The nickel-iron battery has an iron anode, an NiO(OH) cathode, and a KOH electro

Question

The nickel-iron battery has an iron anode, an NiO(OH) cathode, and a KOH electrolyte. This battery uses the following half-reactions and has an E value of 1.37 V at 25C: Fe(s)+2OH(aq)Fe(OH)2(s)+2eNiO(OH)(s)+H2O(l)+eNi(OH)2(s)+OH(aq)

A. Calculate G (in kilojoules) and the equilibrium constant K for the cell reaction at 25C.

B. What is the cell voltage at 25 C when the concentration of KOH in the electrolyte is 5.0 M?

C. How many grams of Fe(OH)2 are formed at the anode when the battery produces a constant current of 0.255 A for 37.0 min ? mFe(OH)2=?

D. How many water molecules are consumed in the process? Nwater =?

Explanation / Answer

standard free energy G = - nFE o

                                          = 2 (electrons) x 96500 F x 1.37 v = -264.41 kj

we have G = -2.303 RT log Kc

Kc is equilibrium constant., = -264.41 / -2.303 RT = 50.1145

where R = 0.00831 J/DEGREE. KELVIN. T =273K

so Kc = anti logarthim of - 50. 1145 = 7.656 x 10 50

B) by Nernst equation = E = Eo - 0.0591/n {( log Fe(OH)2 / (OH)2 )}

concntration hydroxide = 2x concentration of KOH

but information about concentration of IRON is not given.

C) By faradays 1 law , Weight of Fe(OH)2 = quantity of current x electro chemical equivalent

                                                                    = 0.2555 A x 37x 60 seconds x 90/ 96500 x 2

MW of Fe = 56 , electro chemial eqivalent = MW/ valency x 96500 = 0.2645 grams of ferrous hydroxide is formed

as per redox equation 56 grams of Fe needs 2 moles of water.

         for 0.2645 grams of Fe needs (0.2645 / 56) x 2 moles = 0.0568 x 10 23 molecules of water

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