogen undergoes redox reactions with many metals at high temperature.
              \na) Write the reaction between hydrogen with sodium.
              \nb) Comment, whether the product formed, is covalent compound or ionic compound.
              \nc) Which is the reducing agent in this reaction?
              \nAnswer:
              \n1. 2Na + H2<\/sub> \u2192 2NaH
              \n2. Ionic compound is formed. When alkali metals react with hydrogen ionic hydrides are formed.
              \n3. Na is the reducing agent.<\/p>\n

              Question 13.
              \n1. Is it possible to keep copper sulphate solution in zinc pot? Why?
              \n2. Assign oxidation numbers of the underlined elements.
              \ni) NaH2<\/sub>\\(\\underline { P } \\)O4<\/sub>
              \nii) NaH\\(\\underline { S } \\)O4<\/sub>
              \nAnswer:
              \n1. No. Zn being more reactive will displace Cu from CuSO4<\/sub>. Thus Cu will be deposited on the vessel.<\/p>\n

              2. i) NaH2<\/sub>\\(\\underline { P } \\)O4<\/sub>
              \n+1 +(+1 \u00d7 2) + x +(-2 \u00d7 4) = 0
              \n1 + 2 + x – 8 = 0
              \nx – 5 = 0
              \nx = +5
              \nii) NaH\\(\\underline { S } \\)O4<\/sub>
              \n+1 + 1 + x +(-2 \u00d7 4) = 0
              \n+1 + 1 + x – 8 = 0
              \nx – 6 = 0
              \nx = +6<\/p>\n

              Question 14.
              \nIdentify the substance oxidised, reduced, oxidising agent and reducing agent in the reaction:
              \n2Cu2<\/sub>O + Cu2<\/sub>S \u2192 6Cu + SO2<\/sub>
              \nAnswer:
              \n\"Plus
              \nIn this reaction, Cu is reduced from +1 state to zero. oxidation state and S is oxidised from -2 state to +4 state. Cu2<\/sub>O helps S in Cu2<\/sub>S to increase its oxidation number. Therefore, Cu(l) is the oxidising agent. S of Cu2<\/sub>S helps Cu both in Cu2<\/sub>S itself and Cu2<\/sub>O to decrease its oxidation number. Therefore, S of Cu2<\/sub>S is the reducing agent.<\/p>\n

              Question 15.
              \nExplain the following in terms of electron transfer concept:<\/p>\n

                \n
              1. Oxidation<\/li>\n
              2. Reduction<\/li>\n
              3. Oxidising agent<\/li>\n
              4. Reducing agent<\/li>\n<\/ol>\n

                Answer:<\/p>\n

                  \n
                1. Oxidation: Loss of electron(s) by any species.<\/li>\n
                2. Reduction: Gain of electron(s) by any species.<\/li>\n
                3. Oxidising agent: Any species which accepts electrons).<\/li>\n
                4. Reducing agent: Any species which donates electron^).<\/li>\n<\/ol>\n

                  Question 16.
                  \nRepresent the following compounds using Stock notation:
                  \nCu2<\/sub>O, SnCl4<\/sub>, MnO, Fe2<\/sub>O3<\/sub>, V2<\/sub>O5<\/sub>
                  \nAnswer:
                  \n\"Plus<\/p>\n

                  Question 17.
                  \nIn a redox reaction, oxidation and reduction occur simultaneously.
                  \na) Write the classical concept of oxidation and reduction.
                  \nb) Identify the species undergoing oxidation and reduction in the following reaction:
                  \nH2<\/sub>S(S) + Cl2<\/sub>(g) \u2192 2HCl(g) + S(s)
                  \nAnswer:
                  \n1. Oxidation:
                  \naddition of oxygen\/electronegative element to a substance or removal of hydrogen\/ electropositive element from a substance.<\/p>\n

                  Reduction:
                  \nremoval of oxygen\/electronegative element from a substance or addition of hydrogen\/ electropositive element to a substance.<\/p>\n

