Basic conditions are different, some of my recent answers show balancing of basic conditions if you want some examples.) Justify that the following reactions are redox reactions: Question 9. Another method for balancing redox reactions uses half-reactions. What is meant by cell potential? Question 3. In principle, O can have a minimum O.N. Since the electrode potentials of halogens decrease in the order: F2 (+2.87V) > Cl2 (+1.36V) > Br2 (+1.09V) > I2 (+0.54V), therefore, their oxidising power decreases in the same order. Answer: HCl gets oxidised. First Write the Given Redox Reaction. (b) When cone. H2O2 is getting reduced it acts as an oxidising agent. Answer: (i) In aqueous solution, AgNO3 ionises to give Ag+(aq) and NO3– (aq) ions. Here, each K atom as lost one electron to form K+ while F2 has gained two electrons to form two F– ions. Balance the following redox reaction in basic conditions. Further, it may be noted that whenever any half reaction equation is multiplied by any integer, its electrode potential is not multiplied by that integer. Question 20. (b) (i) galvanization (coating iron by a more reactive metal) Thus, it is a redox reaction and more specifically, it is a disproportionation reaction. The following reaction, written in net ionic form, records this change. (b) Cs. (a) an oxidizing agent (b) a reducing agent Therefore, 02 is the limiting reagent and hence calculations must be based upon the amount of 02 taken and not on the amount of NH3 taken. The oxidation number of two iodine atoms forming the I2 molecule is zero while that of iodine forming the coordinate bond is -1. Name the best reducing agent. However, when the mixture contains bromide ion, the initially produced HBr being a strong reducing agent than HCl reduces H2S04to S02 and is itself oxidised to produce red vapour of Br2. (c) Identify the element that exhibits both +ve and -ve oxidation states. First Write the Given Redox Reaction. of three I atoms, atoms in Kl3 are 0, 0 and -1 respectively. and because of the presence of d-orbitals it also exhibits +ve oxidation states of +3, +5 and +7. Question 10. Justify-giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic add is the best reductant. (d) 10. Answer: (i) KMnO4 ; K(+l); Mn(+7), 0(-2) Write a balanced ionic equation for the reaction. Question 18. (i) The reaction involves decomposition of cyanogen, (CN)2 in the alkaline medium to cyanide ion, CN and cyanate ion,CNO–. Question 13. MnO2 + Cu^2+ ---> MnO4^- + Cu^+ chemistry. MnO₄ ----- MnO₂ [Reduction] I⁻ -----I₂ [Oxidation] Step3. Balance the following equations. Question 5. of C2 = 3 (+1) + x + 1 (-1) = 0 or x = -2 C2 is, however, attached to one OH (O.N. A redox reaction is nothing but both oxidation and reduction reactions taking place simultaneously. (a) Hg2(Br03)2 (b) Br – Cl (c) KBrO4 (d) Br2 At anode there is loss of electrons. (b) Which are the negative and positive electrode? The path of reactions (a) and (b) can be determined by using  H20218 or D20 in reaction (b), Question 1. Solution for Balance the following redox reaction in acid: MnO4 – (aq) + C2O4 2– (aq) → Mn2+ (aq) + CO2 (g) a. Question 15. DON'T FORGET TO CHECK THE CHARGE. Starting with the correctly balanced half reactions write the overall net ionic reactions. Answer: Question 22. (a) 3. Since the EMF for the above reaction is positive, therefore, the above reaction is feasible. (a) 6CO2(g) + 12H2O(l) ————-> C6H12O6(s) + 6H2O(l) + 6O2(g) Although oxidation potential of H2O molecules is higher than that of Cl– ions, nevertheless, oxidation of Cl–(aq) ions occurs in preference to H2O since due to overvoltage much lower potential than -1.36 V is needed for the oxidation of H2O molecules. (c) N2H4is getting oxidised it is reducing agent. Overall reaction: 2Fe3+ (aq) + 2I–(aq) ——-> 2Fe2+  (aq) + I2(s); E° = + 0.23 V Answer: (a) Cr is getting oxidised and Mn04“ is getting reduced. What is meant by electrochemical series? To fix this issue, you must add a negative charge to the equation to balance the charges. AgN03(aq) ——–> Ag+(aq) + NO3– (aq) Therefore, CuO is reduced to Cu but H2 is oxidised to H20. Ag(s) ———–> Ag+(aq) + e–; E° = -0.80 V …(iii) Write balanced chemical equation for the following reactions: (i) Permanganate ion (MnO 4 –) reacts with sulphur dioxide gas in acidic medium to produce Mn 2 + and hydrogensulphate ion. No widgets added. Question 4. (a) Mercury (II) chloride, (b) Nickel (II) sulphate, (c) Tin (IV) oxide, (d) Thallium Reducing power goes on increasing whereas oxidising power goes on dcreasing down the series. (c) Cl2O7(g) + H2O2(aq) ———-> ClO2–(aq) + O2(g) + H+ Therefore, BCl3 is reduced while LiAlH4 is oxidised. (c) Following the steps as in part (a), we have the oxidation half reaction as: Fe 2+ (aq) → Fe 3+ (aq) + e-And the reduction half reaction as: H 2 O 2(aq) + 2H + (aq) + 2e- → 2H 2 O (l) Multiplying the oxidation half reaction by 2 and then adding it to the reduction half reaction, we have the net balanced redox reaction as: Ag2+ + e– ————–> Ag+ Arrange the following metals in the order in which they displace each other from the solution of their salts.Al, Cu, Fe, Mg and Zn. Ag+(aq) +e–———-> Ag(s); E° = +0.80 V …(i) Complete and balance the equation for this reaction in basic solution? Here the oxygen of peroxide, which is present in -1 state is converted to zero oxidation state in O2 and decreases to -2 oxidation state in H20. Answer: Question 8. Answer:  O.N. (b) HCHO is oxidised, Ag+   is reduced.Ag+  is oxidising agent whereas HCHO is reducing agent. (iii) individual reaction at each electrode. #MnO4^-) = Mn^(2+) + 4O# You can see in the reaction that oxygen is used to make water and no oxygen is let which is #O_2# thus 4 oxygen atoms can produce 4 water molecules. takes place. Question 16. 20 g of 02 will produce NO =120/160 x 20 = 15 g. Question 26. To do so, Eq. Answer: (a) It may be noted that for oxidation reactions, i.e., Eq. (c) Because it decomposes to give nascent oxygen. In HNO2 oxidation number of nitrogen is +3, it can decrease or increase with range of-3 to +5, hence it can act as both oxidising and reducing agent. a. MnO4- + SO2 Mn2+ + HSO4- The reaction occurs in acidic solution. Thus, cyanogen is simultaneously reduced to cyanide ion and oxidised to cyanate ion. from -1 to zero. S02(g) + 2H2O(l) ——–> HS04– (aq) + 3H+(aq) +2e–                        …(i) Further among HCl and HF, HCl is a stronger reducing agent than HF because HCl reduces MnO2 to Mn2+ but HF does not. Consider the elements: Cs, Ne, I, F Answer: The skeletal equation is: Question 24. Thus, this is a redox reaction. Considering the equation above, we have 2 hydrogen (H) with the total charge +1[Refer the charges of the elements in the above table] and 2 oxygen (O) with the total charge -2 on the L.H.S and 2 hydrogen (H) with total charge +2 and only 1 oxygen (O) with the total charge -2 on the R.H.S. The Half-Reaction Method . Answer: (a) The oxidation number of nitrogen in HNO3 is +5 thus increase in oxidation number +5 does not occur hence HNO3 cannot act as reducing agent but acts as an oxidising agent. 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