Question

A voltaic cell that uses the reaction PdCl42−(aq)+Cd(s) → Pd(s)+4Cl−(aq)+Cd2+(aq) has a measured standard cell potential of +1.03 V. You may want to reference (Pages 860 - 867) Section 20.4 while completing this problem. Part A Write the half-cell reaction at the cathode.

Answer 1

Answer: The half-cell reaction occurring at cathode is $PdCl_4^{2-}(aq)+2e^-\rightarrow Pd(s)+4Cl^-(aq)$

Explanation:

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

Reduction reaction is defined as the reaction in which a chemical specie accepts electrons. The oxidation state of the chemical specie reduces.

The given balanced chemical equation is:

$PdCl_4^{2-}(aq)+Cd(s)\rightarrow Pd(s)+4Cl^-(aq)+Cd^{2+}(aq)$

The half cell reaction occurring at cathode follows:

$PdCl_4^{2-}(aq)+2e^-\rightarrow Pd(s)+4Cl^-(aq)$

Hence, the half-cell reaction occurring at cathode is given above.