Consider the concentration cell consisting of two Cu2+(aq)|Cu(s) half-cells with different molaritie — Electrochemistry Chemistry Question
Problem Context
Consider the concentration cell consisting of two Cu2+(aq)|Cu(s) half-cells with different molarities. The [Cu2+] in the two half-cells are 2.0 M and 2.5 × 10–2 M, respectively.
[Cu2+(aq) + 2e– → Cu(s) E˚ = 0.34 V]
State the E˚ value for the concentration cell and explain why it has this value.
Model Answer
E0 is 0 because the 2 half-cells have the same material.
Calculate the cell potential, E, for the cell with the two concentrations given.
Model Answer
E = 0.056V
Identify the anode of this cell and explain your reasoning.
Model Answer
The anode is the half-cell with [Cu2+] = 2.5×10-2. Oxidation occurs more readily in this half-cell because the [Cu2+] is lower.
For each half-cell, predict whether the [Cu2+] will increase or decrease as the cell operates.
Model Answer
[Cu2+] will increase in the half-cell with [Cu2+]init = 2.5×10-2M and decrease in half-cell with [Cu2+] = 2.0M
For the half reaction V3+(aq) + e– → V2+(aq), E˚ = –0.26 V.
i. Write a balanced equation for the reaction in a standard voltaic cell made with the V3+(aq)/V2+(aq) and Cu2+(aq)/Cu(s) half-cells and calculate the E˚ value for this cell.
Model Answer
2V2+ + Cu2+ → 2V3+ + Cu E0 = 0.34V − (-0.26V) E0 = 0.60V
ii. Identify the Cu2+(aq)| Cu(s) half-cell (2.0 M or 2.5 × 10–2 M) that would yield the greater E value in combination with the V3+(aq)/V2+(aq) half-cell. Explain.
Model Answer
Because Cu2+ is undergoing reduction in this cell (and Cu2+ is a reactant) the higher [Cu2+](2.0M) will give the greater E value.
iii. Write an expression that could be used to calculate the E value for specific [Cu2+], [V2+], and [V3+].