Electrowinning-current efficiency of zinc and its influencing factors

Current efficiency and its influencing factors In production practice, 1 mol of zinc is not precipitated by the 1 Faraday charge on the cathode. This is because while the precipitated zinc metal, as well as impurities and hydrogen evolution, the cathode deposit and dissolution of the oxide, and the drain electrode and the short-circuit losses. Therefore, the problem of effective current utilization, that is, current efficiency, is proposed.
In the industry, the actual precipitation of the actual amount of zinc in the electrowinning process and the percentage of zinc that should theoretically be precipitated by the same amount of electricity represent the current efficiency. The calculation method is as follows:

Actual material mass current efficiency = ——————————— × 100%
The mass of matter that should theoretically be precipitated is ηi = m/(q · I · t · n)
Where ηi - cathode current efficiency, %;
M——the actual production of zinc precipitation in t time, g;
I——through the current between the anode and cathode, A;
T——electric accumulation time, h;
N——the number of electric accumulators;
q - the electrochemical equivalent of zinc, 1.220 g / (A · h).
In the production practice, the current efficiency of zinc electrowinning is different due to different specific conditions. In modern wet zinc smelting industry, the current efficiency fluctuates between 85% and 95%.
Slot voltage and power consumption A tank voltage The tank voltage is the value of the voltage drop between adjacent anodes and anodes in the grid. It can be expressed as the measured voltage drop between each pair of anodes and cathodes. However, in the production practice, since the number of electrowinning tanks is large, the number of anodes and cathodes is more, and the voltage drop between each pair of anodes and cathodes varies depending on the specific conditions, so the measurement method is not used, but The voltage drop of the conductive plate is subtracted from the total voltage supplied to all series of electrical cells, divided by the total number of slots on the series circuit, and the resulting quotient is the cell voltage. The formula is expressed as:

V ₁ -V ₂
V _slot = ————
N

Where V-slot voltage;
V ₁ - the total voltage of all series electric accumulators;
V ₂ - the voltage drop of the conductive plate;
N - the total number of electrolytic cells.
The voltage of an electric accumulator consists of the theoretical decomposition voltage of zinc sulfate, the overvoltage, the voltage drop of the electrolyte solution, and the voltage drop caused by the contact resistance of the wiring, the resistance of the plate, the anode mud resistance, etc. For: [next]

V-groove = E ₊ - E _- + IR _pole + IR _fluid + IR _mud + IR _connection

Wherein ( E ₊ - E _- ) - electrode polarization potential;
IR _pole - plate resistance voltage drop;
IR _liquid - electric _effluent resistance voltage drop;
IR _mud - anode mud resistance voltage drop;
IR _connection - contact resistance voltage drop.
The electrode polarization potential (E ₊ - E _- ) includes the theoretical decomposition voltage and overvoltage of zinc sulfate. The polarization potential of the electrowinning liquid containing H ₂ SO ₄ of 115 g/L and Zn ²⁺ of 54 g/L at a current density of 500 A/m ² at 40 ° C is now discussed .
In the solution, the concentration of Zn ²⁺ is C _{Zn ²⁺} = 0.826mo1/L, and the activity coefficient is 0.053, then a _{Zn ²⁺} =0.053 × 0.826=0:0438. According to the formula of polarization potential, the cathode process should be Have:

E _Zn = E ^Ó© _{Zn ²⁺} +RT/nFlga _{Zn ²⁺} - η _Zn
=0.763+0.031051g(0.0438)- η _Zn
=-0.805 - η _Zn

In the formula, η _Zn represents a precipitation overvoltage of zinc, which is known to be equal to 0.03 V under given conditions.
Therefore, it can be obtained:

η _Zn = -0.805 - 0.03 = -0.835V

In the same solution, the H ⁺ concentration is C _{H ⁺} = 2.345mo1/L, and the OH ^- ion concentration C _OH ^- = K _w /C _{H ⁺} = 3 × 10 ^-14 /2.345 = 1.279 × 10 ^-14 mol/L , the activity coefficient is equal to 0.75, then a _OH - = 0.75 × 1.279 × 10 ^-14 = 9.60 × 10 ^-15 , therefore, the polarization potential of the anode process is:

