(Solved): A Short Problem The electrowinning of cobalt from a sulfate solution is normally carried out at a c ...

A Short Problem The electrowinning of cobalt from a sulfate solution is normally carried out at a current density of $300\mathrm{A}/{\mathrm{m}}^2$ and a cell voltage of $5.0\mathrm{V}$ with a current efficiency of $85\%$. It was found that the current efficiency increased to $90\%$ if the current density was increased to $350\mathrm{A}/{\mathrm{m}}^2$ but the cell voltage also increased to 5.3V. You are asked to assess with justification which option requires the least electrical energy/unit mass of metal produced. Problem It has been suggested that the current efficiency for the electrowinning of zinc from acidic sulfate solutions is reduced due to the simultaneous reduction of oxygen (produced by the anodic reaction) at the cathode. - Estimate the limiting cathodic current density (in $\mathrm{A}/{\mathrm{m}}^2$ ) due to the reduction of oxygen if the concentration of dissolved oxygen is $10^{-3}$ $\mathrm{mol}/\mathrm{L}$ and the mass transfer coefficient at the cathode is $2\times 10^{-3}\mathrm{cm}/\mathrm{s}$. - If the total current density is $400\mathrm{A}/{\mathrm{m}}^2$, calculate the cathodic C.E. - Calculate the annual cost of the electrical power required to reduce the oxygen in a plant producing 100000 t/annum of zinc under the above conditions in cells operating at $4\mathrm{V}$. Atomic mass of $\mathrm{Zn}$ : $65$ Electrical power cost: \$0.15/kwh.