7H2O + 2 Cr³+ + 3 Cl2 = Cr2O,²- + 6 Cl¯ + 14H+ e° = 0.03 V cell a. AG° = kJ K = Cu²+ (aq) + Ba(s) = Ba²+ (aq) + Cu(s) ɛ° = 3.24 V b. AG° = kJ K = Cl2 (g) + 2 Br¯ (aq) = Br2 (g) + 2 Cl¯ (aq) e° = 0.27 V

Open with Calculate ECELL and indicate whether the cell is galvanic or electrolytic. 8. Zn | 0.0364 M Zn+2 || 6.09 x 10-3 M T1+3, 0.0620 M TI+| Pt 9. Cu | 0.0601 M Cu+2 || 0.1350 M I-, AgI(s)| Ag 10. Pt, 1.00 atm H2 | 0.438 M NH3, 0.379 M NH4+ || SHE

Calculate the Gibbs free energy (AG°) of the reaction below. Is this reaction spontaneous or nonspontaneous? Ag (s) + Fe*(aq) = Ag* (aq) + Fe2*(aq) E° (V) 3+ Fe *(aq) + e Fe2"(aq) 0.77 Ag +e Ag(s) 0.80 A/

Calculate the standard free energy, ∆G°, for a galvanic cell represented by the following notation: M| M+||N3+|N     with reduction potential of 0.36 V for M and 1.05 V for N, respectively.  (Note that M and N are fake metals).  Calculate the value of ∆G° in J. F = 96500 C/mol e- Enter your answer in scientific notation with 2 decimal places and be sure to include the correct sign. Example:  -2.34e+6 or 1.23e-5. Note e is used for "x10" in this notation! Do NOT type "x 10"!

Consider the following reaction: Cd(s) + Fe2+(aq) → Cd2+(aq) + Fe(s) E0 (Fe2+ / Fe) = −0.4400 V,  E0 (Cd2+ / Cd) = −0.4000 V Calculate the emf for this reaction at 298 K if [Fe2+] = 0.60 M and [Cd2+] = 0.010 M.

Cd(s) + Fe2+(aq)  →  Cd2+(aq) + Fe(s) E o      (Fe2+ / Fe) = −0.4400 V, E o      (Cd2+ / Cd) = −0.4000 V Calculate the emf for this reaction at 298 K if [Fe2+] = 0.70 M and [Cd2+] = 0.010 M.

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.

What is the value of K for this aqueous reaction at 298 K? A + B 2C+ D AG° = 19.27 kJ/mol K =

just a&b

Consider the cell     Pb(s)│PbX2(s)│X¯(aq) (AX-=0.05) ║ Y¯(aq) (AY-=0.06)│AgY(s)│Ag(s) PbX2 (s) + 2e¯ → Pb (s) +2X¯       -0.391 V AgY (s) + e¯  → Ag (s) + Y¯           0.131 V What is the potential observed under the stated conditions (Ax-=0.05, AY-=0.06)? Make sure to include units in your answer.