Question

please show work so i understand, thank you

Balance the following redox equations by the ion-electron half-reaction method: (a) (acid solution): U4+ + MnO4 → UO2+ + Mn2 (b) (acid solution): Zn (s) + NO, Zn2+ + N,O (g) (c) (basic solution): HPO; + MnO4 → P04 + MnO42

Answer 1

24 Solution. Bylon-electron half reaction method. aj Cacid solution) A + Allatom a balanced o Oxygen add water H + Hydrogen add Ht c Charge add eão Utt +Msou uo 2 + + Meht l. Ut - VO22+ +4H+. po tero 2. MnOu tout Mnou tight 5600 Mnet 4446 Real x.. + Eq 2] X2 And then we get. 504+2M noy**2H20 - > Balancedegen Cacitie Solution). 5 VO2 +2M net +444 50%+ + 2 Monouso + 2H₂O 500, 2+ no tamnt 4447 - this shown acilie i Scanned with CamScanner

bj (acidit). - Zn + Noz 2n24t Ng og zn Zn2+ + 2e to Txu 2. 2NO3 +18* +Be=N, 045440 - add both and get balanced- 42m +2N03.+2047 -> 42n+2+ N, 0 7 "+5Hgo y basic solution 1. HP O² + Mnoua - Pouson tMm O, 200 lo Hoogeto - poy são +34teco - +H₂O 2-1 Mn Oystero MnO, 2007 12 - add: 1 HP03250 +2 MnO yoof Heo - Poy350 L3H+ in acidic medium, +2 MMOR L balanced Scanned with CamScanner

for basic medium. add both side oho. 8o that Ht doh → H₂O IHPO2 +2 MnO4 1204 > Poy 342MMOR ' +2H2O So that in basic medium Aposton +MnOy +307 -- Poy 3.5" 72 M2647_24, 0) Balanced. Scanned with CamScanner