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Chapter 17: Q31P (page 430)

Suppose that the diffusion current in a polarogram for reduction of ${Cd}^{2 +}$ at a mercury electrode is $14 渭础$ If the solution containsof 25mLof $0.50 {mMCd}^{2 +}$ what percentage of ${Cd}^{2 +}$ is reduced in the 3.4 min required to scan from $- 0.6 to - 1.2 V$ ?

Short Answer

Expert verified

The percentage of cadmium ion which can be reduced in polarogram is 0.12%.

Step by step solution

01

Define Molarity.

At any point in an electric circuit the charge is the product of current I and time.

q = l . t

The number of moles of electronsreactreacts in a reaction with a given timeis

Reacted moles $= \frac{l . t}{nF}$

02

Determine the Molarity of Cadmium.

Given,

$l = 14 脳 10^{- 6} A t = 3.4 \min = 204 \sec$

Number of moles of electrons is given as

role="math" $= \frac{(14 脳 10^{- 6} A) (204 \sec)}{96485 C / mol} = 2.9 脳 10^{- 8} mol$

The reduced cadmium ion is calculated as $= \frac{2.9 脳 10^{- 8} mol}{2} = 1.48 脳 10^{- 8} mol$

The 25mL of cadmium ions with has $= \frac{25 脳 0.5 脳 10^{- 3}}{1000} = 1.25 脳 10^{- 5} moles$

Reduced \% of cadmium ions $= \frac{1.48 脳 10^{- 8} moles}{1.25 脳 10^{- 5} moles} 脳 100$

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Most popular questions from this chapter

(a) At what cathode potential will Sb(s)deposition commence from $0.010 {MSbO}^{+}$ solution at pH 0.00? Express this potential versus S.H.E. and versus $Ag | AgCI$ . (Disregard overpotential, about which you have no information.)

${SbO}^{+} + 2 H^{+} + 3 e^{-} 鈫� Sb (s) + H_{2} O E 掳 = 0.208 V$

(b) What percentage of $0.10 {MCu}^{2 +}$ could be reduced electrolytically to Cu(s)before $0.010 {MSbO}^{+}$ in the same solution begins to be reduced at pH 0.00?

Write the chemical reactions that show that 1 mol of I₂ is required for 1 mol of H₂O in a Karl Fischer titration.

${cd}^{2 +}$ was used as an internal standard in the analysis of ${Pb}^{2 +}$ by square wave polarography. ${Cd}^{2 +}$ gives a reduction wave at -0.60 V and ${Pb}^{2 +}$ gives a reduction wave at 鈥�0.40 V. It was first verified that the ratio of peak heights is proportional to the ratio of concentrations over the whole range employed in the experiment. Here are results for known and unknown mixtures:

The unknown mixture was prepared by mixing $25.00 (卤 0.05) mL$ of unknown (containing only ${Pb}^{2 +}$ ) plus $10.00 (卤 0.05) mL$ of $3.23 (卤 0.01) 脳 10^{- 4} {MCd}^{2 +}$ and diluting to $50.00 (卤 0.05) mL$ .

(a) Disregarding uncertainties, find $[{Pb}^{2 +}]$ in the undiluted unknown.

(b) Find the absolute uncertainty for the answer to part (a).

Consider the following electrolysis reactions.

Cathode: $H_{2} O (l) + e^{-} 鈬� \frac{1}{2} H_{2} (g, 1.0 b a r) + O H^{-} (a q, 0.10 M)$

Anode: $B r^{-} (a q, 0.10 M) 鈬� \frac{1}{2} B r_{2} (l) + e^{-}$

Calculate the voltage needed to drive the net reaction if current is negligible.
Suppose that the cell has a resistance of $2.0 惟$ and a current of 100 mA. How much voltage is needed to overcome the cell resistance? This is the ohmic potential.
Suppose that the anode reaction has an overpotential of 0.20 V and that the cathode overpotential is 0.40 V. What voltage is needed to overcome these effects combined with those of parts (a) and (b)?
Suppose that concentration polarization occurs $[O H^{-}]_{s}$ . at the cathode surface increases to 1.0 M and $[B r^{-}]_{s}$ at the anode surface decreases to 0.010 M. What voltage is needed to overcome these effects combined with those of (b) and (c)?

What are the advantages of using a microelectrode for

voltametric measurements?

See all solutions

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