1(a)-Complete and balance the following redox equation thatoccurs in acidic solu
ID: 718645 • Letter: 1
Question
1(a)-Complete and balance the following redox equation thatoccurs in acidic solution using the set of smallest whole-numbercoefficients. What is the sum of all the coefficients in the equation?PbO2(s) + Cl^- ? Pb^2+ + Cl2(g) (acidic solution). Please show all work
(b) Complete and balance the following redox equation with the smallest set of whole numbers possible. What is the sum of the coefficients? (Remember the coefficients equal to one.)
HI + HNO3 ? I2 + NO (acidic solution) .Please show all work
(c) Complete and Balance the following redox equation.
MnO4^- + H^+ + Br^- ? Mn^2+ + Br2 + H2O (acidic Solution).
The sum of the smallest whole number coefficients is ? PLEASE SHOW ALL WORK
Explanation / Answer
1.(a)-Complete and balance the following redox equation thatoccurs in acidic solution using the set of smallest whole-numbercoefficients. What is the sum of all the coefficients in the equation? PbO2(s) + Cl^- Pb^2+ + Cl2(g) (acidic solution). ANSWER: PbO2(s) + Cl^- Pb^2+ + Cl2(g) (acidic solution) Step 1: Balancing Oxidation reaction: PbO2(s) Pb^2+ Add H2O on right side, PbO2(s) Pb^2+ + H2O We have to add H+ on left side (because acidic condition) while balance the elements, 4H+ + PbO2(s) Pb^2+ + 2H2O Balance the Charges with use of e-s , 4H+ + PbO2(s)+ 2e- Pb^2+ + 2H2O -------------(1) Step 2: Balance the reduction reaction. 2Cl- Cl2(g). Balance the charges with use of e-s, 2Cl- Cl2(g) + 2e- ------------------------------------(2) Step 3: Add equation (1) and (2) and cancel out the spectator ions, 4H+ + PbO2(s)+2Cl- -----------> Pb^2+ +Cl2(g) +2H2O Delta n = (Sum of coefficient of product- Sum of coefficient of reactant). Delta n = ( 1+1+2)-(4+1+2)= -3 The sum of the smallest whole number coefficients is -3 1.(b) Complete and balance the following redox equation with the smallest set of whole numbers possible. What is the sum of the coefficients? (Remember the coefficients equal to one.) HI + HNO3 -------> I2 + NO (acidic solution) ANSWER: Step 1: Balancing Oxidation reaction: HI I2 We have to add H+ on right side (because acidic condition) while balance the elements, HI I2 + 2H+ Balance the charges with use of e-s, 2HI I2 + 2H+ +2e- Multiple this equation by 3, 6HI 3 I2 + 6H+ +6e- ----------------------------(1) Step 2: Balance the reduction reaction. Add H2O on right side, HNO3 -------> NO +H2O We have to add H+ on left side (because acidic condition) while balance the elements, 3H+ + HNO3 -------> NO + 2H2O Balance the charges with use of e-s, 3H+ + HNO3 + 3e- -------> NO + 2H2O Multiple this equation by 2, 6H+ + 2HNO3 + 6e- -------> 2NO + 4H2O ------------------(2) Step 3: Add equation (1) and (2) and cancel out the spectator ions, 2HNO3 + 6HI 3 I2 + 2NO + 4H2O Delta n = (Sum of coefficient of product- Sum of coefficient of reactant). Delta n = ( 3+2+4)-(2+6)= 1 The sum of the smallest whole number coefficients is +1. 1.(c) Complete and Balance the following redox equation. MnO4^- + H^+ + Br^-1 -------> Mn^2+ + Br2 + H2O (acidic Solution). The sum of the smallest whole number coefficients is ? ANSWER: MnO4^- + H^+ + Br^-1 -------> Mn^2+ + Br2 + H2O (acidic Solution). Step 1: Balancing Oxidation reaction: MnO4^- Mn^2+ Add H2O on right side, MnO4^- Mn^2+ +H2O We have to add H+ on left side (because acidic condition) while balance the elements, 8H+ + MnO4^- Mn^2+ + 4H2O Balance the Charges with use of e-s , 8H+ + MnO4^- + 6e- Mn^2+ + 4H2O ------------------------(1) Step 2: Balance the reduction reaction. 2HBr Br2 + 2H+ + 2e- Multiple this equation by 3, 6HBr 3Br2 + 6H+ + 6e- ------------------------------------------------(2) Step 3: Add equation (1) and (2) and cancel out the spectator ions, 2H+ + MnO4^- +6HBr -------> Mn^2+ + 4H2O + 3 Br2 Delta n = (Sum of coefficient of product- Sum of coefficient of reactant). Delta n = (1+4+3)- ( 2+1+6)= -1 The sum of the smallest whole number coefficients is -1.Related Questions
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