C2O(g)+NO3 ()CO2(g)+NO(g) C2042am→C02g C2042am→2C02g C2042am)→2C02g+2e NO3q→NOg NO3a→NOg+2H2Ou NO3e+4Hem→NOg+2H,0o NO3(a+4 H'+3-NO)+2 H2Oo) 3C2042m→6C02g+6e 2 NO3g+8 H)+6e-2 NO(g)+4 H2O) 3 C204()+2 NO3(g)+8H'()6CO2(8)+2NO(g)+4 H2O() (b)The half reaction Sn+2eSn has a standard reduction potential of +0.15 V.Assuming a temperature of 25C.and that [Sn=1.00 M,calculate [Sn when the reduction potential of this half reaction is0.00 V. Q=ISn"l [Sn“a】 -6-ho 1.00 thus,0.00V=0.15V-8.314 JK-mol'(298K)n 2(96487Cmo1-) 1.00 -2(96487Cmol-） (8.314 JK-mol-(298KI1.68 R9-law [Sn“m】=8.43×10M 33 C2O4 -2 (aq) + NO3 − (aq) → CO2(g) + NO(g) C2O4 -2 (aq) → CO2(g) C2O4 -2 (aq) → 2 CO2(g) C2O4 -2 (aq) → 2 CO2(g) + 2 e- NO3 - (aq) → NO(g) NO3 - (aq) → NO(g) + 2 H2O(l) NO3 - (aq) + 4 H+ (aq) → NO(g) + 2 H2O(l) NO3 - (aq) + 4 H+ (aq) + 3 e- → NO(g) + 2 H2O(l) 3 C2O4 -2 (aq) → 6 CO2(g) + 6 e- 2 NO3 - (aq) + 8 H+ (aq) + 6 e- → 2 NO(g) + 4 H2O(l) _____________________________________________________ 3 C2O4 -2 (aq) + 2 NO3 - (aq) + 8 H+ (aq) → 6 CO2(g) + 2 NO(g) + 4 H2O(l) (b) The half reaction Sn+4 (aq) + 2 e− → Sn+2 (aq) has a standard reduction potential of + 0.15 V. Assuming a temperature of 25o C, and that [Sn+2 (aq)] = 1.00 M, calculate [Sn+4 (aq)] when the reduction potential of this half reaction is 0.00 V. 2 (aq) 4 (aq) 0 1 1 1 4 (aq) 1 1 4 1 1 (aq) 4 (aq) [Sn ] Q [Sn ] RT E E ln(Q) nF 8.314 J K mol (298K) 1.00 thus, 0.00V 0.15V ln 2(96487 C mol ) [Sn ] 1.00 2(96487 C mol ) ln 0.15J C 11.68 [Sn ] 8.314J K mol (298K) 1.00 118600 [Sn ] [Sn + + − − − + − − + − − + = = − = − ⎛ ⎞ − = − ⎜ ⎟ = ⎝ ⎠ = 4 6 (aq) ] 8.43 10 M + − = ×