Chemistry 9701 · AS & A Level · Electrochemistry

Electrochemistry — practice question

An electrochemical cell is made up from two half-cells: • a $\text{Br}_2/\text{Br}^-$ half-cell • an $\text{Mn}^{3+}/\text{Mn}^{2+}$ half-cell
(a)[1]

State the electrode material for each half-cell. $\text{Br}_2/\text{Br}^-$ half-cell $\text{Mn}^{3+}/\text{Mn}^{2+}$ half-cell

(b(i))[2]

Apply the Nernst equation to determine the electrode potential, $E$, of the $\text{Mn}^{3+}/\text{Mn}^{2+}$ half-cell in these conditions.

(b(ii))[1]

Calculate the value of $E_{\text{cell}}$ under these conditions.

(b(iii))[2]

Write an equation for the overall cell reaction.

(c)[5]

An aqueous copper(II) sulfate solution is electrolysed with copper electrodes. A current of $1.50\,\text{A}$ is passed for $3.00\,\text{hours}$. $5.09\,\text{g}$ of copper is deposited on the cathode. The charge on one electron is $-1.60 \times 10^{-19}\,\text{C}$. The relative atomic mass of copper is $63.5$. Use these data to calculate an experimentally determined value for the Avogadro constant, $L$. Give your answer to three significant figures.

(d)[2]

Explain why magnesium metal is not obtained by electrolysis of dilute aqueous magnesium sulfate. Your response should use data from the Data Booklet.

Worked solution & mark scheme

This 13-mark question has a full step-by-step worked solution and mark scheme. One marking point: Platinum for both half-cells

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