Question

The half-reactions that occur in ordinary alkaline batteries can be written as Cathode:MnO2(s)+H2O(l)+e−→MnO(OH)(s)+OH−(aq)Anode:Zn(s)+2OH−(aq)→Zn(OH)2(s)+2e− In 1999, researchers in Israel reported a new type of alkaline battery, called a "super-iron" battery. This battery uses the same anode reaction as an ordinary alkaline battery but involves the reduction of FeO2−4 ion (from K2FeO4) to solid Fe(OH)3 at the cathode. Part A Use the following standard reduction potential and any data from Appendixes C and D to calculate the standard cell potential expected for an ordinary alkaline battery: MnO(OH)(s)+H2O(l)+e−→Mn(OH)2(s)+OH−(aq) E∘ = -0.380V. E = V

Answer 1

Remember that each species will have a specific reduction potential. Remember that this is, as the name implies, a potential to reduce. We use it to compare it (numerical) with other species.

Note that the basis if 2H+ + 2e- -> H2(g) reduction. Therefore E° = 0 V

All other samples are based on this reference.

Find the Reduction Potential of each reaction (Tables)

Cathode:MnO2(s)+H2O(l)+e−→MnO(OH)(s)+OH−(aq) E = -0.380

Anode:Zn(s)+2OH−(aq)→Zn(OH)2(s)+2e− E = 1.249

The most positive has more potential to reduce, it will be reduced

The most negative will be oxidized, since it will donate it selectrons

For total E°cell potential:

E°cell = Ered – Eox

E|cell = -0.380 - 1.249 = -1.63 V