Question

A battery is constructed from these two half reactions:

            i.   VO²⁺   +   2 H⁺ + e-   → V³⁺   + H₂O        E^o = 0.337 V vs SHE

            ii.   TiF₆^{2 ̶}   +   4 e-   →   Ti(s) + 6 F ^̶                           E^o = ̶ 1.191 V

(a) (0.2 pt) Which half reaction is in the anode compartment?   Circle one:     vanadium rxn      titanium rxn

(b) Which half reaction is in the cathode compartment? Circle one:    vanadium rxn      titanium rxn

(c) Write the balanced chemical equation for the overall battery reaction.

(d) Show the shorthand notation for representing this particular battery:

(e) Calculate the E^o of the battery.

Answer 1

Given :

Cell

VO²⁺   +   2 H⁺ + e-   -- > V³⁺   + H₂O        E^o = 0.337 V vs SHE

TiF₆^{2 ̶}   +   4 e-  -- > Ti(s) + 6 F ^̶                           E^o = ̶ 1.191 V

a). We use standard electrode potential to determine the anode and cathode.

The anode is the one on which oxidation reaction takes place and cathode is the one on which reduction takes place.

Anode has more negative standard reduction potential. So here Titanium rxn is an anode and Vanadium rxn is cathode.

b) As we have discussed there is oxidation takes place is anode compartment and reduction takes place in cathode compartment.

The reason is an anode is always above the cathode in electrochemical series and so it has potential to loose an electron than cathode.

c) To show balanced chemical equation we add oxidation and reduction half by making their electrons equal.

4VO²⁺   +   8 H⁺ +4 e-   -- > 4V³⁺   + 4H₂O       

TiF₆^{2 ̶}   +   4 e-   --- > Ti(s) + 6 F ^̶                

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4 VO²⁺ + 8 H⁺ + TiF₆^{2 ̶}    --- > 4 V³⁺   + 4 H₂O + Ti(s) + 6 F ^̶

  So the balanced equation is

4 VO²⁺ + 8 H⁺ + TiF₆^{2 ̶}    --- > 4 V³⁺   + 4 H₂O + Ti(s) + 6 F ^̶

d).

Short hand notation :

Ti | TiF6^2- (aq) || VO2²⁺ (aq) | V

e)

E₀ calculation:

E⁰_cell = Ecathode – E anode

= 0.337 V – ( 1.191 ) V

= 1.528 V

E_cell⁰ = 1.528 V