Chapter 17: Q12P (page 428)

What cathode potential (versus S.H.E.) is required to reduce 99.99%of cd(II) from a solution containing 0.10Mcd (II) in 1,0M ammonia if there is negligible current? Consider the following reactions and assume that nearly all (II) is in the form $Cd {({NH}_{3})}_{4}^{2 +}$
localid="1663647104121" $C d^{2 +} + 4 N H_{3} ⇌ C d {(N H_{3})}_{4}^{2 +} β_{4} = 3.6 \times 10^{6} C d^{2 +} + 2 e^{-} ⇌ C d (s) E ° = - . 402 V$

Short Answer

Expert verified

The potential of cathode is E=-0.744 V

Step by step solution

Electrogravimetric Analysis

Separation along with the quantification of metals are mainly observed in electrogravimetric analysis.
Electrolysis of the analyte solution is done during this procedure.

Determine the lead lactate

Given data:

$[C d (I I)] = 0.10 M [N H_{3}] = 1 M β_{4} = 3.6 \times 10^{6} E ° = - 0.402 V E = ?$

First we need to calculate $[C d^{2 +}]$

$β_{4} = \frac{[C d {(N H_{3})}_{4}^{2 +}]}{[C d^{2 +}] {[N H_{3}]}^{4}}$

99.99% Of Cd is reduced, so the concentration of $C d {(N H_{3})}_{4}^{2 +}$ is then:

$[C d {(N H_{3})}_{4}^{2 +}] = [C d (I I)] - ([C d (l l)] . 0.9999) [C d {(N H_{3})}_{4}^{2 +}] = 0.10 M - (0.10 M . 0.9999) [C d {(N H_{3})}_{4}^{2 +}] = 1 \times 10^{- 5} M$

The concentration of $C d^{2 +}$ is:

$[C d^{2 +}] = \frac{[C d {(N H_{3})}_{4}^{2 +}]}{{[N H_{3}]}^{4} . β} [C d^{2 +}] = \frac{1 \times 10^{- 5} M}{1^{4} M \times 3.6 \times 10^{6}} [C d^{2 +}] = 2.78 \times 10^{- 12} M$

The potential of cathode is

$E (c a t h o d) = E ° - \frac{0.05916}{2} . \log (\frac{1}{[C d^{2 +}]}) E (c a t h o d) = - 0.402 V - \frac{0.05916}{2} . \log (\frac{1}{[2.78 . 10^{- 12}]}) E (c a t h o d) = - 0.744 V$