30. Calculate the emf of the following cell at 298 K: Al(s)/Al3+ (0.15M)//Cu2 (0.025M) /Cu(s) (Given E (Al3 /Al) -1.66 V, E (Cu2/Cu) 0.34V, log 0.15 = -0.8239, log 0.025 = -1.6020) %3D

Q: Part A What mass of Ni+ is produced in solution by passing a current of 67.0 A for a period of 14.0…

A:

Q: For a Chromium-Nickel cell [Cr3+ + 3e- --> Cr (-0.73)] & [Ni2+ +2e- --> Ni (-0.23)] 1M for both…

A:

Q: Electrochemistry. Consider the redoxreaction, for which E 8. Co S 1.10 v. Zn(s) Cuaq) - Zn (aq)…

A: Oxidation state or oxidation number of an atom refers to the charge on that atom that it would have…

Q: Calculate the Eell for the cell reaction at 25 °C. Take note, not E°.el- 3 Zn (s) + 2 Cr²+ (aq) 3…

A:

Q: Calculate the emf of the following cell at 25°C: Fe+ (a = 0.01), Fe2+ (a = 0.0001), Cut (a =…

A: Solution:- Provided data - Fe3+ = (a = 0.01), Fe2+ = (a = 0.0001), Cu+ = (a = 0.01) E°, Fe3+/Fe2+ =…

Q: The E° cell 0.135 V for the reaction B1,(s) + 5Cr20,2(aq)+ 34H (aq) – 6103 (aq) + 10Cr+(aq) + 17H20…

A: 1st we balanced the equation to find out what number of electrons involved in this reaction. then we…

Q: An iron(II) ion can act as both an oxidizing agent and a reducing agent. Hence, it can react with…

A:

Q: Use the standard reduction potentials at 25° C in Table 18.1 in Tro, Fridgen and Shaw, and calculate…

A:

Q: A concentration cell was made using a silver metal electrode in a saturated solution of AgI for the…

A:

Q: For the reaction: Pd2+ + Ni → Ni²+ + Pd , AG° is –239 kJ/mole. What is the standard reduction…

A: The tendency possessed by ions to get reduced in any reaction is termed as its "standard reduction…

Q: (i) Write the half-cell reaction, the net reaction and cell EMF of the following cell which operated…

A: “Since you have asked multiple questions, we will solve the first question for you. If youwant any…

Q: Porous Cu- disk Cu Cu" Cu" The concentration cell shown above employs the reaction Cu2+ + 2 e Cu E…

A: A concentration cell is composed of the same metals with different concentrations.

Q: An electrochemical cell is constructed with a Zn (s) electrode in contact with a 0.0520 M solution…

A: Given: Concentration of Zn2+ = 0.0520 M Concentration of Ag+ = 0.00420 M And temperature = 5.0 oC

Q: A concentration cell was made using a silver metal electrode in a saturated solution of AgCl for the…

A: There are two half reactions. One is cathodic where the reductions happens and the other one is…

Q: Write the cell diagram and calculate the e.m.f of a cell made by coupling D and E.

A: in the cell diagram (Cell representation ) left side ,Oxidation occur while right side reduction…

Q: What is the emf of the following cell at 25 °C? Ni(s) |Ni2*(1.00 M)||Fe3*(0.00500 M), Fe2*(0.010…

A:

Q: The following reaction proceeds spontaneously under standard conditions because E°cell > 0 (which…

A:

Q: For the galvanic cell at 298K 2Cr(s) + 3NI2*(aq) --> 2Cr3+(ag) + 3Ni(s) E°cell = 0.51 V What is the…

A: Oxidation is the process that involved the loss of electrons and oxidized to cation.  Hence when a…

Q: Many applications of electrochemistry have been in chemical analysis. Suppose a Fe|Fe+ concentration…

A:

Q: Calculate the emf of the following cell at 298 K: Al(s)/Al3+ (0.15M)//Cu2+ (0.025M) /Cu(s) Given Eo…

A: Welcome to bartleby !  Introduction :  We have to calculate Ecell of the reaction .

Q: Calculate the equilibrium constant, ?, for the reaction shown at 25 °C.…

A: Given Fe3+(aq)+3e−⟶Fe(s)   E = −0.04 V H3BO3 (s)+3H3O+(aq)+3e−⟶B(s)+6H2O(l) E = -0.8698 V  T = 298 K

Q: Calculate the maximum work that can be obtained from the Daniell cell Zn/Zn 2* (aq)// Cu2+/ Cu given…

A:

Q: The equilibrium constant, Kc, was found to be 1.3 x 10^3 at 24.6 C for the reaction.…

A: Here we are required to find the reduction potential of half reaction of X.

