Question

help me with 1 and 2 please

1.6) Balance the following redox equation in basic solution. I recommend you work out the problem on the back side of the paper. Scan both sides if needed NO2 (aq) + Al(s) + NH3(aq) + Al(OH)4 (aq) 2.(6) What is the emf of the following cell under standard conditions? 5H2O2 (aq) + 2Mn*- (aq) + 2MnO4 (aq) + 2H20 (aq) + 6H* (aq)

Answer 1

2 we have to balance the given reaction - NO, (09) + Al(s) — NHz (@q) + Al(OH)199) oxidation fait cell reachon: Alls) + 406709) → AB (OH), (aq) +32 Reduction half cell reaction:- O • No (aq) + 5 H20faq) + 6€ NH(aq) + 701H (aq) By doing seq 20 x2} + equ f we have 24L(S) + 80H (90) 248(04), (99) +66 No, (aq) + 5 120(e)+685 + NHtz (aq) + 704 "(aq) 2 Al(s) + No, laz) + 547 01)+ OH-(09]+ 241 (04), (012) + NH3 (99) Hence the required balanced eqon is No, (aq) + 2Al(s) + oH+ (aq) + 51, 0 kq) = 2A1(04), (99) + N H3 (99) 2) Gireen that 542O2 (aq) + 2 mm (ar) → 2 MM 04 (aq) + 2H₂0 (04) + 6H4 (99) We have to calculate the emf of the cell at the standard conditions. • Oxidation half cell san :- Milan 4 H₂0 (24) MnO4 (aq) + 8 + (aq) & ge- & Reduction half cell sxe - H2O2 + 2H+ + 20 - 2H₂O (ar) x5 (09) (aq) . The balanced reaction is [5 H2O2 (ore) + 2 Mn 24 (ar) = 2MMO (99) + 2H20lia) + 6H* (ar)

Now, the standard emf of the half cells are Emoa (MM 2t = 1.514 and 402 / 420 = 1.781) rence, the standard conf of the cell is teen - € 4202 /4,0 – E Mmoa (amer/ ou = 1.78 V -151V => bocell = 0.270/ the standard emf of the cell is So 0.27V,