Given the following standard reduction potentials: Mg+2 = Mg(s) + 2e- E = (-) 2.37 V 2e- + I2(s) = 2I-   E = (+) 0.53V Calculate the cell potential for the following reaction Calculate the ∆G for the reaction described in the problem

The anode compartment of a galvanic cell consists of a Zn anode immersed in 0.05M ZnCl2 The cathode compartment consists of a Cu cathode immersed in an 0.10M CuCl2 solution. Using the cell potentials found on page 815 of your textbook, calculate the cell potential for this reaction. Assume T = 298K.

Calculate the G in kJ for the following cell: Al(s)|Al3+(0.0035M)||Fe3+(0.55M)|Fe(s)  Eocell=+1.59 V Al(s)|Al3+(0.0035M)||Fe3+(0.55M)|Fe(s) Group of answer choices -473 -374 -158 -734 -658   Calculate Ecell in volts. Cu(s)|Cu2+(3.00x10-4)||Ag+(0.15M)|Zn(s)  Eocell=+0.620 V 0.700 0.544 0.944 0.644 0.800

Calculate the cell potential of a cell composed of a cathode consisting of a Co wire immersed in a 1.44 M Co²+ solution and an anode consisting of a Co wire immersed in a 5.60×10-5 M Co²+ solution. Assume the temperature of the solutions is 298 K. Cell potential: V

The free energy change for the following reaction at 25 °C, when [Zn²+] = 2.66×10³ M and [Cr³*] = 1.12 M, is 28.3 kJ: 3Zn2*(2.66x10-3 M) + 2Cr(s)→3Zn(s) + 2Cr³†(1.12 M) AG = 28.3 kJ What is the cell potential for the reaction as written under these conditions? Answer: V Would this reaction be spontaneous in the forward or the reverse direction?

An aqueous solution of an unknown salt of ruthenium is electrolyzed by a current of 3.00 A passing for 25.0 min. If 1.572 g Ru is produced at the cathode, what is the charge on the ruthenium ions in solution? Charge =|

A chemist designs a galvanic cell that uses these two half-reactions: half-reaction standard reduction potential 2+ Zn (aq)+2e - Zn(s) :-0.763 V (red |Cro (aq)+4 H2O(1)+3e Cr(OH)3(s)+5OH (aq) :-0.13 V = - (red Answer the following questions about this cell. Write a balanced equation for the half-reaction that happens at the cathode. x10 Write a balanced equation for the half-reaction that happens at the anode. Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous as written.

The free energy change for the following reaction at 25 °C, when [Cu2+] = 2.57x10-³ M and [I₂] = 1.10 M, is 88.8 kJ: 2Cu²+(2.57x10-3 M) + 2.001 (aq)—2Cu+ (aq) + 1₂(1.10 M) AG= 88.8 kJ What is the cell potential for the reaction as written under these conditions? Answer: V Would this reaction be spontaneous in the forward or the reverse direction?

Calculate the theoretical cell potential of the following cells. If the cell is short-circuited, indicate the direction of the spontaneous cell reaction. Zn/Zn²+ (0.1000 M) // Co²+(5.87 x 104 M)/ Co Ez2+ = -0.763V E Co2+ = -0.277V O +0.419 V spontaneous reaction +0.422 V spontaneous reaction O +0.421 V spontaneous reaction +0.420 V spontaneous reaction

+ The Nernst Equation The Nernst equation is one of the most important equations in electrochemistry. To calculate the cell potential at non-standard-state conditions, the equation i E=E 2.303 RT log10 Q where E is the potential in volts, Eo is the standard potential in volts, R is the gas constant, T is the temperature in kelvins, 12 is the number of moles of electrons transferred, Fis the Faraday constant, and is the reaction quotient. Using the common reference temperature, 25 °C or 298 K, the equation has the form E = ED (0.0592) log Q The reaction quotient has the usual form Q [products [reactants A table of standard reduction potentials gives the voltage at standard conditions, 1.00 M for all solutions and 1.00 atm for all gases. The Nernst equation allows for the calculation of the cell potential E at other conditions of concentration and pressure. Part A For the reaction 2Co²+ (aq) + 2Cl(aq) +2Co²+ (aq) + Ch₂(g). E 0.483 V what is the cell potential at 25°C if the…