AG = AG° + RT · In(Q) ΔΕ ΔΕ. - (E) · In(Q) Half Reaction (Note: All given as reduction) E° (V) 02 (g) + 4 H*(aq) + 4 e 2z° (aq) 2 H20 (1) 1.229 |Z2 (s) + 2 e 3+ (aq) + 3 e 0.426 A' → A (s) 0.292 2 H20 (1) + 2 e- G2+ (aq) + 2 e H2 (g) + 2 OH¯ (aq) - 0.828 G (s) - 1.245 M2+ (aq) + 2 e M (s) - 1.893 A student constructs a voltaic electrochemical cell with two metal electrodes [metal G and metal A] in their respective aqueous nitrate solutions [G(NO3)2 and A(NO3)3]. Use this information, as well as the reduction potentials in the table above to complete each statement below. The metal electrode A is... o not changing in mass. o increasing in mass. o decreasing in mass. Question 4 The metal solution A(NO3)3 is... o increasing in concentration. o not changing in concentration. o decreasing in concentration.

AG = AG° + RT · In(Q) ΔΕ ΔΕ. - (E) · In(Q) Half Reaction (Note: All given as reduction) E° (V) 02 (g) + 4 H*(aq) + 4 e 2z° (aq) 2 H20 (1) 1.229 |Z2 (s) + 2 e 3+ (aq) + 3 e 0.426 A' → A (s) 0.292 2 H20 (1) + 2 e- G2+ (aq) + 2 e H2 (g) + 2 OH¯ (aq) - 0.828 G (s) - 1.245 M2+ (aq) + 2 e M (s) - 1.893 A student constructs a voltaic electrochemical cell with two metal electrodes [metal G and metal A] in their respective aqueous nitrate solutions [G(NO3)2 and A(NO3)3]. Use this information, as well as the reduction potentials in the table above to complete each statement below. The metal electrode A is... o not changing in mass. o increasing in mass. o decreasing in mass. Question 4 The metal solution A(NO3)3 is... o increasing in concentration. o not changing in concentration. o decreasing in concentration.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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7
Calculate AG° f( in kJ) or the reaction of iron(II) ions with permanganate ions. (F =96500 C) MnO4 (aq) + 8H*(aq) + 5e → Mn2+(aq) + 4H20(1) E° = 1.51 V Fe3+ (aq) + e → Fe2*(aq) E°= 0.77 V
please answer first 3 parts (a) , (b) and (c) only
Reduction potential of Mg2+ is -2.37 V. Reduction Potential of Fe2+ is 0.44 V. Based on the reaction below what is the electrical energy? Mg(s) + Fe2+(aq) + Fe(s) Fe2+(aq) = 0.18 M ; Mg2+(aq) = 0.53M
Consider the following half-reactions: Half-reaction E° (V) 2+ Hg"(aq) + 2e → Hg(1) 0.855V Co**(aq) + 2e → Co(s) -0.280V 2+ Fe"(aq) + 2e → Fe(s) |-0.440v (1) The strongest oxidizing agent is: enter formula (2) The weakest oxidizing agent is: (3) The weakest reducing agent is: (4) The strongest reducing agent is: (5) Will Hg*(aq) oxidize Fe(s) to Fe2*(aq)?| (6) Which species can be oxidized by Co2*(aq)? If none, leave box blank.
Standard Electrode Potentials at 25 °C Reduction Half-Reaction O Cu (aq) O H,Ozlaq) O Br O 10₂ (aq) Fyl) +20 H₂O₂(aq) + 2 H(aq) +2e" PbOg(s) +4 H(aq) +50, (aq) + 2 e Mn0 (aq) +4 H(aq) + 3 e MnO, (aq) +8H(aq) + 5 e Au³(aq) + 3 e PbOy(s) +4 H(aq) + 2e¯ C+20 00, (aq) +14 H(aq) + 6 e 00+ 4H*(aq) + 4 MnO₂ (s) +4 H(aq) +2e¯ 10₂(aq) + 6H(aq) +5e Big(0+20 VO₂ (aq) + 2 H(aq) + e¯ NO₂ (aq) + 4H(aq) + 3 e CO₂(0)+ Ag (ag)+e Fe(aq) + e 0)+2 H(aq) +2 e Mn0, (aq) + e (8) +26 Cu (aq) + e -2 (aq) -2 H₂O(0) Which of the following is the strongest oxidizing agent? -PbSO4(s) + 2 H₂O(0) →→MnO₂ (s) + 2 H₂O(0) Mn²(aq) + 4H₂O(1) - Au(s) ·Pb²+ (aq) + 2 H₂O(0) - 2 Cr (aq) 2 Cr³+ (aq) + 7 H₂O(1) → 2 H₂O( -2(aq) -Cu(s) Mn²+ (aq) + 2 H₂O(0) ₂(aq) + 3 H₂0(0) -2 Br (aq) - Vo²+ (aq) + H₂O( -NO(g) + 2 H₂O(1) — CIO; (aq) -Ag(s) -Fe²+ (aq) -) HyOz(ag) MnO₂ (aq) 1.00 0.96 0.95 0.80 0.77 0.70 0.56 0.54 0.52 E° (V) 2.87 1.78 1.69 1.68 1.51 1.50 1.46 1.36 1.33 1.23 1.21 1.20 1.09
Consider the following half-reactions: Half-reaction E° (V) Cl2(g) + 2e" → 2CI'(aq) 1.360v Ni?*(aq) + 2e- Ni(s) |-0.250V Mn2*(aq) + 2e → Mn(s) -1.180V (1) The strongest oxidizing agent is: enter formula (2) The weakest oxidizing agent is: (3) The weakest reducing agent is: (4) The strongest reducing agent is: (5) Will Mn2*(aq) oxidize Cl'(aq) to Cl2(g)? (6) Which species can be reduced by Ni(s)? If none, leave box blank.
