Question

Be sure to answer all parts.

Calculate E o , E, and ΔG for the following cell reactions:

(a) Mg(s) + Sn2+(aq) ⇌ Mg2+(aq) + Sn(s) where [Mg2+] = 0.045 M and [Sn2+] = 0.065 M

E ^o = _________ V

E = ____________ V

ΔG = __________kJ

(b) 3Zn(s) + 2Cr3+(aq) ⇌ 3Zn2+(aq)+ 2Cr(s) where [Cr3+] = 0.080 M and [Zn2+] = 0.0055 M

E ^o = __________ V

E = ____________ V

ΔG = ___________kJ

Answer 1

For such type of problems, applying Nernst equation;

E_cell = E^o_cell - {0.0591/n}log([product] /[reactant])

Where [product] and [reactant] are the concentration of products and reactants respectively in the reaction.

E^o_cell is the standard potential of the reaction.

The half cell reactions and their standard reduction potentials in the 1st problem are

Mg Mg²⁺ + 2 e^- , E^o_Mg²⁺/Mg = -2.37 Volts

Sn Sn²⁺ + 2 e^-    , E^o_Sn²⁺/Sn = -0.14 volts

Therefore,

E^o_cell = E^o_cathode - E^o_anode

_{The half cell reaction having more positive reduction potential acts as cathode and which has less, acts as anode.}

_Therefore,

E^o_cell = - 0.14 - (-2.37) = -0.14+2.37 = 2.23 volts

And, E_cell = 2.23 - {0.0591/2}*log(0.045/0.065)

E_cell = 2.23 - 0.005 = 2.225 volts

Finally,

G = - nE_cellF

Where 'n' is the no. of electron trasfer in the reaction

F is faraday constant having value equal to 96500.

Putting in the equation,

G = - 2*2.23*96500 = - 430.39 kJ

2). Similar procedure for 2nd problem,

Two half cell reactions and there standard reduction potentials are,

Zn²⁺ + 2e^-   Zn, E^o_Zn²⁺/Zn = - 0.76 volts

Cr³⁺ + 3e^- Cr, E^o_Cr³⁺/Cr = - 0.74 volts

Then, similarly as above,

E^o_cell = - 0.74-(-0.76) = 0.02 volts

For E_cell = 0.02 - (0.0591/6)*log(0.0055)²/(0.080)²

E_cell = 0.031 volts.

And G = - nE_cellF = - 6*0.031*96500 = - 17.95 kJ