Question 21.4: Calculating an Unknown Ehalf-cell° from Ecell° Problem A vol......
Calculating an Unknown E^\circ_{\text{half-cell}} from E^\circ_{\text{cell}}
Problem A voltaic cell houses the reaction between aqueous bromine and zinc metal:
Br_2(aq) + Zn(s) ⟶ Zn^{2+}(aq) + 2Br^−(aq) E^\circ_{\text{cell}} = 1.83 VCalculate E^\circ_{\text{bromine}}, given E^\circ_{\text{zinc}} = −0.76 V.
Plan E^\circ_{\text{cell}} is positive, so the reaction is spontaneous as written. By dividing the reaction into half-reactions, we see that Br_2 is reduced and Zn is oxidized; thus, the zinc half-cell contains the anode. We use Equation 21.3 and the known E^\circ_{\text{zinc}} to find E^\circ_{\text{unknown}} (E^\circ_{\text{bromine}}).
E^\circ_{\text{cell}} = E^\circ_{\text{cathode(reduction)}} − E^\circ_{\text{anode(oxidation)}} (21.3)
Solution Dividing the reaction into half-reactions:
Br_2(aq) + 2e^− ⟶ 2 Br^−(aq) E^\circ_{\text{unknown}} = E^\circ_{\text{bromine}} = ? V
Zn(s) ⟶ Zn^{2+}(aq) + 2e^− E^\circ_{\text{zinc}} = −0.76 V
Calculating E^\circ_{\text{bromine}}:
E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} − E^\circ_{\text{anode}} = E^\circ_{\text{bromine}} − E^\circ_{\text{zinc}}
E^\circ_{\text{bromine}} = E^\circ_{\text{cell}} + E^\circ_{\text{zinc}} = 1.83 V + (−0.76 V) = 1.07 V
Check A good check is to make sure that calculating E^\circ_{\text{bromine}} − E^\circ_{\text{zinc}} gives E^\circ_{\text{cell}}: 1.07 V − (−0.76 V) = 1.83 V.
Comment Keep in mind that, whichever half-cell potential is unknown, reduction is the cathode half-reaction and oxidation is the anode half-reaction. Always subtract E^\circ_{\text{anode}} from E^\circ_{\text{cathode}} to get E^\circ_{\text{cell}}. (We use E^\circ_{\text{bromine}} in Follow-up Problem 21.4A.)