Given the following half-reactions Mn2+(aq) + €2- -› Mn(s); E°= - 1.18 V Cu2+(ac) + €2- -Cu(s): E° = 0.34V Cr-0-2-(aq) + 14H*(aq) + €6- - 2Cp+ (aq) + 7H20(1); E° = 1.33V select the best reducing agent.

R = 8.314 mol·K F = 96,485 mot AG = AG° + RT · In(Q) ΔΕΔΕ-).In(O) Half Reaction (Note: All given as reduction) E° (V) 02 (g) + 4 H*(aq) + 4 e¯ → 2 H20 (1) 1.229 2 z" (aq) Z2 (s) + 2 e 3+ (aq) + 3 е 0.426 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 - 1.893 → M (s) 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. Consider the same cell from the above prompt. Calculate AG° in kJ/mol for this galvanic cell. Report your answer with 4 significant figures. You do not need to report units with your answer. If your value is negative, make sure to include a "-" symbol.

Examine the following half-reactions and select the strongest oxidizing agent among the species listed. Cr2*(aq) + 2e Cr(s); E° = -0.913 V Fe2*(aq) + 2e Sr2+(aq) + 2e Co2*(aq) + 2e O a. Cr2*(aq) %3D Fe(s); E° = -0.447 V Sr(s); E° = -2.89 V %3D * Co(s); E° = -0.28 V 2+, O b. Sr2+(aq) O C. Fe2+(aq) O d. Co2*(aq)

is 0.085 correct or should it be 0.116? I got two answers and i’m not sure.

For the galvanic cell: Zn(s) + 2TI*(aq) 2TI(s) + Zn2*(aq) What is E°cel? Standard Electrode Potentials: Zn2*(aq) + 2e Zn(s) E° = -0.76 V TI*(aq) + eTI(s) E° = -0.34 V O 0.42 V O -0.42 V O 1.1 V O -1.1 V

5a only

Half-Reaction E° (V) 2H* (aq) + 2e¯ - H2 (9) 0.000 Ag* (aq) + e¯ → Ag (s) 0.7996 2H20 (I) + 2e¯ - H2 (g) + 20OH" (aq) -0.8277 AI3+ (aq) + 3e¯ → Al (s) -1.676 H2O2 (aq) + 2H* (aq) + 2e¯ → 2H20 (1I) 1.776 Au+ (aq) + e¯→ Au (s) 1.692 I2 (s) + 2e¯ → 21¯ (aq) 0.5355 Au3+ (aq) + 3e¯ → Au (s) 1.498 2103- (aq) + 12H* (aq) + 10e¯ → I2 (s) + 6H2O (1) Mg2+ (aq) + 2e - Mg (s) 1.195 Ba2+ (aq) + 2e¯ → Ba (s) -2.912 -2.372 Br2 (1) + 2e¯ → 2B1¯ (aq) 1.066 Mn2+ (aq) + 2e" → Mn (s) -1.185 Ca2+ (aq) + 2e¯ → Ca (s) -2.868 Mno2 (s) + 4H+ (aq) + 2e¯ → Mn2+ (aq) + 2H20 (1) 1.224 Cl2 (g) + 2e¯ → 2C1- (aq) 1.35827 Mno4- (aq) + 8H* (aq) + 5e¯ → Mn2+ (aq) + 4H2O (1) 1.507 Co2+ (aq) + 2e¯ → Co (s) -0.28 Mnog- (aq) + 2H20 (I) + 3e¯ → MnO2 (s) + 40H¯ (aq) 0.595 Co3+ (aq) + e¯ → Co2+ (aq) 1.92 HNO2 (aq) + H+ (aq) + e¯ → NO (g) + H2O (I) 0.983 Cr2+ (aq) + 2e¯ → Cr (s) -0.913 N2 (g) + 4H2O (1) + 4e¯ → 40H" (aq) + N2H4 (aq) -1.16 Cr3+ (aq) + 3e¯ → Cr (s) -0.744 NO3- (aq) + 4H* (aq) + 3e¯ → NO (g) + 2H2O (1) 0.957 Cr3+…

Balance in acid-medium. Thanks

Consider the following half-reactions: Half-reaction E° (V) Br2(l) + 2e-  2Br-(aq) 1.080V Pb2+(aq) + 2e-  Pb(s) -0.126V Fe2+(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 Fe2+(aq) oxidize Br-(aq) to Br2(l)? (yes/no) (6) Which species can be oxidized by Pb2+(aq)?

Two half-reactions of an electrochemical cell are given below:Mn04–(aq) + 8H+(aq) + 5e– ——-> Mn2+(aq) + 4H2O(l), E° = + 1.51 VSn-2(aq) ——–> Sn4+(aq) + 2e–, E° = + 0.15 VConstruct the redox equation from the standard potential of the cell and predict the reaction is reactant favoured or product favoured.