0

answers

0

watching

108

views

23 Nov 2019

Fe(s)-->Fe2+(aq) + 2e- E standard=0.45v (anode)
O2(g)+4H=(aq)+4e-->2H2O(l) E standard=1.23v (cathode)
2Fe3+(aq)+4H2O(l)-->Fe2O3*H2O(s)+6H+(aq)

What is the cell potential for the entire cell at a pH=7, a partialpressure of O2(g) of 0.18 bar, and an [Fe2+]=0.00150M?

0

answers

0

watching

108

views

For unlimited access to Homework Help, a Homework+ subscription is required.

Join us on Discord

Chemistry Study Group

Related textbook solutions

Basic Chemistry

ISBN: 9780134138046

Principles of Chemistry Molecular Approach

ISBN: 9780134112831

Principles of Chemistry Molecular Approach

3rd Edition, 2014

ISBN: 9780321971944

Chemistry: Structure and Properties

ISBN: 9780134293936

Chemistry: A Molecular Approach

ISBN: 9780321809247

Chemistry: A Molecular Approach

ISBN: 9780134874371

Principles of Chemistry: A Molecular Approach

ISBN: 9780134895741

Chemistry: The Central Science

14th Edition, 2017

ISBN: 9780134414232

Related questions

Write the half-reactions and the products for the reactions that occur in the electrolysis of a saturated solution of magnesium sulfate using inert electrodes. Choose at least one answer. should be 3-4 right answers a. 2Cl- ? 2Cl2(g) + 2e- (anode) b. 2H2O(l) ? O2(g) + 4H+(aq) + 4e- (anode) c. 2SO42-(aq) ? S2O82-(aq) + 2e- (anode) d. Ni(s) ? Ni2+(aq) + 2e- (anode) e. Ni2+(aq) + 2e- ? Ni(s) (cathode) f. Mg2+(aq) + 2e- ? Mg(s) (cathode) g. 2H2O (l) + 2e-? H2(g) + 2OH - (aq) (cathode) h. No net reaction. However, metal is transfer from the anode to the cathode. i. Cl2(g) j. H2(g) k. O2(g) l. Mg(s) m. Ni(s) n. S2O82-(aq) o. OH -(aq)

