An electrochemical cell based on the reaction; M(s) + Cu2+(aq) → M2+(aq) + Cu(s) E˚ = 1.52 V is cons — Electrochemistry Chemistry Question
Problem Context
An electrochemical cell based on the reaction;
M(s) + Cu2+(aq) → M2+(aq) + Cu(s) E˚ = 1.52 V
is constructed using equal volumes of solutions with all substances in their standard states.
Use the value of the reduction potential of Cu2+(aq) (E˚ = 0.34 V) to determine the standard reduction potential for the reaction;
M2+(aq) + 2e– → M(s)
Model Answer
E˚cell = E˚ox + E˚red and E˚ox = 1.52 V – 0.34 V = 1.18 V
so E˚red(M2+ + 2e– → M) = –1.18 V
The cell is allowed to discharge until the [Cu2+] = 0.10 M. Find
i. the M2+ concentration in moles per liter,
Model Answer
If Cu2+ decreases to 0.10 M then M2+ must increase to 1.90 M
ii. the cell potential, E.
Model Answer
= 1.52 – 0.01285 ln(19) = 1.52 – 0.0378 = 1.48 V
50 mL of distilled water is added to each cell compartment of the original cell. Compare the potential of the cell after the addition of water with the potential of the original cell. Explain your answer.
Model Answer
The E of the cell with dilute solutions will be the same as the original E˚. Because the solutions are diluted by the same amount and the ions have the same coefficients (from the balanced chemical equation), Q in the Nernst equation is 1, and lnQ = 0.