Question

Please use the following to answer the below:

A fictional galvanic (voltaic) cell consists of an electrode composed of a metal, M, in a 1.0 mol/L M²⁺ ion solution and a second electrode composed of an unreactive platinum metal (Pt) in a 1.0 mol/L X^– ion solution, connected by a salt bridge and an external wire at 25.0 ⁰C. Consider the two entries below from a fictional table of standard reduction potentials.

Reduction Equation	Standard reduction potentials (E⁰reduction)
M2+ (aq) + 2e–   → M (s)	– 0.30 V
X2 (g) + 2e–   → 2 X– (aq)	+1.49 V

a) Determine the following in this cell.

            Metal serving as anode electrode          _________________

            Metal serving as cathode electrode       _________________

Oxidized species: _________________

Reduced species: _________________

Oxidizing agent:   _________________

Reducing agent:    _________________

b) Write the balanced net ionic equation for this cell. Include state notations.                    

c) Calculate the standard emf (Eº_cell) for this cell.

d) Write the cell diagram notation for the cell.                                                                            

e) Calculate the Standard Free Energy (ΔGº) for this cell. (Faraday’s Constant = 96500 J/(V.mol).                                                       

  

f) Will the equilibrium constant (K) for this cell be ˃˃1 or ˂˂1?   

Answer 1

Solution : Two dectro de consist of of a metal (M) in 1.0 mol/lit M2+ solution, and the other one is composed of Pt metal in a 1.0 mol/lit x-solution. Halt cell reactions, M2+ (09) +2e - → M(S) E' -030 valt and X₂(g) +2e 2x- (aq) E' - +1.49 volt. more the value of reduction potential indicates its ability to reduce. Therefore, in cathode reduction occured = +1.49 v Cathode X2(g) 420 > 2x- (aq) and in anode oxidation occured M2+ (aq) tac. ..E anode M(S) anode 20.30v anode ya) metal serving as and Pt metal serving as cathode 个 metal is oxidized to m2+. (oxidised state) (reduced state). and ya is reduced to X2 oxidizing agent : reducing agent Metal um (6) Net ionic equation, X2(g) + M(s) → 2x (aq) + M M2+ (aq). ) E° E E° cathode = and de 1.49-(-0.30) 1.79v

(d ) Cell diagram: M/ M2 + 11x, 1 x (Pt) anode ( Cathode salt bridge Gº = en FE cell n> 11 no. of electron tramper, (2x 96500x 1.79) Inol there n=2) - 345.47X103 (Answer) I more! as we know &Gº= RT Ink = nFE eu :: RT INK nFE Ka equilibrium contant. or, luk NFE call RT 2 x 96500 x 1.79 8.314 x 298 Ink - 139.44 K or, 4.94 ( Answer) equilibrian contant MI