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

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 .

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