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13 Dec 2019

5. Alkaline batteries operate in a potassium hydroxide liquid electrolyte which allows current to flow at room temperature. The two half-cell reactions, written as reduction reactions, are given below along with their standard potentials in the alkaline solution. Zn(OH)2+ 2e Zn 20H Vo 1.25 V vo 0.30 V 2Mno2 2H20 2e 2 MnC (OH) 20H a) What material is the anode in an AA battery? b) What is the valence of the cathode cation after reduction? c) Write the overall reaction for the battery. d) Calculate the value of deltaG (in kJ/mol) for the overall reaction. e) Calculate the value of the equilibrium constant at room temperature. f Calculate deltaG for Mno(OH) formation at room temperature. (Hint: Assume that Zn(OH)2 decomposes to ZnO and H20 without changing the voltage (deltaV) of the overall reaction). Values for deltaG at room temperature are -237.14 kJ/mol for water, -320.476 kJ/mol for zinc oxide, and -465.138 kJ/mol for manganese dioxide.

I just need help with part F

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Related questions

Mercuric oxide dry-cell batteries are often used where a flat discharge voltage and long life are required, such as in watches and cameras. The two half-cell reactions that occur in the battery are

HgO(s) + H2O(l) + 2 e-β†’ Hg(l) + 2 OH-(aq)

Zn(s) + 2 OH-(aq)β†’ ZnO(s) + H2O(l) + 2 e-

a) Write the overall cell reaction. (b) The value of EΒ°red for the cathode reaction is +0.098 V. The overall cell potential is +1.35 V. Assuming that both half-cells operate under standard conditions, what is the standard reduction potential for the anode reaction? (c) Why is the potential of the anode reaction different than would be expected if the reaction occurred in an acidic medium?

One Class Solution:-

a) Hg(s) +Zn (s) β†’ Hg(l) + ZnO(s)

b) E0=E0(cathode)-E0(anode)=0.098-1.35= -1.25V

c) Er red. is different from Zn 2+ (aq) + 2e-β†’ Zn(s), (-0.76 V) because in the battery the process happens in the presence of base and Zn2+ is stabilized as ZnO(s).

Stabilization of a reactant in a half-reaction decreases the driving force, so Er is more negative.

You are asked to construct a voltaic cell that would simulate an alkaline battery by providing an electrical output of 1.50 VV at the beginning of its discharge. After you complete it, your voltaic cell will be used to power an external device that draws a constant current of 0.50 AA (amperes) for 2.0 hh. You are given the following supplies: electrodes of each transition metal from manganese to zinc; the chloride salts of the +2 transition metal ions from Mn2+Mn2+ to Zn2+Zn2+ (MnCl2MnCl2, FeCl2FeCl2, CoCl2CoCl2, NiCl2NiCl2, CuCl2CuCl2, and ZnCl2ZnCl2); two 100 mLmL beakers; a salt bridge; a voltmeter; and wires to make electrical connections between the electrodes and the voltmeter. The standard reduction potentials for the divalent metal ions are as follows: Half-reactionMn2+(aq)Fe2+(aq)Co2+(aq)Ni2+(aq)Cu2+(aq)Zn2+(aq)++++++2eβˆ’2eβˆ’2eβˆ’2eβˆ’2eβˆ’2eβˆ’β†’β†’β†’β†’β†’β†’Mn(s)Fe(s)Co(s)Ni(s)Cu(s)Zn(s)E∘(V)βˆ’1.18βˆ’0.44βˆ’0.28βˆ’0.250.34βˆ’0.76 Since you were tasked with constructing a cell that produces a potential of 1.50 VV, you decide to fine tune your concentrations to minimize the error. If you used the same concentration for each electrolyte, your voltaic cell would produce a potential of 1.52 VV. Therefore, you should adjust the concentration ratio that would lower your cell potential by 0.02 VV. You and a fellow researcher decide that you should prepare two 100 mLmL solutions for the voltaic cell. Your fellow researcher already prepared one of the electrolyte solutions for you: 0.228 MM CuCl2 CuCl2. At what concentration should you prepare the MnCl2MnCl2 solution such that your initial voltaic cell potential is 1.50 VV? Assume that the temperature is consistently 25 ∘C∘C. If the battery is permitted to discharge for a period of two hours at a constant current of 0.50 AA, what will be the concentration of the manganese(II) ion in the anode electrolyte solution at the end of the discharge? Similarly, what will be the concentration of the copper(II) ion in the cathode electrolyte solution at the end of the discharge? Given the nature of how a cell is constructed, you would usually deplete the ions at the cathode before completely oxidizing the anode (since the latter would break the circuit). This is because the complete oxidation of the anode would likely only occur if you were using an extremely thin wire of the metal. In this case, let us assume you constructed the voltaic cell using sufficiently sized rods for each metal. This means that the total time you can run your cell will depend on the initial copper ion concentration. Calculate the maximum run time for your voltaic cell if it runs at 0.5 AA and your initial concentration of CuCl2(aq)CuCl2(aq) was 0.228 M'

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