                  2.Oxidised species:
                  \nH2<\/sub>S. This is because a more electronegative element, Cl is added to H or a more electro positive element, H has been removed from S.<\/p>\n

                  Reduced species:
                  \nCl. This is due to addition of more electropositive element H to it.<\/p>\n

                  Plus One Chemistry Redox Reactions Three Mark Questions and Answers<\/h3>\n

                  Question 1.
                  \nAn equation is given below:
                  \nHNO3<\/sub>+ l2 \u2192 HlO3<\/sub> + NO2<\/sub> + H2<\/sub>O<\/p>\n

                    \n
                  • Find the oxidising agent and reducing agent.<\/li>\n
                  • Balance the equation using half reaction method.<\/li>\n<\/ul>\n

                    Answer:
                    \nOxidising agent = HNO3<\/sub>
                    \nReducing agent = l2<\/sub>
                    \nSkeletal equation:
                    \n\"Plus
                    \nBalancing the charge on the half-reactions by adding electrons and equalising the number of electrons:
                    \n\"Plus<\/p>\n

                    Question 2.
                    \n1. Define redox reactions.
                    \n2. Predict whether the following reaction is a redox reaction or not? Justify.
                    \nCr2<\/sub>O7<\/sub>2-<\/sup> + H2<\/sub>O \u2192 2CrO4<\/sub>2-<\/sup> + 2H+<\/sup>
                    \nAnswer:
                    \n1. Redox reactions are those reactions in which reduction and oxidation occur simultaneously. These reactions involve change in oxidation state of the interacting species.<\/p>\n

                    2. No.
                    \nBecause no element undergoes change in oxidation number.<\/p>\n

                    Question 3.
                    \na) Find out the oxidising agent and reducing agent in the following reaction:
                    \nCu(s) + 2Ag+<\/sup>(aq) \u2192 Cu2+<\/sup>(aq) + 2Ag(s)
                    \nb) Balance the following redox reaction in acid medium using oxidation number method.
                    \nCr2<\/sub>O7<\/sub>2-<\/sup> + Fe2+<\/sup> \u2192 Cr3+<\/sup> + Fe3+<\/sup>
                    \nAnswer:
                    \n1. Oxidising agent – Ag
                    \nReducingAgent – Cu<\/p>\n

                    2. Assigning oxidation number:
                    \n\"Plus<\/p>\n

                    Question 4.
                    \nCalculate the oxidation number of sulphur, chromium and nitrogen in H2<\/sub>SO4<\/sub>, Cr2<\/sub>O7<\/sub>2-<\/sup> and NO3<\/sub>–<\/sup>.
                    \nAnswer:
                    \nH2<\/sub>SO4<\/sub>
                    \n(2 \u00d7 +1) + x+ (4 \u00d7 -2) = 0
                    \n+2 + x – 8 = 0
                    \nx – 6 = 0
                    \nx = +6<\/p>\n

                    Cr2<\/sub>O7<\/sub>2-<\/sup>
                    \n2x + 7 \u00d7-2 = -2
                    \n2x = -2 + 14
                    \n2x = 12
                    \n\u2234 x = +6<\/p>\n

                    NO3<\/sub>–<\/sup>
                    \nx + 3 \u00d7 -2 = -1
                    \nx = -1 + 5
                    \nx = +4<\/p>\n

                    Question 5.
                    \n1. Assign oxidation numbers
                    \n(i) P in NaH2<\/sub>PO4<\/sub>
                    \n(ii) Mn in KMnO4<\/sub>
                    \n(iii) B in NaBH4<\/sub>
                    \n(iv) S in H2<\/sub>SO4<\/sub>
                    \n2. Identify the oxidising and reducing agents in the following reaction:
                    \nCuO + H2<\/sub> \u2192 Cu + H2<\/sub>O
                    \nAnswer:
                    \n1. i) NaH2<\/sub>PO4<\/sub>
                    \nNa+1<\/sup>H2<\/sub>+1<\/sup>PO4<\/sub>-2<\/sup>
                    \n1+2 + x- 8 = 0
                    \n3 + x – 8 = 0
                    \nx – 5 = 0
                    \nx =+ 5<\/p>\n

                    ii) K+1<\/sup>MnO4<\/sub>-2<\/sup>
                    \n1+ x – 8 = 0
                    \nx – 7 = 0
                    \nx = +7<\/p>\n

                    iii) Na+1<\/sup>BH4<\/sub>+1<\/sup>
                    \n1 + x + 1 \u00d7 4 = 0
                    \nx + 5 = 0
                    \nx = -5<\/p>\n

                    iv) H2<\/sub>+1<\/sup>SO4<\/sub>-2<\/sup>
                    \n2 + x – 8 = 0
                    \nx – 6 = 0
                    \nx = +6<\/p>\n

                    2.
                    \n\"Plus<\/p>\n

                    Question 6.
                    \nA copper rod is dipped in silver nitrate solution.<\/p>\n

                      \n
                    1. What are the observations?<\/li>\n
                    2. Write the displacement reaction.<\/li>\n
                    3. Identify the species getting oxidised and reduced.<\/li>\n<\/ol>\n

                      Answer:<\/p>\n

                        \n
                      1. The colour of the solution changes to blue. Silver is deposited on the copper rod.<\/li>\n
                      2. Cu(s) +2AgNO3<\/sub>(aq) \u2192 Cu(NO3<\/sub>)2<\/sub>(aq) + 2Ag(s)<\/li>\n
                      3. Oxidised species – Cu Reduced species – Ag+<\/sup><\/li>\n<\/ol>\n

                        Question 7.
                        \n1. Identify the oxidising and reducing agent in the reaction:
                        \nCuS + O2<\/sub> \u2192 Cu + SO2<\/sub>
                        \n2. Determine the oxidation number of the underlined element in the following:
                        \n\"Plus
                        \nAnswer:
                        \n\"Plus<\/p>\n

                        Question 8.
                        \n1. Identify the substance oxidised, substance reduced, oxidising agent and reducing agent in the reaction:
                        \nCl2<\/sub> + 2l–<\/sup> \u2192 2Cl–<\/sup> +l2<\/sub><\/p>\n

                        2. Calculate the oxidation number of underlined elements in the following compounds:
                        \ni) K2<\/sub>\\(\\underline { Cr } \\)2<\/sub>O7<\/sub>
                        \nii) H\\(\\underline { H } \\)O3<\/sub>
                        \nAnswer:
                        \n1. Cl2<\/sub> is reduced, therefore Cl2<\/sub> is the oxidising agent. I’ is oxidised, therefore I” is the reducing agent.<\/p>\n

                        2. i) K2<\/sub>\\(\\underline { Cr } \\)2<\/sub>O7<\/sub>
                        \n(+1 \u00d7 2) + 2x +(-2 \u00d7 7) = 0
                        \n+2 + 2x – 14 =0
                        \n2x – 12 =0
                        \n2x = 12
                        \nx = +6
                        \nii) H\\(\\underline { H } \\)O3<\/sub>
                        \n(+1 \u00d7 1) + x + (-2 \u00d7 3) = 0
                        \n1+ x – 6 = 0
                        \nx – 5 = 0
                        \nx = +5<\/p>\n

                        Question 9.
                        \nDetermine oxidation number of the elements underlined in each of the following.
                        \n\"Plus
                        \nAnswer:
                        \n\"Plus<\/p>\n

                        Plus One Chemistry Redox Reactions Four Mark Questions and Answers<\/h3>\n

                        Question 1.
                        \nPermanganate ion (MnO4<\/sub>–<\/sup>) reacts with bromide ion (Br–<\/sup>) in basic medium to give manganese dioxide
                        \n(MnO2<\/sub>) and bromate ion (BrO3<\/sub>–<\/sup>).
                        \na) Write the balanced ionic equation for this reaction.
                        \nb) Identify the oxidising agent and reducing agent in this reaction.
                        \nAnswer:
                        \n\"Plus<\/p>\n

                        Question 2.
                        \nA redox reaction involves oxidation and reduction.
                        \na) What do you understand by electrode potential?
                        \nb) Define a redox couple.
                        \nc) Explain the set-up for Daniell cell with a diagram.
                        \nd) Write the electrode reactions and overall cell reaction which occur in the Daniel cell.
                        \nAnswer:
                        \na) The potential difference between metal and its own ion is called electrode potential.<\/p>\n

                        b) A redox couple is defined as the combination of oxidised and reduced forms of a substance taking part in an oxidation or reduction half reaction.<\/p>\n

                        c) Take copper sulphate solution in a beaker and put a copper strip. Take zinc sulphate solution in another beaker and put a zinc rod. The two redox couples are represented as Zn2+<\/sup>\/Zn and Cu2+<\/sup>\/Cu. Put the beaker containing copper sulphate solution and beaker containing zinc sulphate side by side. Connect two solution by a salt bridge. The Zn and Cu rods are connected by a metalic wire with a provision for ammeter and switch. Transfer of electrons now does not take place directly from Zn to Cu2+<\/sup>, but through metallic wire. The electricity from solution in one beaker to solution in the other beaker flows by the migration of ions through the salt bridge.
                        \n\"Plus<\/p>\n

                        Question 3.
                        \nRedox reactions are those in which oxidation and reduction takes place. Explain the different types of redox reactions with suitable examples.
                        \nAnswer:
                        \nCombination Reactions: The reactions in which two substances combine together to form a new compound are called combination reactions. These can be denoted as A+ B \u2192 C where either A or B or both A and B should be in the elemental form.
                        \n\"Plus
                        \nDecomposition reactions:
                        \nThe reactions in which a compound breaks up into two or more substances at least one of which is in elemental form are called decompositions reactions.
                        \n\"Plus<\/p>\n

                        Displacement reactions:
                        \nThe reactions of the type X + YZ \u2192 XZ + Y in which an atom or ion in a compound is displaced by an ion (atom) of another element, such that X and Y are in elemental form are called displacement reactions. They are of two categories:
                        \n1. Metal displacement reactions: Reactions in which a more electropositive metal displaces a less electropositive metal from its compound.
                        \n\"Plus<\/p>\n

                        2. Non-metal displacement reactions: These are reactions in which a non-metal is displaced by another metal or non-metal.
                        \n\"Plus
                        \nDisproportionation reactions: These are special type of redox reactions in which an element in one oxidation state is simultaneously oxidised and reduced. Here one of the reactants should contain an element that should exist in at least three oxidation states. The element in the form of reacting substance is in the intermediate oxidation state; and both higher and lower oxidation states of that element are formed in the reaction.
                        \ne.g. The decomposition of hydrogen peroxide.
                        \n\"Plus
                        \nHere the oxygen of peroxide, which is present in -1 state, is converted to zero oxidation state in O2<\/sub> and to -2 state in H2<\/sub>O.<\/p>\n

                        Plus One Chemistry Redox Reactions NCERT Questions and Answers<\/h3>\n

                        Question 1.
                        \nFluorine reacts with ice and results in the change :
                        \n\"Plus
                        \nJustify that this reaction is a redox reaction.
                        \nAnswer:
                        \nIn the given reaction O.N. of F2<\/sub> changes from zero to -1 in HF and HOF whereas O.N. of oxygen change from -2 in H2<\/sub>O to zero in HOF. Thus, F2<\/sub> is reduced, whereas oxygen is oxidised and, therefore, it is a redox reaction.<\/p>\n

                        Question 2.
                        \nWrite formulas for the following compounds:<\/p>\n

                          \n
                        1. Mercury (II) chloride<\/li>\n
                        2. Nickel (II) sulphate<\/li>\n
                        3. Tin (IV) oxide<\/li>\n
                        4. \u00a0Thallium (I) sulphate<\/li>\n
                        5. Iron (III) sulphate<\/li>\n
                        6. Chromium (III) oxide<\/li>\n<\/ol>\n

