Question

1). What is the value of the equilibrium constant at 25 ^oC for the reaction between the pair:

Sn(s) and Pb²⁺(aq) to give Pb(s) and Sn²⁺(aq)

Use the reduction potential values for Sn²⁺(aq) of -0.14 V and for Pb²⁺(aq) of -0.13 V

2). What is the value of ΔG^o in kJ at 25 ^oC for the reaction between the pair:

Sn(s) and Pb²⁺(aq) to give Pb(s) and Sn²⁺(aq)

Use the reduction potential values for Sn²⁺(aq) of -0.14 V and for Pb²⁺(aq) of -0.13 V

Answers: 1): 2

2): -1.93

I keep getting 5 for one and -3.8 for 2. Please show me the work for these :/

Answer 1

1)

Eo(Sn2+/Sn(s)) = -0.14 V

Eo(Pb2+/Pb(s)) = -0.13 V

As per given reaction/cell notation,

cathode is (Pb2+/Pb(s))

anode is (Sn2+/Sn(s))

Eocell = Eocathode - Eoanode

= (-0.13) - (-0.14)

= 0.01 V

here, number of electrons being transferred, n = 2

Eo = (2.303*R*T)/(n*F) log Kc

At 25 oC or 298 K, R*T/F = 0.0592

So, Eo = (0.0592/n)*log Kc

0.01 = (0.0592/2)*log Kc

log Kc = 0.3378

Kc = 2.2

Answer: 2 (after rounding to 1 significant figure)

2)

number of electrons being transferred, n = 2

F = 96500.0 C

use:

ΔG = -n*F*E

= -2*96500.0*0.01

= -1930 J/mol

= -1.93 KJ/mol

Answer: -1.93