Question

Calculate the cell potential, the equilibrium constant, and the free energy change for

Ba(s)+Mn2+(aq,1M)→Ba2+(aq,1M)+Mn(s)
Given the following E degree values:
Ba2+(aq)+2e→Ba(s), E degree=−2.90V
Mn2+(aq)+2e→Mn(s),E degree=0.80V

Answer 1

Step-by-step explanation:

Ba(s) + Mn²⁺ (aq,1M) → Ba²⁺ (aq,1M) + Mn(s)

Ba⁺²(aq) +2e → Ba(s) , E° = −2.90 V

Mn⁺²(aq) +2e → Mn(s), E⁰ =0.80 V

Anode reaction :

Ba(s) → Ba⁺²(aq) +2e E° = −2.90 V

Cathode reaction :

Mn⁺²(aq) +2e → Mn(s) E⁰ =0.80 V

Cell potential = Ecathode - Eanode

Ecell = .80 - ( - 2.90 )

Ecell = 3.7 V .

equilibrium constant ( K ) :

Ecell = .059 log K / n

n = 2

3.7 = .059 log K / 2

log K = 125.42

K = 2.63 x 10¹²⁵ .

Free energy change :

ΔG = - n F Ecell

= - 2 x 96500 x 3.7

= 714100 J

= 7.141 x 10⁵ J .