Answer to Question #258126 in General Chemistry for Wolf@OT7

Question

Calculate the equilibrium constant for redox reactions that could occur in the following situations and use that value to explain whether or not any reaction will be observed. A piece of iron is palced in a 1.0 M solutions of NiCl2(aq). A copper wire is palced in a 1.0 M solution of Pb (NO3)2 (aq).

Answer 1

(a) Piece of iron dipped in 1 M solution of nickel chloride:

When a piece of iron is placed in 1 M solution of nickel chloride, iron will try to replace nickel ions from the solution, which would cause iron to form iron(II) ions while nickel would be deposited by getting reduced to metallic form.

The cell reaction is as shown below:

Fe(s) + Ni²⁺(aq) → Fe²⁺(aq) + Ni(s)

Here iron metal is oxidized to iron(II) metal ion. So it represents reaction at anode.

And nickel(II) metal ion is reduced to nickel metal. So it represents reaction at cathode.

The standard reduction potential values for both are given as:

$E^0_{red} = -0.25 \;V \\ E^0_{ox} = -0.44 \;V \\ E^0_{cell} = E^0_{red} -E^0_{ox} \\ = (-0.25) -(-0.44) \\ = 0.19 \;V$

Calculation of equilibrium rate constant:

Number of electrons involved,

n=2

R, Gas Constant

R=8.314 kJ/mol K

T = 298 K

$log K = \frac{nFE^0}{2.303RT} \\ = \frac{2 \times 96485 \times 0.19}{2.303 \times 8.314 \times 10^3 \times 298} \\ = 6.42 \times 10^{-3}$

Taking anti-log, we get,

K = 3 \times 10^6

Thus, value of equilibrium rate constant for this reaction is $3 \times 10^6$

The cell potential for this reaction is greater than zero so it will be spontaneous and fast.

(b) Wire of copper dipped in 1 M solution of lead(II) nitrate:

When a copper wire is placed in 1 M solution of lead(II) nitrate, copper will try to replace lead ions from the solution which would cause copper to form copper(II) ions while lead would be deposited by getting reduced to metallic form.

The cell reaction is as shown below:

C(s) + Pb²⁺(aq) → Cu²⁺(aq) +Pb(s)

Here copper metal is oxidized to copper(II) metal ion. So it represents reaction at anode.

And lead(II) metal ion is reduced to lead metal. So it represents reaction at cathode.

The standard reduction potential values for both are given as:

$E^0_{red} = -0.126 \;V \\ E^0_{ox} = 0.337 \;V \\ E^0_{cell} = E^0_{red} -E^0_{ox} \\ = (-0.126) -(0.337) \\ = -0.463 \;V$

Number of electrons involved,

n=2

R, Gas Constant

R=8.314 kJ/mol K

T = 298 K

$log K = \frac{nFE^0}{2.303RT} \\ = \frac{2 \times 96485 \times (-0.463)}{2.303 \times 8.314 \times 10^3 \times 298} \\ = -1.5 \times 10^{-3}$

Taking anti-log, we get,

$K = 2 \times 10^{-16}$

Thus, value of equilibrium rate constant for this reaction is $2 \times 10^{-16}$

The cell potential for this reaction is near to zero so it will be spontaneous but not so fast.