A solution is made by adding some CuSO4•5H2O to water, followed by enough sulfuric acid to make the — Electrochemistry Chemistry Question
Problem Context
A solution is made by adding some CuSO4•5H2O to water, followed by enough sulfuric acid to make the pH = 1.00. Two platinum electrodes are then placed in the solution (which has a total volume of 0.500 L) and the solution is electrolyzed at a constant current of 0.120 A, separately capturing any gases that are evolved at the two electrodes.
Initially, gas is evolved at the anode, but no gas is evolved at the cathode. However, after 10.0 min of electrolysis, gas evolution begins at the cathode as well, and eventually the total volume of gas evolved at the cathode is equal to the total volume of gas evolved at the anode.
Half-Reaction Eº, V Half-Reaction Eº, V
Cu+(aq) + e– → Cu(s) 0.518 Cu2+(aq) + 2 e– → Cu(s) 0.337
O2(g) + 2 H+(aq) + 2 e– → H2O2(aq) 1.780 O2(g) + 4 H+(aq) + 4 e– → 2 H2O(l) 1.229
2 H+(aq) + e– → H2(g) 0.000
Write the balanced reaction that takes place initially in this electrolytic cell.
Model Answer
2 Cu2+(aq) + 2 H2O(l) → 2 Cu(s) + O2(g) + 4 H+(aq)
How many moles of copper(II) sulfate pentahydrate were initially added to the solution?
Model Answer
Gas evolution begins at the cathode when the Cu(II) is exhausted from the solution. 0.120 A × 600. s = 72.0 C
72.0 C / 96500 C mol-1 = 7.46 × 10-4 mol electrons
Since each mol of Cu(s) deposited requires 2 mol electrons, there must have initially been (7.46 × 10-4 mol)/2 = 3.73 × 10-4 mol CuSO4
What is the initial cell potential for this reaction at 298 K? (In air, the partial pressure of O2(g) is 0.20 bar and the partial pressure of H2(g) may be taken to be 10-4 bar.)
Model Answer
E° = 0.337 V – 1.229 V = –0.892 V
From the Nernst equation, E = Eº – (RT/nF)ln([PO2][H+]4/[Cu2+]2)
= –0.892 V – (0.0591/4)log([0.20][0.1]4/[7.46 × 10-4]2) = –0.915 V
As the electrolysis proceeds (before t = 10.0 min), will this cell potential become more positive, more negative, or remain unchanged? Explain your answer.
Model Answer
As electrolysis proceeds, the reactants are depleted and the products increase in concentration. These will both contribute to an increasingly negative cell potential.
What gas is evolved at the anode? What gas is evolved at the cathode (after t = 10.0 min)?
Model Answer
Oxygen is evolved at the anode and hydrogen at the cathode.
At what time will the total volumes of evolved gases at the two electrodes be equal?
Model Answer
One mole of O2 is evolved for every four electrons, while one mole of H2 is evolved for every two moles of electrons. Thus, once H2 evolution begins, in ten minutes as much volume of H2 will be evolved as O2 is evolved in twenty minutes. In other words, the two volumes will be equal at t = 20.0 min.