Show transcribed image text Using standard reduction potentials from the ALEKS Data tab, calculate the standard reaction free energy Î Go for the following redox reaction. Round your answer to 3 significant digits. 3Fe2 (aa)+Noj (a)+4H (aa)-3Fe (a)+ NO )+2H,00) kJ
Half-Reaction 2.866 192 1.776 1692 1.507 1.498 1.35827 F2 (g)+2e >2F (aq) Co3+ (aq) + e-â Co2+ (aq) H2O2 (aq) + 2H+ (aq) + 2e-â 2H2O (1) Au+ (aq) + e-â Au (s) MnO4" (aq) + 8H+ (aq) + 5e-â Mn2+ (aq) + 4H2O (1) Au3+ (aq) + 3e-â Au (s) Cl2 (g) + 2e-â 2Cl-(aq) O2 (g)+4H+ (a)4e2H2O() MnO2 (s) + 4H+ (aq) + 2e-â Mn2+ (aq) + 2H2O (l) 21O3 (aq) +12H (aq)+10e2 (s) 6H20 () Br2)2e2Br (aq) VO2+ (aq) + 2H+ (aq) + e-â VO2+ (aq) + H2O (l) HNO2 (aq) + H+ (aq) + e-â NO (g) + H2O (l) NO3 (a) 4H+ (aq) 3e NO (g) 2H20 () Ag+ (aq) + e-â Ag (s) 1.229 1.224 1.195 1.066 0.991 0.983 0.957 0.7996
Fe3+ (aq)eFe2+ (acq) 02 (g) + 2H+ (aq) + 2e-â H2O2 (aq) MnO4 (aq) 2H2O ()3eMnO2 (s40H (aq) l2 (s)+ 2e 2 (aq) Cu (aq)Cu (s) 02 (g) + 2H2O (l) + 4e-â 4OH-(aq) Cu2+ (aq) +2e> Cu (s) HSO4 (aq)3Ht (aq)2eH2SO3 (aq)+ H2O (0) Cu2+ (aq) + e-â Cu+ (aq) Sn4+ (aq) + 2e-â Sn2+ (aq) 2H+ (aq) + 2e-â H2 (g) Fe3+ (aq) 3e Fe (s) Pb2+ (aq) + 2e-â Pb (s) CrO42-(aq)4H20( 3e Cr(OH)3 (s)+50H- (aq) Sn2+ (aq)+2e> Sn (s) Ni2+ (aq)+2eNi (s) Co2+ (aq) + 2e-â Co (s) PbSO4 (s) + H+ (aq) + 2e-â Pb (s) + HSO4-(aq) 0.695 0.595 0.5355 0.521 0.401 0.3419 0.172 0.153 0.151 0.000 -0.037 -0.1262 -0.13 -0.1375 -0.257 -0.28 -0.3588
CrO42-(aq)4H20(3e Cr(OH)3 (s)+50H- (aq) Sn2+ (aq)+2e> Sn (s) Ni2+ (aq) + 2e, Ni (s) Co2+ (aq) + 2e-â Co (s) PbSO4 (s)H (aq)2ePb (sHSO4 (aq) Cr3+ (aq) + e-â Cr2+ (aq) Fe2+ (aq) +2e Fe (s) Cr3+ (aq) + 3e-â Cr(s) Zn2+ (aq)+2e Zn (s) 2H20 () 2eH2 (g) + 20H (aq) Cr2+ (aq)+2e Cr (s) N2 (g) 4H2O)+4e 40H (aq) N2H4 (aq) Mn2+ (aq)+ 2e Mn (s) Al3+ (aq)+3e Al (s) Sc3+ (aq)3eSc (s) Mg2+ (aa)+ 2e Mg (s) Nat (aq)eNa (s) Ca2+ (aq) + 2e-â Ca (s) Ba2+ (aq) + 2e-â Ba (s) -0.13 0.1375 -0.257 -0.28 0.3588 -0.407 -0.447 -0.744 0.7618 0.8277 -0.913 1.185 1.676 -2.077 2.372 2.71 2.868 2.912

Using standard reduction potentials (Appendix E), calculate the standard emf for each of the following reactions:

(a) Cl₂ (g) + 2I^- _(aq)→ 2Cl^- _(aq)+ I_2(s)

(b) Ni(s) + 2 Ce⁴⁺(aq) →Ni²⁺(aq) + 2 Ce³⁺(aq)

(c) Fe(s) + 2 Fe³⁺(aq)→3 Fe²⁺(aq)

(d) 2 NO₃^-(aq) + 8 H⁺(aq) + 3 Cu(s)→ 2NO(q) + 4H₂O(l) + 3 Cu²⁺(aq)

One Class Solution:-

a)CI₂(g) +2e-→2Cl^-(aq) E^o = 1.359 V

2I^- _(aq)→+ I_2(s)+ 2e- E^o = 0.536 V, E⁰= 1.359-0.536=0.823V

b) Ni(s) →Ni²⁺(aq) + 2e- anode E^o = –0.28V

2[Ce⁴⁺(aq) +e- →2 Ce³⁺(aq)] cathode E^o =1.61 V

E^o = 1.61-(-0.28)=1.89 V

c) Fe(s) + 2 Fe³⁺(aq)→3 Fe²⁺(aq) E^o =?

Fe(s) → Fe²⁺(aq) +2e- E^o = -0.440V

Fe³⁺(aq) +e- → Fe²⁺(aq) E^o = 0.771V

E=0.771-(-0.440)= 1.211 V

(d) 2 NO₃^-(aq) + 8 H⁺(aq) + 3 Cu(s)→ 2NO(q) + 4H₂O(l) + 3 Cu²⁺(aq)

2 NO₃^-(aq) + 8 H⁺(aq) + 3e-→ 2NO(q) + 4H₂O(l) E^o = 0.96V

Cu(s)→ Cu²⁺(aq) +2e- E^o = V+0.377

E^o =0.96-0.377= 0.583V

Weekly leaderboard

Homework Help 3,900,000

Chemistry 830,000

Start filling in the gaps now

New to OneClass?Sign up