Question

Two primary batteries are based on the cell reactions as follows:

2 MnO2 + Zn + 2NH4Cl -> Mn2O3 + Zn(NH3)2Cl2 + H2O
and
2Li + 2SO2 -> Li2S2O4
that have free energies of -229 Kj/mol of Zn and -294 Kj/mol of Li, respectively. Typical batteries have a capacity of 2 Ah (ampere hours). Estimate the battery voltages and the weights of the two metals needed to deliver this capacity. Do your calculations reveal any advantages for Li battery? (Atomic weights of Zn and Li are 65 and 7, respectively.)

Answer 1

The voltage of the MnO₂ and Zn-based primary battery is approximately 1.19 V, calculated from a given standard free energy change.

Step-by-step explanation:

To calculate the voltage of the primary battery based on the cell reaction 2 MnO2 + Zn + 2NH4Cl -> Mn2O3 + Zn(NH3)2Cl2 + H2O, the free energy (ΔG) is -229 kJ/mol Zn. Using the formula ΔG = -nFE to calculate the cell voltage (E), where n is the number of moles of electrons transferred, F is the Faraday constant (96,485 C/mol), and E is the cell voltage. Knowing 1 J = 1 C·V, we can solve for E. Assuming two electrons are exchanged (since Zn becomes Zn2+), we have E = ΔG / (nF) = (-229*1000) / (2*96485) V ≈ 1.19 V.