Identify the cathode in this diagram - HSC - SSCE Chemistry - Question 19 - 2003 - Paper 1

To write the net redox equation, we identify the half-reactions for both the oxidation and reduction processes:

1. Oxidation (at the anode):

   $\text{Pb} \rightarrow \text{Pb}^{2+} + 2e^-$

2. Reduction (at the cathode):

   $\text{Ag}^+ + e^- \rightarrow \text{Ag}$

To combine the half-reactions, we need to balance the number of electrons:

• Multiply the reduction reaction by 2:

  $2\cdot \left(\text{Ag}^+ + e^- \rightarrow \text{Ag}\right) \Rightarrow 2\text{Ag}^+ + 2e^- \rightarrow 2\text{Ag}$

Now, combining both half-reactions gives:

$\text{Pb} + 2\text{Ag}^+ \rightarrow \text{Pb}^{2+} + 2\text{Ag}$

Cell Potential Calculation: The standard cell potential (E^*) can be calculated using standard reduction potentials:

• E_ox for Pb: -0.13 V
• E_red for Ag: +0.80 V

The cell potential is given by:

$E^* = E_{red} - E_{ox} = 0.80V - (-0.13V) = 0.93V$

Thus, the net redox equation is:

$\text{Pb} + 2\text{Ag}^+ \rightarrow \text{Pb}^{2+} + 2\text{Ag}$

And the cell potential E^* is 0.93 V.