E _O2 = E ^Ó© _O2 - RT/F1ga _OH ^- + η ₀₂
=1.272+ η ₀₂

Where η ₀₂ represents the precipitation overvoltage of oxygen on the anode lead under electrowinning conditions and is known to be equal to 0.993V. Therefore, it can be obtained

Ñ„O ₂ = 1.272 + 0.993 = 2.265V

Based on the above calculation results, the polarization potential difference V _{pole of the cathode and cathode} can be obtained. [next]

V _pole = 2.265 - (-0.835) = 3.10V

Electrolyte can rely on ion conduction, but it is much more resistant than metal conductors. When the current passes through the effusion, it will inevitably cause a voltage drop. Its magnitude is proportional to the current density, the distance between the anode and the cathode, and the resistivity of the effusion. It can be expressed by the following formula:

V _liquid = IR _liquid = J·pL

Where V _liquid - electric _effluent resistance voltage drop, V;
J——cathode current density, A/mz;
P——electrical resistivity, Ω · m;
L - the distance between the anode and the cathode, m.
The table below shows the specific resistance values ​​of acidic zinc sulfate solutions of different compositions at 40 °C . From the value of the known resistivity, the voltage drop of the solution between the two electrodes can be calculated. Usually, the V _{liquid is} between 0.4 and 0.6.

Lead- silver anode plates, rods and conductive heads all have a certain resistance, resulting in a voltage drop, typically around 0.02V. Aluminum cathode plate, the conductive rod has a certain resistance, but also the voltage drop, is also about about 0.02V. The cathode-anode contact has a contact voltage drop at the contact point of about 0.03V. Since such contact conductive heads are tens of millions in industrial production, it is of great practical significance to strive to reduce the contact voltage drop for energy conservation. In actual operation, it must be noted that each contact point conducts well. As the electrolytic deposition progresses, the anode surface inevitably generates anode mud (Mn0 ₂ ), which consumes a part of the voltage, which is between 0.02 and 0.08 V. In order to reduce the voltage consumed by the formation of the anode mud, measures for periodically cleaning the anode surface of the anode surface are taken in the production, but the first day after the brushing of the anode often leads to turbidity of the electrolyzed liquid, which leads to an increase in the lead content of the precipitated zinc.
The sum of the above five resistance voltages is the tank voltage of the electric storage tank. It is between 3.2 and 3.6V. The cell voltage is determined by the current density, the acidity and temperature of the effusion, and the distance between the electrodes, in addition to the contact resistance. Therefore, the way to reduce the cell voltage is to reduce the resistivity of the electrolyte, shorten the distance between the electrodes, and reduce the voltage loss at the contact point. [next]
B. Electrical energy efficiency Electrical energy efficiency is the ratio of the amount of electrical energy required to produce a certain amount of metal in electrowinning production, which is theoretically necessary to be actually consumed.

The theoretically necessary electrical energy e _O of a certain mass of matter precipitated
η _e = ——————————————————— × 100%
Precipitating the electrical energy actually consumed by the same mass of matter e

Where η _e - electrical energy efficiency, %.
Since electric energy = electricity × voltage,

e _O = amount of electricity necessary to deposit metal I ^ө t × theoretical decomposition voltage V- _groove
e = total power through the cell It × slot voltage V _slot
^I ө · t · V _reason ^I ө · V _Li
η _e = ————— × 100% = ————
I · t· V _slot I · V _slot

Where I ^ө /I = Ηi is the current efficiency. _Li V / V _groove = η _V, that is, the voltage efficiency.

That is, electrical energy efficiency (η _e ) = current efficiency (η _i ) × voltage efficiency (η _V )

In order to improve the efficiency of electric energy, in addition to improving the current efficiency, the voltage efficiency is also increased. The improvement method is to reduce the electrolyte resistance, appropriately increase the temperature of the electrolyte, shorten the pole pitch to reduce the polarization of the electrode, and reduce the voltage of the tank.

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