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Chapter 17: Q2P (page 427)

How many hours are required for 0.100 mol of electrons to flow through a circuit if the current is 1.00 A?

Short Answer

Expert verified

Total hours requires for electrons to flow through a circuit is t = 2.68h

Step by step solution

01

Step1: Determination of electrons:

Electrons move through a wire from the negative end to the positive end.
The resistor uses the energy of the electrons around the wire and slows down the flow of electrons.

02

Calculation of electrons:

Given data,

$n (e^{-}) = 0.1 moll = 1 At = ?$

The charge of electrons is calculated

$n (e^{-}) = \frac{q}{n . F}$

The is 1, so the q is,

$q = n (e^{-}) . F q = 0.1 mol . 96485 C / mol q = 9648.5 C$

03

Step 3: Ascertaining the flow of current in hours:

Now we can calculate using the formula,

$q = l . t$

Where is,

$t = \frac{q}{l} = \frac{9648.5 C}{1 A} t = 9648.5 s = 2.68 h$

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Most popular questions from this chapter

$T i^{3 +}$ is to be generated in $0.10 M H C l O_{4}$ for coulometric reduction of azobenzene.

$\begin{aligned} {TiO}^{2 +} + 2 H^{+} + e^{-} ⇌ {Ti}^{3 +} + H_{2} O E^{0} = 0 .100 V \\ 4 {Ti}^{3 +} + C_{6} H_{5} N = {NC}_{6} H_{5} + 4 H_{2} O \to 2 C_{6} H_{5} {NH}_{2} + 4 {TiO}^{2 +} + 4 H^{+} \end{aligned}$

At the counter electrode, water is oxidized, and \(\mathrm{O}_{2}\) is liberated at a pressure of \(0.20\) bar. Both electrodes are made of smooth Pt, and each has a total surface area of $1.00 c m^{2}$ . The rate of reduction of the azobenzene is 25.9nmol/s , and the resistance of the solution between the generator electrodes is $52.4 Ω$ .

Calculate the current density $(A / m^{2})$ at the electrode surface. Use Table 17-1 to estimate the overpotential for $O_{2}$ liberation.
Calculate the cathode potential (versus S.H.E.) assuming that role="math" localid="1668356673323" $[Ti {O^{2 +}}_{surface}] = {[{TiO}^{2 +}]}_{bulk} = 0.050 M$ and $[T i^{3 +}]_{s u r f a c e} = 0.10 M$ .
Calculate the anode potential (versus S.H.E.).
What should the applied voltage be?

Consider the cyclic voltammogram of the ${Co}^{3 +}$ compoundrole="math" localid="1663646447735" $Co {(B_{9} C_{2} H_{11})}_{2}^{-}$ . Suggest a chemical reaction to account for each wave. Are the reactions reversible? How many electrons are involved in each step? Sketch the sampled current and square wave polarograms expected for this compound.

Cyclic voltammogramofrole="math" localid="1663646461802" $Co {(B_{9} C_{2} H_{11})}_{2}^{-}$ . [Data from W. E. Geiger, Jr., W. L. Bowden, and N. El Murr, "An Electrochemical Study of the Protonation Site of the Cobaltocene Anion and of Cyclopentadienylcobalt(I) Dicarbollides," Inorg. Chem. 1979,18,2358.]

For a rotating disk electrode operating at sufficiently great potential, the redox reaction rate is governed by the rate at which analyte diffuses through the diffusion layer to the electrode (Figure 17-15b). The thickness of the diffusion layer is

$δ = 1.61 D^{1 / 3} V^{1 / 6} ω^{- 1 / 2}$

whereis the diffusion coefficient of reactant $(m^{2} / s), v$ is the kinematic viscosity of the liquid

$(= v i s c o s i t y / d e b s i t y = (m^{2} / s), v$ and $ω$ is the rotation rate (radians/s) of the electrode. There are $2 π$ radians in a circle. The current densityis localid="1655441451764" $(A / m^{2})$ is

localid="1655441445229" $C u r r e n t d e b s i t y = 0.62 n F D^{2 / 3} v^{- 1 / 6} ω^{1 / 2} C_{0}$

where nis the number of electrons in the half-reaction, Fis the Faraday constant, and localid="1655441459070" $C_{0}$ is the concentration of the electroactive species in bulk solution localid="1655441466748" $(m o l / m^{3}, not m o l / L)$ Consider the oxidation oflocalid="1655441474339" $F e {(C N)}_{6}^{4 -}$ in a solution of localid="1655441479067" $10.0 m M K_{3} F e {(C N)}_{6} + 50.0 m M K_{4} F e {(C N)}_{6}$ at +0.90V(versus S.C.E.) at a rotation speed oflocalid="1655441490849" role="math" $2.00 \times 10^{3}$ revolutions per minute. 27The diffusion coefficient oflocalid="1655441497131" $F e {(C N)}_{6}^{4 -} is 2.5 \times 10^{- 9} m^{2} / s$ and the kinematic viscosity islocalid="1655441503345" $1.1 \times 10^{- 5} m^{2} / s$ Calculate the thickness of the diffusion layer and the current density. If you are careful, the current density should look like the value in Figure 17-16b.

Chlorine has been used for decades to disinfect drinking water. An undesirable side effect of this treatment is reaction with organic impurities to create organochlorine compounds, some of which could be toxic. Monitoring total organic halide (designated TOX) is required for many water providers. A standard procedure for TOX is to pass water through activated charcoal, which adsorbs organic compounds. Then the charcoal is combusted to liberate hydrogen halides:

$Organic halide (RX) \overset{o_{2} / 800 ° C}{\to} {CO}_{2} + H_{2} O + HX$

HX is absorbed into aqueous solution and measured by coulometric titration with a silver anode:

$X^{-} (aq) + Ag (s) \to AgX (s) + e^{-}$

When 1.00 L of drinking water was analyzed, a current of 4.23 mA was required for 387 s. A blank prepared by oxidizing charcoal required 6 s at 4.23 mA. Express the TOX of the drinking water as $μ$ mol halogen/L. If all halogen is chlorine, express the TOX as $μ g CI / L$ .

Ions that react with $A g^{+}$ can be determined electrogravimetrically by deposition on a silver working anode:

$A g (s) + X^{-} \to A g X (s) + e^{-}$

(a) What will be the final mass of a silver anode used to electrolyze 75.00 mL of 0.02380 M KSCN if the initial mass of the anode is 12.4638 g?

(b) At what electrolysis voltage (versus S.C.E.) will AgBr(s) be deposited from 0.10M Br? (Consider negligible current flow, so that there is no ohmic potential, concentration polarization, or overpotential.)

(c) Is it theoretically possible to separate 99.99% of0.10M Klfrom0.10MKBr by controlled-potential electrolysis?

See all solutions

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