Question 21.3: Using Ehalf-cell° Values to Find Ecell° Problem Balance the ......
Using E^\circ_{\text{half-cell}} Values to Find E^\circ_{\text{cell}}
Problem Balance the following skeleton ionic reaction, and calculate E^\circ_{\text{cell}} to decide whether the reaction is spontaneous:
Mn^{2+}(aq) + Br_2(l) ⟶ MnO_4^−(aq) + Br^−(aq) [acidic solution]
Plan We balance the skeleton reaction (see Sample Problem 21.1). Then, we look up the E^\circ_{\text{half-cell}} values in Appendix D and use Equation 21.3 to find E^\circ_{\text{cell}}. If E^\circ_{\text{cell}} is positive, the reaction is spontaneous.
E^\circ_{\text{cell}} = E^\circ_{\text{cathode(reduction)}} − E^\circ_{\text{anode(oxidation)}} (21.3)
Solution Balancing the skeleton reaction:
2[Mn^{2+} + 4H_2O ⟶ MnO_4^− + 8H^+ + 5e^−] [oxidation]
5[Br_2 + 2e^− ⟶ 2Br^−] [reduction]
2Mn^{2+}(aq) + 5Br_2(l) + 8H_2O(l) ⟶ 2MnO_4^−(aq) + 10Br^−(aq) + 16H^+(aq) [overall]
Using E^\circ_{\text{half-cell}} values to find E^\circ_{\text{cell}}:
E^\circ_{\text{cell}} = E^\circ_{\text{cathode(reduction)}} − E^\circ_{\text{anode(oxidation)}} = 1.07 V − 1.51 V
= −0.44 V;\text{ not spontaneous}
Check Rounding to check the calculation: 1.00 V − 1.50 V = −0.50 V. A negative E^\circ_{\text{cell}} indicates a nonspontaneous reaction.