Question

A galvanic cell is constructed in which a ${\mathbf{\text{Pt|Fe}}}^{\mathbf{\text{2+}}}{\mathbf{\text{Fe}}}^{\mathbf{\text{3+}}}$half-cell is connected to a${\mathbf{\text{Cd}}}^{\mathbf{\text{2+}}}\mathbf{\text{|Cd}}$half-cell.

(a) Referring to Appendix E, write balanced chemical equations for the half-reactions at the anode and the cathode and for the overall cell reaction.

(b) Calculate the cell potential, assuming that all reactants and products are in their standard states.

Short answer

1. Balanced equation for the half-cell reaction at the anode –

${\text{Fe}}^{\text{+2}}\text{(aq)}\to {\text{Fe}}^{\text{+3}}{\text{(aq)+e}}^{\text{-}}\frac{1}{2}$

Balanced equation for the half-cell reaction at the cathode –

${\text{Cd}}^{\text{+2}}{\text{(aq)+2e}}^{\text{-}}\to \text{Cd(aq)}$

The overall cell reaction is –

${\text{2Fe}}^{\text{+2}}{\text{(aq)+Cd}}^{\text{+2}}\text{(aq)}\to {\text{2Fe}}^{\text{+3}}\text{(aq)+Cd(aq)}$

1. Assuming that all reactants and products are in their standard states, the cell potential is ${\text{E}}_{\text{cell}}^{\text{°}}\text{=0.3674V}$.

Step-by-step solution

Concept Introduction

An oxidation process in which electrons are lost or a reduction reaction in which electrons are gained is known as a half-cell reaction. The processes take place in an electrochemical cell, where electrons are lost at the anode via oxidation and consumed at the cathode via reduction.

Balanced Equation for half-cell Reaction

The balanced cathode half-cell reaction for the reduction of$\text{Cd}$–

${\text{Cd}}^{\text{+2}}{\text{(aq)+2e}}^{\text{-}}\to \text{Cd(aq)}$

The anode half-cell reaction for the oxidation of$\text{Fe}$–

${\text{Fe}}^{\text{+2}}\text{(aq)}\to {\text{Fe}}^{\text{+3}}{\text{(aq)+e}}^{\text{-}}$

The overall reaction is –

${\text{2Fe}}^{\text{+2}}{\text{(aq)+Cd}}^{\text{+2}}\text{(aq)}\to {\text{2Fe}}^{\text{+3}}\text{(aq)+Cd(aq)}$

Therefore, the overall reaction is${\text{2Fe}}^{\text{+2}}{\text{(aq)+Cd}}^{\text{+2}}\text{(aq)}\to {\text{2Fe}}^{\text{+3}}\text{(aq)+Cd(aq)}$

Calculation for Cell Potential

The formula for Cell Potential is –

${\text{E}}_{\text{cell}}^{\text{°}}{\text{=E}}_{\text{cathode}}^{\text{°}}{\text{-E}}_{\text{anode}}^{\text{°}}\mathrm{...}\left(1\right)$

The cell potential for Cathode is:${\text{E}}_{\text{cathode}}^{\text{°}}\text{=-0.4026V}$.

The cell potential for Anode is:${\text{E}}_{\text{anode}}^{\text{°}}\text{=0.77V}$.

Plugging in the values in Equation –

$\begin{array}{c}{\text{E}}_{\text{cell}}^{\text{°}}\text{=-0.4026-(-0.77V)}\\ \text{=0.3674V}\end{array}$

Therefore, the value for the cell potential is obtained as ${\text{E}}_{\text{cell}}^{\text{°}}\text{=0.3674V}$.