ElectrochemistryHard

Question

The reactions taking place in the dry cell are:

Anode: Zn → Zn2+ + 2e−

Cathode: 2MnO2 + 2NH4+ + 2e− → Mn2O3 + 2NH3+ H2O

The minimum mass of reactants, if a dry cell is to generate 0.25A for 9.65 h, are (Mn = 55, Zn = 65.4) (neglect any other chemical reactions occurring in the cell)

Options

A.2.943 g Zn
B.7.83 g MnO2
C.1.62 g NH4+
D.3.915 g MnO2

Solution

Moles of electron involved = $\frac{0.25 \times 9.65 \times 3600}{96500} = 0.09$

∴ Mass of Zn involved = $\frac{0.09}{2} \times 65.4 = 2.943\text{ gm}$

Mass of MnO2 involved = 0.09 × 87 = 7.83 gm

Mass of NH4+ involved = 0.09 × 18 = 1.62 gm

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