A certain metal M forms a soluble sulfate salt MSO, Suppose the left half cell of a galvanic cell apparatus is filled with a 3.0
0 M solution of MSO, and the ight half cell with a 30.0 mM solution of the same substance. Electrodes made of M are dipped into both solutions and a voltmeter is connected between them. The temperature of the apparatus is held constant at 35.0 °C lf right 410 Which electrode will be positive? What voltage will the voltmeter show?
1 answer:
Answer:
E = 0.062 V
Explanation:
(a) See the attached file for the answer
(b)
Calculating the voltage (E) using the formula;
E = - (2.303RT/nf)log Cathode/Anode
Where,
R = 8.314 J/K/mol
T = 35°C = 308 K
F- Faraday's constant = 96500 C/mol,
n = number of moles of electron = 2
Substituting, we have
E = -(2.303 * 8.314 *308/2*96500) *log (0.03/3)
= -0.031 * -2
= 0.062V
Therefore, the voltmeter will show a voltage of 0.062 V
You might be interested in
Answer:
3.65 x 10¹⁰ electrons
Explanation:
we'll apply the following equation for electric field of a point charge on a spherical conductor
where E is the electric field
k is a constant of the value 8.99 x 10⁹ Nm²/C²
r is the radius of the spherical conductor
q is the total charge in the sphere
Given diameter d =41.0cm, radius r = 20.5cm = 0.205m (convert cm to m)
Electrical field E = 1250 N/C
we are asked to determine how many excess electrons must be added to the surface of the sphere to produce this electric field
q = <u>E x r²</u>
k
q = <u>1250 N/C x 0.205m</u>²
8.99 x 10⁹ Nm²/C²
q = 5.84 x 10⁻⁹ C
this is the total charge in the sphere
To determine the number of electrons, we can divide the charge q by the charge on an electron e (1.6 x 10⁻¹⁹C)
n = <u>5.84 x 10⁻⁹ C </u>
1.6 x 10⁻¹⁹C
n = 3.65 x 10¹⁰ electrons
Therefore, to apply an electric field of magnitude 1250 N/C, the isolated spherical conductor must contain 3.65 x 10¹⁰ electrons
When NAD becomes NADH, it is being reduced and gaining chemical energy.
Nicotinamide adenine dinucleotide (NAD), a coenzyme, can exist in two forms, NAD⁺ (oxidized) and NADH (reduced form).
Electrons and protons released in catabolism reactions are attached to NAD⁺. The conversion of NAD⁺ to NADH is important reaction for production of ATP during the cellular respiration.
Reduction is lowering oxidation number because element, ion or compound gain electrons.
Chemical equation for reaction of reduction of NAD⁺ (see picture below):
NAD⁺ + 2e⁻ + H⁺ → NADH
Nicotinamide adenine dinucleotide (NAD) is made of two nucleosides joined by pyrophosphate.
More about reduction :brainly.com/question/25334331
#SPJ4
