Question

Given: Cr3+(aq)+3e−⇌Cr(s); E°=−0.74 V Pb2+(aq)+2e−⇌Pb(s); E°=−0.13  V What is the equilibrium constant (K) at  25°C  for...

Given:

Cr3+(aq)+3e−⇌Cr(s); E°=−0.74 V

Pb2+(aq)+2e−⇌Pb(s); E°=−0.13  V

What is the equilibrium constant (K) at  25°C  for the following cell reaction?

2Cr(s)+3Pb2+(aq)→3Pb(s)+2Cr3+(aq)

Select one:

a. 7.4 ×10^61

b. 2.1×10^26

c. 2.9×10^44

d. 8.33x10^56

Homework Answers

Answer #1

we have:

Eo(Cr3+/Cr(s)) = -0.74 V

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

here:

cathode is (Pb2+/Pb(s))

anode is (Cr3+/Cr(s))

The chemical reaction taking place is

Pb2+(aq) + Cr(s) --> Pb(s) + Cr3+(aq)

Eocell = Eocathode - Eoanode

= (-0.13) - (-0.74)

= 0.61 V

here, number of electrons being transferred, n = 6

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

0.61 = (0.0592/6)*log Kc

log Kc = 61.8243

Kc = 6.673*10^61

Answer: a

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