Question

3. Calculate the redox potential (in volts) for a LiMnO₂/graphite battery.

(Redox reaction)

MnO₄^- + 8H + 5e^- → Mn²⁺ (aq) + 4H₂O (+1.51 V)

MnO₂ + 4H +2e^- →  Mn²⁺ (aq) + 2H₂O (+1.22 V)

xLi⁺ + C6 + e^- → LixC₅ (-3.00 V)

1. Write balanced equations for the following processes: a. The reaction of potassium with water. C. Thermal decomposition of sodium azide. d. The reaction of potassium peroxide with water. e. Calcium hydride reacting with water. f. The reaction between ethanol and lithium hydride. 2. What are products of the following reaction? (note: the reactants are not necessarily balanced) KO2+H2O → NaHCO3 + heat → NaNH2 + H20 2. LiC.Hg + C2H5OH → 3. Calculate the redox potential (in volts) for a LiMnO/graphite battery? +1.51V MnO4 +8H+ 5e-Mn2+ (aq) + 4H20 MnO2 + 4H+ 2e → Mn2(aq) + 2H2O xLit+C6+ e- LixCs +1.22V -3V Answer:

Answer 1

Question 3.

The redox potential (E^o_cell) = E^o_cathode - E^o_anode

In the given cell, MnO2 is the anode, whereas graphite is the cathode.

Therefore, E^o_cell = -3 V - (1.22 V) = -4.22 V

Answer 2

E^o_cell= E^oreduction - E^o_oxidation

_{Oxidation Reaction:-}

xLi⁺ + C₆ + e^- → LixC₅  (This is not the correct oxidation reaction)

Reversing the equation we get:-

LixC₅ → xLi⁺ + C₆ + e^-  E^o_oxidation= + 3 V

_{Reduction Reaction:-}

MnO₂ + 4H⁺ +2e^- →  Mn²⁺ (aq) + 2H₂O (l)  E^o_reduction= 1.22V

E^o_cell = 1.22 -3 V

= -1.78 V