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Romashka [77]
4 years ago
10

The half-cell is a chamber in the voltaic cell where one half-cell is the site of the oxidation reaction and the other half-cell

is the site of the reduction reaction. type the half-cell reaction that takes place at the anode for the iron-silver voltaic cell. indicate the physical states of atoms and ions using the abbreviation (s), (l), or (g) for solid, liquid, or gas, respectively. use (aq) for an aqueous solution. do not include phases for electrons. express your answer as a chemical equation.
Chemistry
1 answer:
marysya [2.9K]4 years ago
4 0

Answer:

\boxed{\rm \text{Fe(s) $\rightleftharpoons$ Fe$^{2+}$(aq) + 2e$^{-}$}}

Explanation:

The half-cell reduction potentials are

Ag⁺(aq) +   e⁻ ⇌ Ag(s)     E° =  0.7996 V

Fe²⁺(aq) + 2e⁻ ⇌ Fe(s)     E° = -0.447    V

To create a spontaneous voltaic cell, we reverse the half-reaction with the more negative half-cell potential.

The anode is the electrode at which oxidation occurs.

The equation for the oxidation half-reaction is

\boxed{\rm \textbf{Fe(s) $\rightleftharpoons$ Fe$^{2+}$(aq) + 2e$^{-}$}}

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Answer:

a) 8.33 ml of the original stomach acid is neutralized

b) 191.67 ml of the stomach acid was neutralized

c)  249.68 ml acid would be neutralized by the original tablet

Explanation:

a) how much of the stomach acid had been neutralized in the 25 mL sample wich was titrated?

25.5 ml of a NaOH solution is equivalent to 25.00 ml of the original stomach acid

8.5 ml NaOH * (25.00 ml original stomach acid / 25.5 ml NaOH) = 8.33 ml original stomach acid

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It takes 8.5 ml NaOH to neutralize 8.33 ml original acid (this is the answer for question 1)

This means the antacid neutralized = 200 ml - 8.33 ml = 191.67 ml

c) how much stomach acid would have been neutralized by the original 5.6832 g tablet

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5.6832g antacid * (191.67 ml acid / 4.3628 g antacid) = 249.68 ml acid

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So, what does that tell you?

In order to increase the temperature of 1 g of aluminium by 1∘C, you need to provide it with 0.90 J of heat.

But remember, this is how much you need to provide for every gram of aluminium in order to increase its temperature by 1∘C. So if you wanted to increase the temperature of 10.0 g of aluminium by 1∘C, you'd have to provide it with

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