Q: d) For the electrochemical cell Cr(s) | Cr2O72-(aq) || Cu²*(aq) | Cu(s) the two half reactions have…

A: According to the guidelines, I am posting answer for the first question. posst other question…

Q: A solution containing Pt* is electrolyzed with a current of 5.80 A. How long will it take to plate…

A: Moles of Pt4+ to be electrolyzed: = 99% × (0.20 L)  × 0.025 M= 0.99 × (0.20)  × 0.025 mol= 0.00495…

Q: Q1) Calculate the e.m.f for the cell at 27°C Pt.| F2 (g, 0.6bar) |F (aq.0.2M) || F (aq.0.03M) |F2…

A: First of all we have to write both equation at cathode and anode...  Formula used :-  Ecell = E°cell…

Q: A solution containing Pt** is electrolyzed with a current of 3.00 A. How long will it take to plate…

A: Introduction : given : Current=3 A  Volume = 0.50 L  Molarity=0.010M

Q: A galvanic cell consists of a Cr+/Cr half-cell with unknown [Cr³+] and a Ni²+/Ni half-cell with…

A: GALVANIC CELL: It is a type of an electrochemical cell which is utilized to provide electricity by…

Q: At 25 °C, calculate the voltage of the cell 2Ag*(aq, 0.10M) + Cu(s) → Cu²*(aq, 0.010M) + 2Ag(s) If…

A:

Q: Based on the information in the table of standard reduction potentials below, calculate ΔGo for the…

A: the redox reaction of bromine with potassium will be  2 K + Br2 -------> 2 KBr  Hence from the…

Q: At 25°C calculate the voltage of the cell 2Ag*(aq)(0.010M) + Cu(s) Cu2+lag) (0.10M) + 2Ag(s) !3! If…

A: Consider the oxidation and reduction reaction as:  Oxidation: Cus→Cu2+aq+2e- Reduction: 2Ag+aq+2e-→2…

Q: A galvanic cell is constructed as follows. One half-cell consists of a platinum wire immersed in a…

A:

Q: candard reduction potentials to calculate the equilibrium constant for the reaction: Ni2+(aq) +…

A:

Q: What is [Fe2*] if E = 1.64 V, [Zn2+] = 0.25 M, and [Fe3+] = 0.75 M at 25 °C for the below voltaic…

A: To solve the above problem, we need to use the Nernst equation and then solve for concentration of…

Q: Calculate the cell potential for the galvanic cell in which the reaction Ti(s) + With(so much) =…

A:

Q: 29) Calculate the emf of the cell at 25.0 °C given by the cell diagram: Ni(s) | Ni²* (0.0750 M) ||…

A:

Q: Given that standard reduction potentials at 298.15 K are E° Ag+ /Ag = +0.800 V and E° Cu?"/Cu =…

A: When Ag foil is soaked into Cu(NO3)2 solution, for Ag to dissolve the reaction involved would be:…

Q: Using tne following half- cell values from the two equations Fe*(aq) + 2e Fe(s); Eo =-0.44 V Cu*(aq)…

A:

Q: Fe(s) I (Fe2+(0.100 M) || Pd2+(1.0 x 10-5 M) I Pd(s) given that the standard reduction potential for…

A:

Q: Calculate the Ecell between two Cu2+/Cu half-cells, one of which contains 0.001 M Cu2+ and the other…

A: Ecell is defined as the potential difference between the two half cells. The concentration of two…

Q: concentration cell similar to the one shown is composed of two Mn ectrodes and solutions of…

A:

Q: The standard reduction potential (E°) for the conversion of Cr,0 to Cr* at 25°C in an aqueous…

A: Cr2O7 2- + 14H+ + 6e- ------------------> 2Cr3+  + 7H2O            E0(Cr6+/Cr3+)   = 1.33 V  +6…

Q: 5(a) Use the standard reduction potentials at 25° C in Table 18.1 in Tro, Fridgen and Shaw, and…

A: Given,  Mn + Cd2+ → Mn2+ + Cd the standard emf E° of an electrochemical cell for the reaction is:

Q: A concentration cell was made using a silver metal electrode in a saturated solution of AgBr for the…

A:

Q: Calculate the emf of the cell : (Ans. 0.2706 V) OCr/Cr** (0.1M)//Fe**(0.01M)/Fe*). Given, E°cr/Cr =…

A:

Q: For the cell diagram below, what is the equilibrium constant K for the reaction at 25 °C? * Pt(s) |…

A: We are given a cell notation of an electrochemical cell. With standard reduction potential values of…

Q: Calculate the emf of the Daniell cell at 25.0°C when the concentrations of Cuso4 and ZnSOo4 are…

A: Answer-1 Cell reaction will be: Zn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s) half cell reduction potential for Zn…

Q: For a Chromium-Nickel cell [Cr3+ + 3e- --> Cr (-0.73)] & [Ni2+ +2e- --> Ni (-0.23)] 1 M for both:…

A: Please look for detailed solution in step 2 .

Q: 1) For the following combination of electrodes and solutions, indicate, 0.340 V a) Anode, cathode…

A: E° Cu2+/Cu is 0.34 V E° Ag+/Ag is 0.79 V Ag+ gets reduced and Cu gets Oxidised.

Q: 5. A student-produced a plot of Ecei vs. In[Cu2-] which had an equation of y = -0.0977x + 0.0057.…

A: