References] Balance each of the following oxidation-reduction reactions by using the oxidation states method. (Use the...
[References] Balance each of the following oxidation-reduction reactions by using the oxidation states method. (Use the lowest possible coefficients for all reactions.) C.110(9) + 02(9) — CO2(g) + H2O(g) E K(s) + HCl(aq) — K+ (aq) + CI+ (aq) + H2(g) Col+ (aq) + Ni(s) - Co2+ (aq) + Ni2+ (aq) Mg(s) + H2SO4 (aq) MgSO4(aq) + H2(g) 0 Submit Answer Try Another Version 5 item attempts remaining
Balance the following equations. (Use the lowest possible whole-number coefficients. These may be zero.) (a) MnO4-(aq) + Cl-(aq) Mn2+(aq) + Cl2(aq) MnO4- + Cl- + H+ + H2O Mn2+ + Cl2 + H+ + H2O (b) Cr2O72-(aq) + NO2-(aq) Cr3+(aq) + NO3-(aq) Cr2O72- + NO2- + H+ + H2O Cr3+ + NO3- + H+ + H2O (c) Tl2O3(s) + NH2OH(aq) TlOH(s) + N2(g) Tl2O3 + NH2OH + OH- + H2O TlOH(s) + N2 + OH- + H2O (d) CrO42-(aq) + C2O42-(aq) Cr(OH)3(s) + CO2(g) CrO42- + C2O42- + OH- + H2O Cr(OH)3 + CO2 + OH- + H2O
Balance the following equations. (Use the lowest possible whole-number coefficients. These may be zero.) (a) Cr202-(aq) + NO2 (aq) = Cr3+ (aq) + NO3"(aq) Cr20,2- + NO3 + H+ + H2O= Cr3+ + NO3 + H+ + H20 (b) Mn04 (aq) + CH3OH(aq) Æ Mn2+ (aq) + HCO2H(aq) MnO4 + CH3OH + H+ + H2O= Mn2+ + HCO2H + H+ + H2O (c) ClO2(aq) + H2O2(aq) = C102 (aq) + O2(9) C102 + H2O2 + OH + H20 = ClO2...
Balance the following oxidation-reduction equations. The reactions occur in acidic solution. (Use the lowest possible coefficients. Omit states of matter. Add H20 or H to any side of the reaction if it is needed.) a. Pb + Bi0,- + PbO2+ BiS+ Pb + BiO3 + PbO2 + Bi9+ + b. Cr,0,? + Fe2+ + Cr+ + Fe3+ | Cr₂O₂ ² + | Fe²+ + + Fel- c. MnO, +1“ + Mn2+ + Cl2 MnO2 + Cl + Mn + Cl2...
5) Balance the following by the method of reactions: a. SO32- + MnO4 + H+ + SO42- + Mn2+ + H2O b. Oz (aq) + H2O → OH(aq) + O2(8)
Balance the following redox reactions by balancing the half reactions and then combine the half reactions to get the overall balanced redox reaction with the lowest possible whole number coefficients. 1. Consider the following unbalanced redox reaction: MnO2(s) + BrO3−(aq) → MnO4−(aq) + Br−(aq) (a) Balance the corresponding half reactions in acidic conditions using the lowest possible whole number coefficients. (Enter coefficients for one and zero. Blanks will be marked incorrect.) MnO2(s) + H2O(l) + OH−(aq) + H+(aq) + e− → MnO4−(aq) + H2O(l) + OH−(aq) + H+(aq)...
(10 pts) Balance the following redox reactions by first separating the oxidation and reduction half-reactions. a. Cut (aq) + Fe (s) Fe3+ (aq) + Cu(s) b. Cu(s) + HNO3 (aq) Cu2+ (aq) + NO (g) (basic solution) c. NH(aq) + O2(g) → N03 (aq) + H2O(l) (acidic solution) d. Cd(s) + NiO(OH)(s) + Ca(OH)2(s) + Ni(OH)2(s) (Nicad battery) e. The oxidation of iodide ion (1) by permanganate ion (MnO4) in basic solution to yield molecular iodine (12) and manganese(IV) oxide...
Balancing Oxidation–Reduction Reactions (Section)Complete and balance the following equations, and identify the oxidizing and reducing agents:(a) Cr2O72-(aq) + I-(aq)→Cr3+(aq) + IO3-(aq) (acidic solution)(b) MnO4-(aq) + CH3OH(aq) →Mn2+(aq) +HCO2H(aq) (acidic solution)(c) I2(s) + OCl-(aq)→IO3-(aq) + Cl-(aq) (acidic solution)(d) As2O3(s) + NO3-(aq)→H3AsO4(aq) + N2O3(aq) (acidic solution)(e) MnO4-(aq) + Br-(aq)→MnO2(s) + BrO3-(aq) (basic solution)(f) Pb(OH)42-(aq) + ClO-(aq)→PbO2(s) + Cl-(aq) (basic solution)
(1) Identify each of the following half-reactions as either an oxidation half-reaction or a reduction half-reaction. half-reaction identification Cr3+(aq) + 3e — Cr(s) Al(s)—+A13+(aq) + 3e (2) Write a balanced equation for the overall redox reaction. Use smallest possible integer coefficients. <> Submit Answer Retry Entire Group 9 more group attempts remaining When the following half reaction is balanced under acidic conditions, what are the coefficients of the species shown? NO3 + H —— NO + H2O In the above...
Balance the following equations. (Use the lowest possible whole-number coefficients. These may be zero.) Balance the following equations. (Use the lowest possible whole-number coefficients. These may be zero.) (a) Cr2022(aq) + I (aq) = Cr3+ (aq) + 103 (aq) 1 Cr2O72- + 1 1 + 8 H+ + 1 X H20 = 2 Cr3+ + 1 103 + 1 X H+ + 4 H2O (b) CIO3-aq) + As(s) Æ HCIO(aq) + H3A5O3(aq) 3 CIO3 + 4 As + 3 H...