                          Answer:<\/p>\n

                            \n
                          1. Hg(II)Cl2<\/sub><\/li>\n
                          2. Ni(II)SO4<\/sub><\/li>\n
                          3. Sn(IV)O2<\/sub><\/li>\n
                          4. Tl2<\/sub>(I)SO4<\/sub><\/li>\n
                          5. Fe2<\/sub>(III)(SO4<\/sub>)3<\/sub><\/li>\n
                          6. Cr2<\/sub>(III)O3<\/sub><\/li>\n<\/ol>\n

                            Question 3.
                            \nThe compound AgF2<\/sub> is unstable. However, if formed, the compound acts as a very strong oxiding agent. Why?
                            \nAnswer:
                            \nIn AgF2<\/sub>, oxidation state of Ag is + 2 which is very unstable. Since Ag can exist in a stable state of + 1 it quickly accepts an electron to form the more stable + 1 oxidation state.
                            \nAg2+<\/sup> + e–<\/sup> \u2192 Ag+<\/sup><\/p>\n

                            Question 4.
                            \nConsider the reactions:
                            \n\"Plus
                            \nWhy does the same reductant, thiosulphate react differently with iodine and bromine?
                            \nAnswer:
                            \nThe average O.N. of S in S2<\/sub>O3<\/sub>2-<\/sup> is + 2 while in S4<\/sub>O6<\/sub>2-<\/sup> it is + 2.5. The O.N. of S in SO4<\/sub>2-<\/sup> is+6. Since Br2<\/sub> is a stronger oxidising agent that l2<\/sub>, it oxidises S of S2<\/sub>O3<\/sub>2-<\/sup> to a higher oxidation state of + 6 and hence forms SO4<\/sub>2-<\/sup> ion. l2<\/sub>, however, being a weaker oxidising agent oxidises S of S2<\/sub>O3<\/sub>2-<\/sup> ion to a lower oxidation of + 2.5 in S4<\/sub>O6<\/sub>2-<\/sup> ion.<\/p>\n

                            Question 5.
                            \nWhy does the following reaction occur?
                            \nXeO6<\/sub>4-<\/sup>(aq) + 2F–<\/sup>(aq) + 6H+<\/sup>(aq) \u2192 XeO3<\/sub>(s) + F2<\/sub>(g) + 3H2<\/sub>O(I)
                            \nWhat conclusion about the compound Na4<\/sub>XeO6<\/sub> (of which XeO6<\/sub>4-<\/sup> is a part) can be drawn from the reaction?
                            \nAnswer:
                            \nThe balanced equation along with O.N. of the elements above their symbols will be as:
                            \n\"Plus
                            \nIn the equation the, O.N. of Xe decreases from + 8 in XeO6<\/sub>4-<\/sup> to + 6 in XeO3<\/sub> while that of F increases from – 1 in F–<\/sup> to 0 in F2<\/sub>. Therefore, XeO6<\/sub>4-<\/sup> is reduced while F–<\/sup> is oxidised. This reaction occurs because Na4<\/sub>XeO6<\/sub> (0r XeO6<\/sub>4-<\/sup>) is stronger oxidising agent than F2<\/sub>.<\/p>\n

                            We hope the Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions help you. If you have any query regarding Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

                            Plus One Chemistry Chapter Wise Questions and Answers\u00a0Chapter 8 Redox Reactions is part of Kerala Plus One Chemistry Chapter Wise Questions and Answers. Here we have given Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions. Kerala Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions Plus One Chemistry […]<\/p>\n","protected":false},"author":7,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[42728],"tags":[],"yoast_head":"\nPlus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions - A Plus Topper<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/www.aplustopper.com\/plus-one-chemistry-chapter-wise-questions-answers-chapter-8\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Plus One Chemistry Chapter Wise Questions and Answers Chapter 8 Redox Reactions\" \/>\n<meta property=\"og:description\" content=\"Plus One Chemistry Chapter Wise Questions and Answers\u00a0Chapter 8 Redox Reactions is part of Kerala Plus One Chemistry Chapter Wise Questions and Answers. 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