Question

A cell and its emf is given below:
Pt (s) $\mid$ H$_2$(g, 1 bar) $\mid$ H$^+$(aq, 1 M) $\parallel$ Cu$^{2+}$(aq, 1 M) $\mid$ Cu (s)
emf of the cell = +0.34 V
Write the reduction half-reaction at cathode.

Hint:

Always remember:
- Oxidation occurs at the anode (left side of cell notation)
- Reduction occurs at the cathode (right side of cell notation)
Identify the species gaining electrons for the reduction half-reaction.

Answer 1

Step 1: The given electrochemical cell is:
Pt (s) $\mid$ H$_2$(g, 1 bar) $\mid$ H$^+$(aq, 1 M) $\parallel$ Cu$^{2+}$(aq, 1 M) $\mid$ Cu(s)
Step 2: According to standard notation:
- The left-hand side (LHS) represents the anode (oxidation).
- The right-hand side (RHS) represents the cathode (reduction).
Step 3: At the cathode, Cu$^{2+}$ ions in solution accept electrons and get reduced to solid Cu.
Reduction half-reaction at cathode:
\[\text{Cu}^{2+} (aq) + 2e^- \rightarrow \text{Cu} (s)\]