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8090 [49]
3 years ago
11

The half-reaction occurring at the cathode in the balanced reaction shown below is __________?2Ag(s) + 1/2 O2(g) + 2H+(aq) → 2Ag

+(aq) + H2O (l)A. 2H2O (l) → O2 (g) + 4H+ (aq) + 4e-B. 2Ag (s) → 2Ag+ (aq) + 2e-C. 1/2 O2 (g) + 2H+ (aq) + 2e- → H2O (l)D. 1/2 O2 (g) + 2H+ (aq) + 2e- → H2O (l)
Chemistry
1 answer:
Shtirlitz [24]3 years ago
3 0

Answer:

Correct choice are C and D (they are both, the same).

Explanation:

Cathode is the positive(+) electrode where a reduction occurs.

Reduction is the chemical reaction where the oxidation state is reduced.

2Ag(s) + 1/2 O2(g) + 2H+(aq) → 2Ag+(aq) + H2O (l)

A. 2H2O (l) → O2 (g) + 4H+ (aq) + 4e-

B. 2Ag (s) → 2Ag+ (aq) + 2e-

C. 1/2 O2 (g) + 2H+ (aq) + 2e- → H2O (l)

D. 1/2 O2 (g) + 2H+ (aq) + 2e- → H2O (l)

C or D, are ok. They are the same equation.

Oxygen from ground state reduce the oxidation state from 0 to -2

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When it comes to equilibrium reactions, it useful to do ICE analysis. ICE stands for Initial-Change-Equilibrium. You subtract the initial and change to determine the equilibrium amounts which is the basis for Kc. Kc is the equilibrium constant of concentration which is just the ratio of products to reactant. 

Let's do the ICE analysis

      2 NH₃ ⇄ N₂ + 3 H₂
I         0        1.3    1.65
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-------------------------------------
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The variable x is the amount of moles of the substances that reacted. You apply the stoichiometric coefficients by multiplying it by x. Now, we can solve x by:

Equilibrium NH₃ = 0.1 = 0 + 2x
x = 0.05 mol
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For the second part, I am confused with the given reaction because the stoichiometric coefficients do not balance which violates the law of conservation of mass. But you should remember that the Kc values might differ because of the stoichiometric coefficient. For a reaction: aA + bB ⇄ cC, the Kc for this is

K_{C} = \frac{[ C^{c} ]}{[ A^{a} ][ B^{b} ]}

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8 0
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Which statement is true about the products of two reactants that combine chemically?
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I think its C i don't know if i am right
7 0
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The chemical reaction would be as follows:

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