Consider the following half-reactions: Half-reaction E° (V) F2(g) + 2e → 2F*(aq) 2.870V Co2+*(aq) + 2e → Co(s) |-0.28OV 2+ Mg"(aq) + 2e → Mg(s) -2.37OV (1) The weakest oxidizing agent is: enter formula (2) The strongest reducing agent is: (3) The strongest oxidizing agent is: (4) The weakest reducing agent is: (5) Will Mg(s) reduce F2(g) to F"(aq)? (6) Which species can be reduced by Co(s)? If none, leave box blank.
Consider the following half-reactions: Half-reaction ° (V) Br2(1) + 2e → 2Br°(aq) |1.080V Sn²+ (aq) + 2e . Sn(s) |-0.140V Zn2*(aq) + 2e → Zn(s)|-0.763V (1) The strongest oxidizing agent is: enter formula (2) The weakest oxidizing agent is: (3) The weakest reducing agent is: (4) The strongest reducing agent is: (5) Will Br2(0) oxidize Zn(s) to Zn2*(aq)? (6) Which species can be oxidized by Sn2*(aq)? If none, leave box blank.
Give the standard line notation for each cell. (a) 2 IO3 (aq) + 10 Fe2+ (ag) + 10 H* (aq) = 10 Fet (ag) + I2 (aq) + 5 H2O(1) O Pt(s) | Fe?+ (aq), Fe+ (ag) || IO3 (ag), Ht (ag), I2 (ag) | Pt(s) O Pt(s) | Fet (aq), I2 (ag) || Fe2+ (ag), IO3 (ag), H* (ag) | Pt(s) Pt(s) | IO3 (ag), H* (ag), I2 (aq) || Fe2+ (ag), Fet (ag) | Pt(s) Pt(s) | Fe2+ (ag), IO3 (ag), H* (aq) || Fet (aq), I2 (ag) | Pt(s) (b) Zn(s) + 2 Ag* (ag) = Zn?+ (aq) + 2 Ag(s) O Zn(s) | Zn?+ (aq) || Ag* (aq) | Ag(s) O Zn(s) | Ag* (aq) || Zn?+ (ag) | Ag(s) Ag(s) | Ag* (ag) || Zn2+ (aq) | Zn(s) O Ag(s) | Zn?+ (aq) || Ag* (aq) | Zn(s) (c) H2O2 (aq) + 2 H* (ag) + 2 e → 2 H2O(1) E = 1.78 V 02 (9) + 2 H* (ag) + 2 e + H2O2 (ag) E = 0.68 V O Pt(s) | O2(9) | H2O2 (aq), H* (ag) || H2O2 (ag), H* (ag) | Pt(s) Pt(s) | H2O2 (aq), H* (ag) || O2 (9) | H2O2 (ag), H* (ag) | Pt(s) Pt(s) | H2O2 (aq) | O2(9) || H* (aq) | Pt(s) O Pt(s) | H2O2(aq) | Pt(s) (d) Mn2+ (aq) + 2 e → Mn(s) E° = - 1.18 V Fe+ (aq) + 3 e → Fe(s) E - -0.036 V O Mn(s) | Mn²+ (ag)…
114. One type of breathalyzer employs a fuel cell to measure the quantity of alcohol in the breath. When a suspect blows into the breathalyzer, ethyl alcohol is oxidized to acetic acid at the anode: CH;CH,OH(g) + 4 OH (aq) - HC,H;O;(g) + 3 H2O(1) + 4 e ethyl alcohol acetic acid At the cathode, oxygen is reduced: 0,(g) + 2 H,O(1) + 4 e 4 OH (aq) The overall reaction is the oxidation of ethyl alcohol to acetic acid and water. When a suspected drunk driver blows 188 mL of his breath through this breathalyzer, the breathalyzer produces an average of 324 mA of current for 10 s. Assuming a pressure of 1.0 atm and a temperature of 25°C what percent (by volume) of the driver's breath is ethanol?
Balance the REDOX reaction associated with the cell shown below, in acidic medium: 2+ Fe(s) | Fe² || MnO (aq) Mn²) | Pt(s) Μη '(aq) 4(aq)' Enter the number for the coefficient of H₂O in the balanced equation: