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Question 11.5: A 12-V lead–acid battery delivers a current of 60 A. Calcula......

A 12-V lead–acid battery delivers a current of 60 A. Calculate (a) the battery out-put power, (b) the number of moles of sulfuric acid required for anode and cathode reactions, and (c) the sulfuric acid consumption rate at the anode.

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1. Battery output power

P=V\,I=12\ V\times60\ \mathrm{A}=720\ \mathrm{W}

2. At the cathode of the battery, n = 2 mol of electrons are released per mole of sulfuric acid. Thus, the number of moles of sulfuric acid required per unit time for anode and cathode reactions

{ N}_{\mathrm{acid}}\,=\,2I/(n F)\,=\,2\times60\ {\mathrm{A}}/(2\times94,687\ {\mathrm{C/mol}})\,=\,6.34\times10^{-4}\ {\mathrm{mol/s}}

where F = 96,487 C/mol is Faraday’s constant and n = 2 mol electrons/mol sulfuric acid.

3. Mass flow rate of sulfuric acid (molar mass M_{acid} = 98 g/mol) is

m=N_{\mathrm{acid}}M_{\mathrm{acid}}=6.34\times10^{-4}\ \mathrm{mol/s}\times98\ \mathrm{g/mol}=0.06\ \mathrm{g/s}=0.216\ \mathrm{kg/h}

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