Question

A voltaic cell with an aqueous electrolyte is based on the reaction between Cd2+(aq) and Mg(s), producing Cd(s) and Mg2+(aq). Write half-reactions for the anode and cathode and then write a balanced cell reaction. Please include the states of matter in the equations.

Answer 1

half-reactions

cathode : Cd²⁺ (aq) + 2e⁻ ---> Cd (s)

anode :  Mg (s) → Mg²⁺ (aq) + 2e−

a balanced cell reaction

Cd²⁺(aq) + Mg(s)→ Cd(s) + Mg²⁺ (aq)

Further explanation

Cell potential (E °) is the potential difference between the two electrodes in an electrochemical cell.

Electric current moves from a high potential pole to a low potential, so the cell potential is the difference between an electrode that has a high electrode potential (cathode) and an electrode that has a low electrode potential (anode)

$\large {\boxed {\bold {E ^ osel = E ^ ocatode -E ^ oanode}}}$

or:

E ° cell = E ° reduction-E ° oxidation

(At the cathode the reduction reaction occurs, the anode oxidation reaction occurs)

The value of E cells uses a reference electrode which is used as a comparison called the Standard Electrode which is the hydrogen-platinum electrode

In reaction:

Cd²⁺ + Mg → Cd + Mg²⁺

half-reactions

•  at the cathode (reduction reaction) Cd²⁺ (aq) + 2e⁻ ---> Cd (s)

•   at the anode (oxidation reaction) Mg (s) → Mg²⁺ (aq) + 2e−

a balanced cell reaction

Cd²⁺(aq) + Mg(s)→ Cd(s) + Mg²⁺ (aq)

Learn more

The standard cell potential

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Answer 2

Answer : The balanced two-half reactions will be,

Oxidation half reaction (anode) : $Mg(s)\rightarrow Mg^{2+}(aq)+2e^-$

Reduction half reaction (cathode) : $Cd^{2+}(aq)+2e^-\rightarrow Cd(s)$

Thus the overall reaction will be,

$Mg(s)+Cd^{2+}(aq)\rightarrow Mg^{2+}(aq)+Cd(s)$

Explanation :

Voltaic cell : It is defined as a device which is used for the conversion of the chemical energy produces in a redox reaction into the electrical energy. It is also known as the galvanic cell or electrochemical cell.

The given redox reaction occurs between the magnesium and cadmium.

In the voltaic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

The balanced two-half reactions will be,

Oxidation half reaction (anode) : $Mg(s)\rightarrow Mg^{2+}(aq)+2e^-$

Reduction half reaction (cathode) : $Cd^{2+}(aq)+2e^-\rightarrow Cd(s)$

Thus the overall reaction will be,

$Mg(s)+Cd^{2+}(aq)\rightarrow Mg^{2+}(aq)+Cd(s)$