Explain how the salt bridge provides an electrical connection between the two solutions - AQA - A-Level Chemistry - Question 6 - 2018 - Paper 1

The electrode potential for the left-hand electrode can be calculated using the equation: $E_{cell} = E_{right} - E_{left}$ Where:

• $E_{cell} = +0.16 \ V$
• $E_{right} = +0.34 \ V$ Thus, we rearrange to find: $E_{left} = E_{right} - E_{cell} = +0.34 \ V - +0.16 \ V = +0.18 \ V$

The left-hand electrode does not have an electrode potential of +0.34 V because the concentration of Cu extsuperscript{2+} ions in the left half-cell is significantly different from that needed to achieve this potential.

The conventional representation of the cell is: $Cu(s) | Cu^{2+}(aq, 0.15 \ mol \ dm^{-3}) || Cu^{2+}(aq, 1.0 \ mol \ dm^{-3}) | Cu(s)$

The concentration of copper(II) ions in the left-hand electrode increases as copper dissolves from the anode. The EMF of the cell decreases to 0 V because the concentrations of Cu extsuperscript{2+} ions in the two solutions become equal, resulting in no further potential difference.