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

Step 1: Meaning of the terms. The electrode potential is the tendency of an electrode to lose or gain electrons when in contact with a solution of its ions, measured as a voltage. When this is measured at unit concentration (1 M), 298 K and 1 bar against the standard hydrogen electrode (assigned 0 V), it is the standard electrode potential \(E^\circ\).
Step 2: Convert to standard reduction potentials. The data are oxidation potentials, so the corresponding reduction potentials are their negatives: \(E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76\,\text{V}\) and \(E^\circ_{\text{Co}^{2+}/\text{Co}} = -0.28\,\text{V}\).
Step 3: Cathode and anode. The species reduced (\(\text{Co}^{2+}\to\text{Co}\)) is the cathode; the species oxidised (\(\text{Zn}\to\text{Zn}^{2+}\)) is the anode.
Step 4: Apply the reduction-potential formula. \(E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = (-0.28) - (-0.76)\).
Step 5: Compute. \(E^\circ_{cell} = -0.28 + 0.76 = 0.48\,\text{V}\).
\[\boxed{E^\circ_{cell} = +0.48\,\text{V}}\]
Because \(E^\circ_{cell}\) is positive, the cell reaction proceeds spontaneously in the given direction.