Chapter 20 Practice Set

Separate the overall reaction into two half-reactions, one for oxidation and one for reduction: oxidation: Br?(aq)?BrO3?(aq) reduction: MnO4?(aq)?MnO2(s) Balance each half-reaction with respect to mass and charge: Br?(aq)+3H2O(l)?BrO3?(aq)+6H+(aq)+6e? MnO4?(aq)+2H2O(l)+3e??MnO2(s)+4OH?(aq) Multiply each half-reaction by a coefficient to make sure the numbers of electrons in the half-reactions are equal, and add the half-reactions together. Finally, the balanced redox reaction is 2MnO4?(aq)+H2O(l)+Br?(aq)?2MnO2(s)+2OH?(aq)+BrO3?(aq)

Separate the overall reaction into two half-reactions, one for oxidation and one for reduction: oxidation: Ag(s)?Ag(CN)2?(aq) reduction: O2(g)?OH?(aq) Balance each half-reaction with respect to mass and charge: Ag(s)+2CN?(aq)?Ag(CN)2?(aq)+e? O2(g)+2H2O(l)+4e??4OH?(aq) Multiply each half-reaction by a coefficient to make sure the numbers of electrons in the two half-reactions are equal, and add the half-reactions together. Finally, the balanced redox reaction is 4Ag(s)+8CN?(aq)+O2(g)+2H2O(l)?4Ag(CN)2?(aq)+4OH?(aq)

Separate the overall reaction into two half-reactions, one for oxidation and one for reduction: oxidation: Al(s)?AlO2?(aq) reduction: NO2?(aq)?NH3(g) Balance each half-reaction with respect to mass and charge: Al(s)+4OH?(aq)?AlO2?(aq)+2H2O(l)+3e? NO2?(aq)+5H2O(l)+6e??NH3(g)+7OH?(aq) Multiply each half-reaction by a coefficient to make sure the numbers of electrons in the half-reactions are equal, and add the half-reactions together. Finally, the balanced redox reaction is NO2?(aq)+H2O(l)+2Al(s)+OH?(aq)?NH3(g)+2AlO2?(aq)

Separate the reaction into oxidation and reduction half-reactions: oxidation: Mg(s)?Mg2+(aq)+2e? reduction: Ni2+(aq)+2e??Ni(s) Look up the standard electrode potentials for each half-reaction. Add the half-cell reactions together to obtain the overall redox equation: oxidation (anode): Mg(s)?Mg2+(aq)+2e? E?=?2.37V reduction (cathode):Ni2+(aq)+2e??Ni(s) E?=?0.23V Ni2+(aq)+Mg(s) ? Ni(s)+Mg2+(aq) Calculate the standard cell potential by subtracting the electrode potential of the anode from the electrode potential of the cathode: E?cell=E?cathode?E?anode (?0.23V)?(?2.37V) = 2.14V

Separate the reaction into oxidation and reduction half-reactions: oxidation: Ni(s)?Ni2+(aq)+2e? reduction: 2H+(aq)+2e??H2(g) Look up the standard electrode potentials for each half-reaction. Add the half-cell reactions together to obtain the overall redox equation: oxidation (anode): Ni(s) ?Ni2+(aq)+2e? E?=?0.23V reduction (cathode): 2H+(aq)+2e? ? H2(g) E?=0.00V 2H+(aq)+Ni(s) ? H2(g)+Ni2+(aq) Calculate the standard cell potential by subtracting the electrode potential of the anode from the electrode potential of the cathode: E?cell = E?cathode ? E?anode (0.00V)?(?0.23V) = 0.23V

Separate the reaction into oxidation and reduction half-reactions: oxidation: Cu(s) ? Cu2+(aq) +2e? reduction: NO?3(aq)+4H+(aq) +3e? ? NO(g)+2H2O(l) Look up the standard electrode potentials for each half-reaction. Add the half-cell reactions together to obtain the overall redox equation: oxidation (anode): Cu(s)?Cu2+(aq) +2e? E?=0.34V reduction (cathode):NO?3(aq)+4H+(aq)+3e? ? NO(g)+2H2O(l) E?=0.96V Cu(s)+NO?3(aq)+4H+(aq) ? Cu2+(aq)+NO(g)+2H2O(l) Calculate the standard cell potential by subtracting the electrode potential of the anode from the electrode potential of the cathode: E?cell=E?cathode?E?anode (0.96V)?(0.34V)=0.62V

To label the voltaic cell, start by writing out the half-reactions that occur at each half-cell and look up the standard reduction potentials. Anode: Pb(s)?Pb2+(aq)+2e? E?red=?0.13 V Cathode:Cl2(g)+2e??2Cl?(aq) E?red= 1.36 V Since this is a voltaic cell, cell potentials must be assigned to give a positive E?cell . This means that to have a positive cell potential, the cathode needs to have a higher reduction potential value. The electrode where oxidation occurs is the anode (left side) and the electrode where reduction occurs is the cathode (right side). Electrons flow out of the anode into the cathode.

Add the two-half reactions and cancel electrons. Pb(s)+Cl2(g)?Pb2+(aq)+2Cl?(aq)

E?cell = 1.49V

Fe(s)+Mn2+(aq) ? Fe2+(aq)+Mn(s) Mg2+(aq)+Fe(s) ? Mg(s)+Fe2+(aq)

Use the equation relating standard cell potentials and the equilibrium constant, E?cell=0.0592VnlogK where E?cell is the standard cell potential, n is the number of moles of electrons transferred in the redox reaction, and K is the equilibrium constant for the balanced redox reaction at 25 ?C . Rearrange the equation and solve for K : K=10[(n)(E?cell)/0.0592] K= 6.1x10^8

E?cell = ?0.036 V ? (?2.37 V) = 2.33