Question

Consider the half‑reaction for the reduction of Cd 2+ Cd2+ to Cd(s) Cd(s) . Cd 2 +(aq)+2 e − ⟶Cd(s) ? ∘ Cd 2+ /Cd =−0.403 V Cd2+(aq)+2e−⟶Cd(s)ECd2+/Cd∘=−0.403 V Calculate the potential of the cadmium electrode at 25 ∘ C 25 ∘C when immersed in a 0.0560 M 0.0560 M solution of CdBr 2 . CdBr2. ? Cd = ECd= V V Calculate the potential of the cadmium electrode at 25 ∘ C 25 ∘C when immersed in a 0.0330 M NaOH 0.0330 M NaOH solution that is saturated with Cd (OH) 2 . Cd(OH)2. The ? sp Ksp for Cd (OH) 2 Cd(OH)2 is 7.2× 10 −15 . 7.2×10−15. ? Cd = ECd= V V Calculate the potential of the cadmium electrode at 25 ∘ C 25 ∘C when immersed in a 0.0200 M Cd ( NH 3 ) 2+ 4 0.0200 M Cd(NH3)42+ and 0.125 M NH 3 0.125 M NH3 solution. The ? 4 β4 for Cd ( NH 3 ) 2+ 4 Cd(NH3)42+ is 3.63× 10 6 . 3.63×106. ? Cd = ECd= V

ⓘ Assignment Score: 1405/1500 Cx Give Up? 9 Hint Resources 2 Question 15 of 15> Cd +(aq) + 2eCd(s) E0.403 V Cd2+/Cd =V Calculate the potential of the cadmium electrode at 25 'C when immersed in a 0.0560 M solution of CdBr,. Cd= & TOOLS Calculat x10 with Cd(UH)2'The Ksp lor ca(UH, is 7.2 × 10-15 n electrode at 25°C when immersed in a 0.0330 M NaOH solution that is saturated Calculate the potential of the cadmium electrode at 25°C when immersed in a 0.0200 M Cd(NH3r and 0125 M NH, solution. The Pa for Cd(NH, is 3.63 x 10 1080 ace HD

Answer 1

a)

$[CdBr_{2}]=0.0560\: M$

$[Cd^{2+}]=0.0560\: M$

$E_{Cd}=E^{o}_{Cd}-\frac{0.059}{n}log(\frac{1}{[Cd^{2+}]})$

$E_{Cd}=-0.403-\frac{0.059}{2}log(\frac{1}{0.056})$.

$E_{Cd}=-0.4399\: V$

b)

$[OH^{-}]=0.033\: M$

$K_{sp}=[Cd^{2+}][OH^{-}]^{2}$

$7.2\times10^{-15}=[Cd^{2+}][0.033]^{2}$

$[Cd^{2+}]=6.6115\times10^{-12}M$

$E_{Cd}=E^{o}_{Cd}-\frac{0.059}{n}log(\frac{1}{[Cd^{2+}]})$

$E_{Cd}=-0.403-\frac{0.059}{2}log(\frac{1}{6.6115\times10^{-12}})$

$E_{Cd}=-0.7328\: V$