Question

Explain electrode potential and standard electrode potential. The following reaction takes place in a cell: \[ \text{Zn(s)} + \text{Co}^{2+} \rightleftharpoons \text{Co(s)} + \text{Zn}^{2+} \] Calculate the standard cell potential \(E^\circ_{cell}\) of the cell. Given: \(E^\circ_{(\text{Zn} \rightarrow \text{Zn}^{2+})} = 0.76\,\text{V}\) and \(E^\circ_{(\text{Co} \rightarrow \text{Co}^{2+})} = 0.28\,\text{V}\) (oxidation potentials).

Hint:

Zn is the anode (oxidation), Co is the cathode. Convert the given oxidation potentials to reduction potentials and use \(E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}\).

Answer 1

Definitions.
Electrode potential: When a metal electrode is dipped in a solution of its own ions, a potential difference is set up between the metal and the solution due to the tendency of the metal to lose electrons (oxidation) or to gain electrons (reduction). This potential difference is called the electrode potential.
Standard electrode potential \((E^\circ)\): It is the electrode potential measured under standard conditions, that is, ion concentration \(1\,\text{M}\), temperature \(298\,\text{K}\) and pressure \(1\,\text{bar}\), with respect to the standard hydrogen electrode (SHE) whose potential is taken as zero.

Numerical.
Step 1: Identify anode and cathode. In the cell reaction Zn is oxidised (loses electrons) so Zn is the anode; \(\text{Co}^{2+}\) is reduced so cobalt is the cathode.
Step 2: Formula. Using oxidation potentials, \[ E^\circ_{cell} = E^\circ_{oxidation(anode)} + E^\circ_{reduction(cathode)} \] The reduction potential is the negative of the given oxidation potential: \(E^\circ_{reduction(\text{Co}^{2+}/\text{Co})} = -0.28\,\text{V}\).
Step 3: Substitute. \[ E^\circ_{cell} = 0.76 + (-0.28) \]
Step 4: Arithmetic. \[ E^\circ_{cell} = 0.76 - 0.28 = 0.48\,\text{V} \]
\[\boxed{E^\circ_{cell} = 0.48\,\text{V}}\]
The positive value shows the reaction is spontaneous as written.