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Title: In a dry cell, MnO2 occurs in a semisolid electrolyte paste and is reduced at th Post by: randomwolfe on May 9, 2014 In a dry cell, MnO2 occurs in a semisolid electrolyte paste and is reduced at the cathode. The MnO2 used in dry cells can itself be produced by an electrochemical process of which one half-reaction is shown below.
Mn2+(aq) + 2 H2O(l) MnO2(s) + 4 H+(aq) + 2 e - If a current of 26.5 A is used, how many hours are needed to produce 2.75 kg of MnO2? Title: Re: In a dry cell, MnO2 occurs in a semisolid electrolyte paste and is reduced at th Post by: skip5284 on May 9, 2014 Similar question:
In a dry cell, MnO2 occurs in a semisolid electrolyte paste and is reduced at the cathode. The MnO2 used in dry cells can itself be produced by an electrochemical process of which one half-reaction is shown below. Mn2+(aq) + 2 H2O(l) MnO2(s) + 4 H+(aq) + 2 e - If a current of 30.5 A is used, how many hours are needed to produce 1.15 kg of MnO2? Answer: Mn2+(aq) + 2 H2O(l) => MnO2(s) + 4 H+(aq) + 2 e- Moles of MnO2 = mass/molar mass = 1150/86.9368 = 13.228 mol Moles of electrons required = 2 x moles of MnO2 = 2 x 13.228 = 26.456 mol Total charge = moles of electrons x Faraday constant = 26.456 x 96485 = 2.5526 x 106 C Time = total charge/current = 2.5526 x 106/30.5 = 8.3692 x 104 s = 23.2 hours |