The E\u00b0 values for two electrodes are shown - AQA - A-Level Chemistry - Question 13 - 2019 - Paper 3

To find the EMF of the cell, we use the formula:

$E_{cell} = E_{cathode} - E_{anode}$

In this case:

• The cathode is the copper half-cell (Cu²⁺/Cu) with a standard reduction potential of +0.34 V.
• The anode is the iron half-cell (Fe³⁺/Fe) with a standard reduction potential of -0.44 V.

Substituting the values into the equation:

$E_{cell} = 0.34 V - (-0.44 V)$

This simplifies to:

$E_{cell} = 0.34 V + 0.44 V = 0.78 V$

Therefore, the EMF of the cell is +0.78 V.